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#### Question 1:

(i) A(9, 3) and B(15, 11)
The given points are A(9, 3) and B(15, 11).
Then (x1 = 9, y1 = 3) and (x2 = 15, y2 = 11)

(ii) A(7, −4) and B(−5, 1)
The given points are A(7, −4) and B(−5, 1).
Then (x1= 7, y1 = −4) and (x2= −5, y2 = 1)

(iii) A(−6, −4) and B(9, −12)
The given points are A(−6, −4) and B(9, −12).
Then (x1 = −6, y1 = −4) and (x2 = 9, y2 = −12)

(iv) A(1, −3) and B(4, −6)
The given points are A(1, −3) and B(4, −6).
Then (x1= 1, y1 = −3) and (x2 = 4, y2 = −6)

(v) P(a + b, ab) and Q(ab, a + b)
The given points are P(a + b, a − b) and Q(a − b, a + b).
Then (x1 = a + b, y1 = a − b) and (x2 = a − b, y2 = a + b)

(vi) P(a sin α, a cos α) and Q(a cos α, −a sin α)
The given points are P(a sin α, a cos α) and Q(a cos α, −a sin α).
Then (x1 = a sin α, y1 = a cos α) and (x2 = a cos α, y2 = −a sin α)

#### Question 2:

(i) A(9, 3) and B(15, 11)
The given points are A(9, 3) and B(15, 11).
Then (x1 = 9, y1 = 3) and (x2 = 15, y2 = 11)

(ii) A(7, −4) and B(−5, 1)
The given points are A(7, −4) and B(−5, 1).
Then (x1= 7, y1 = −4) and (x2= −5, y2 = 1)

(iii) A(−6, −4) and B(9, −12)
The given points are A(−6, −4) and B(9, −12).
Then (x1 = −6, y1 = −4) and (x2 = 9, y2 = −12)

(iv) A(1, −3) and B(4, −6)
The given points are A(1, −3) and B(4, −6).
Then (x1= 1, y1 = −3) and (x2 = 4, y2 = −6)

(v) P(a + b, ab) and Q(ab, a + b)
The given points are P(a + b, a − b) and Q(a − b, a + b).
Then (x1 = a + b, y1 = a − b) and (x2 = a − b, y2 = a + b)

(vi) P(a sin α, a cos α) and Q(a cos α, −a sin α)
The given points are P(a sin α, a cos α) and Q(a cos α, −a sin α).
Then (x1 = a sin α, y1 = a cos α) and (x2 = a cos α, y2 = −a sin α)

(i) A(5, −12)
Let O(0, 0) be the origin.

(ii) B(−5, 5)
Let O(0, 0) be the origin.

(iii) C(−4, −6)
Let O(0,0) be the origin.

#### Question 3:

(i) A(5, −12)
Let O(0, 0) be the origin.

(ii) B(−5, 5)
Let O(0, 0) be the origin.

(iii) C(−4, −6)
Let O(0,0) be the origin.

Given AB = 5 units
Therefore, (AB)2 = 25 units

Therefore, x = 2 or 8.

#### Question 4:

Given AB = 5 units
Therefore, (AB)2 = 25 units

Therefore, x = 2 or 8.

The given points are .

Hence, the possible values of y are .

#### Question 5:

The given points are .

Hence, the possible values of y are .

The given points are P(x, 4) and Q(9, 10).

Hence, the values of x are 1 and 17.

#### Question 6:

The given points are P(x, 4) and Q(9, 10).

Hence, the values of x are 1 and 17.

As per the question
$AB=AC\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(x-8\right)}^{2}+{\left(2+2\right)}^{2}}=\sqrt{{\left(x-2\right)}^{2}+{\left(2+2\right)}^{2}}$
Squaring both sides, we get
${\left(x-8\right)}^{2}+{4}^{2}={\left(x-2\right)}^{2}+{4}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-16x+64+16={x}^{2}+4-4x+16\phantom{\rule{0ex}{0ex}}⇒16x-4x=64-4\phantom{\rule{0ex}{0ex}}⇒x=\frac{60}{12}=5$
Now,

Hence, x = 5 and AB = 5 units.

#### Question 7:

As per the question
$AB=AC\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(x-8\right)}^{2}+{\left(2+2\right)}^{2}}=\sqrt{{\left(x-2\right)}^{2}+{\left(2+2\right)}^{2}}$
Squaring both sides, we get
${\left(x-8\right)}^{2}+{4}^{2}={\left(x-2\right)}^{2}+{4}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-16x+64+16={x}^{2}+4-4x+16\phantom{\rule{0ex}{0ex}}⇒16x-4x=64-4\phantom{\rule{0ex}{0ex}}⇒x=\frac{60}{12}=5$
Now,

Hence, x = 5 and AB = 5 units.

As per the question
$AB=AC\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(0-3\right)}^{2}+{\left(2-p\right)}^{2}}=\sqrt{{\left(0-p\right)}^{2}+{\left(2-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(-3\right)}^{2}+{\left(2-p\right)}^{2}}=\sqrt{{\left(-p\right)}^{2}+{\left(-3\right)}^{2}}$
Squaring both sides, we get
${\left(-3\right)}^{2}+{\left(2-p\right)}^{2}={\left(-p\right)}^{2}+{\left(-3\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒9+4+{p}^{2}-4p={p}^{2}+9\phantom{\rule{0ex}{0ex}}⇒4p=4\phantom{\rule{0ex}{0ex}}⇒p=1$
Now,

Hence, p = 1 and AB$\sqrt{10}$ units.

#### Question 8:

As per the question
$AB=AC\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(0-3\right)}^{2}+{\left(2-p\right)}^{2}}=\sqrt{{\left(0-p\right)}^{2}+{\left(2-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(-3\right)}^{2}+{\left(2-p\right)}^{2}}=\sqrt{{\left(-p\right)}^{2}+{\left(-3\right)}^{2}}$
Squaring both sides, we get
${\left(-3\right)}^{2}+{\left(2-p\right)}^{2}={\left(-p\right)}^{2}+{\left(-3\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒9+4+{p}^{2}-4p={p}^{2}+9\phantom{\rule{0ex}{0ex}}⇒4p=4\phantom{\rule{0ex}{0ex}}⇒p=1$
Now,

Hence, p = 1 and AB$\sqrt{10}$ units.

Let the point on the x - axis be (x, 0).

Point on the x - axis is ($-$2, 0).

#### Question 9:

Let the point on the x - axis be (x, 0).

Point on the x - axis is ($-$2, 0).

Let P (x, 0) be the point on the x-axis. Then as per the question, we have

Hence, the points on the x-axis are (5, 0) and (17, 0).

#### Question 10:

Let P (x, 0) be the point on the x-axis. Then as per the question, we have

Hence, the points on the x-axis are (5, 0) and (17, 0).

Let P (0, y) be a point on the y-axis. Then as per the question, we have

$⇒36+{y}^{2}-10y+25=16+{y}^{2}-6y+9\phantom{\rule{0ex}{0ex}}⇒4y=36\phantom{\rule{0ex}{0ex}}⇒y=9$
Hence, the point on the y-axis is (0, 9).

#### Question 11:

Let P (0, y) be a point on the y-axis. Then as per the question, we have

$⇒36+{y}^{2}-10y+25=16+{y}^{2}-6y+9\phantom{\rule{0ex}{0ex}}⇒4y=36\phantom{\rule{0ex}{0ex}}⇒y=9$
Hence, the point on the y-axis is (0, 9).

As per the question, we have

$⇒-10x-2y=2x-10y\phantom{\rule{0ex}{0ex}}⇒8y=12x\phantom{\rule{0ex}{0ex}}⇒3x=2y$
Hence, 3x = 2y.

#### Question 12:

As per the question, we have

$⇒-10x-2y=2x-10y\phantom{\rule{0ex}{0ex}}⇒8y=12x\phantom{\rule{0ex}{0ex}}⇒3x=2y$
Hence, 3x = 2y.

The given points are A(6, −1) and B(2, 3). The point P(x, y) is equidistant from the points A and B. So, PA = PB.
Also, (PA)2 = (PB)2

Hence proved.

#### Question 13:

The given points are A(6, −1) and B(2, 3). The point P(x, y) is equidistant from the points A and B. So, PA = PB.
Also, (PA)2 = (PB)2

Hence proved.

Let the required point be P(x, y). Then AP = BP = CP
That is, (AP)2=(BP)2=(CP)2
This means (AP)2=(BP)

Hence, the required point is (3, −1).

#### Question 14:

Let the required point be P(x, y). Then AP = BP = CP
That is, (AP)2=(BP)2=(CP)2
This means (AP)2=(BP)

Hence, the required point is (3, −1).

Given, the points A(4, 3) and B(x, 5) lie on a circle with centre O(2, 3).
Then OA = OB
Also (OA)2 = (OB)2

Therefore, x = 2.

#### Question 15:

Given, the points A(4, 3) and B(x, 5) lie on a circle with centre O(2, 3).
Then OA = OB
Also (OA)2 = (OB)2

Therefore, x = 2.

As per the question, we have

Now

Hence, x = 2 or −6 and .

#### Question 16:

As per the question, we have

Now

Hence, x = 2 or −6 and .

As per the question, we have

Now for $k=-1$

For $k=-3$

Hence, .

#### Question 17:

As per the question, we have

Now for $k=-1$

For $k=-3$

Hence, .

(i) As per the question, we have

$⇒-xa-xb-ay+by=-xa+bx-ya-by\phantom{\rule{0ex}{0ex}}⇒by=bx$
Hence, bx = ay.

(ii)
As per the question, we have

$⇒-10x-2y=2x-10y\phantom{\rule{0ex}{0ex}}⇒8y=12x\phantom{\rule{0ex}{0ex}}⇒3x=2y$
Hence, 3x = 2y.

#### Question 18:

(i) As per the question, we have

$⇒-xa-xb-ay+by=-xa+bx-ya-by\phantom{\rule{0ex}{0ex}}⇒by=bx$
Hence, bx = ay.

(ii)
As per the question, we have

$⇒-10x-2y=2x-10y\phantom{\rule{0ex}{0ex}}⇒8y=12x\phantom{\rule{0ex}{0ex}}⇒3x=2y$
Hence, 3x = 2y.

(i)
Let A(1, −1), B(5, 2) and C(9, 5) be the given points. Then

Hence, the given points are collinear.

(ii)
Let A(6, 9), B(0, 1) and C(−6, −7) be the given points. Then

Hence, the given points are collinear.

(iii)
Let A(−1, −1), B(2, 3) and C(8, 11) be the given points. Then

Hence, the given points are collinear.

(iv)
Let A(−2, 5), B(0, 1) and C(2, −3) be the given points. Then

Hence, the given points are collinear.

#### Question 19:

(i)
Let A(1, −1), B(5, 2) and C(9, 5) be the given points. Then

Hence, the given points are collinear.

(ii)
Let A(6, 9), B(0, 1) and C(−6, −7) be the given points. Then

Hence, the given points are collinear.

(iii)
Let A(−1, −1), B(2, 3) and C(8, 11) be the given points. Then

Hence, the given points are collinear.

(iv)
Let A(−2, 5), B(0, 1) and C(2, −3) be the given points. Then

Hence, the given points are collinear.

The given points are A(7, 10), B(−2, 5) and C(3, −4).

Since, AB and BC are equal, they form the vertices of an isosceles triangle.
Also, (AB)2+(BC)2
and (AC)2 = ${\left(\sqrt{212}\right)}^{2}$ = 212
Thus, (AB)2+(BC)2 = (AC)2
This show that $∆ABC$ is right- angled at B.
Therefore, the points A(7, 10), B(−2, 5) and C(3, −4) are the vertices of an isosceles right-angled triangle.

#### Question 20:

The given points are A(7, 10), B(−2, 5) and C(3, −4).

Since, AB and BC are equal, they form the vertices of an isosceles triangle.
Also, (AB)2+(BC)2
and (AC)2 = ${\left(\sqrt{212}\right)}^{2}$ = 212
Thus, (AB)2+(BC)2 = (AC)2
This show that $∆ABC$ is right- angled at B.
Therefore, the points A(7, 10), B(−2, 5) and C(3, −4) are the vertices of an isosceles right-angled triangle.

The given points are A(3, 0), B(6, 4) and C(− 1, 3). Now

Therefore, A(3, 0), B(6, 4) and C(− 1, 3) are the vertices of an isosceles right triangle.

#### Question 21:

The given points are A(3, 0), B(6, 4) and C(− 1, 3). Now

Therefore, A(3, 0), B(6, 4) and C(− 1, 3) are the vertices of an isosceles right triangle.

$\because \angle B={90}^{\circ }\phantom{\rule{0ex}{0ex}}\therefore A{C}^{2}=A{B}^{2}+B{C}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(5+2\right)}^{2}+{\left(2-t\right)}^{2}={\left(5-2\right)}^{2}+{\left(2+2\right)}^{2}+{\left(2+2\right)}^{2}+{\left(-2-t\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(7\right)}^{2}+{\left(t-2\right)}^{2}={\left(3\right)}^{2}+{\left(4\right)}^{2}+{\left(4\right)}^{2}+{\left(t+2\right)}^{2}$
$⇒49+{t}^{2}-4t+4=9+16+16+{t}^{2}+4t+4\phantom{\rule{0ex}{0ex}}⇒8-4t=4t\phantom{\rule{0ex}{0ex}}⇒8t=8\phantom{\rule{0ex}{0ex}}⇒t=1$
Hence, t = 1.

#### Question 22:

$\because \angle B={90}^{\circ }\phantom{\rule{0ex}{0ex}}\therefore A{C}^{2}=A{B}^{2}+B{C}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(5+2\right)}^{2}+{\left(2-t\right)}^{2}={\left(5-2\right)}^{2}+{\left(2+2\right)}^{2}+{\left(2+2\right)}^{2}+{\left(-2-t\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(7\right)}^{2}+{\left(t-2\right)}^{2}={\left(3\right)}^{2}+{\left(4\right)}^{2}+{\left(4\right)}^{2}+{\left(t+2\right)}^{2}$
$⇒49+{t}^{2}-4t+4=9+16+16+{t}^{2}+4t+4\phantom{\rule{0ex}{0ex}}⇒8-4t=4t\phantom{\rule{0ex}{0ex}}⇒8t=8\phantom{\rule{0ex}{0ex}}⇒t=1$
Hence, t = 1.

The given points are A(2, 4), B(2, 6) and . Now

Hence, the points A(2, 4), B(2, 6) and are the vertices of an equilateral triangle.

#### Question 23:

The given points are A(2, 4), B(2, 6) and . Now

Hence, the points A(2, 4), B(2, 6) and are the vertices of an equilateral triangle.

Let the given points be A(− 3, − 3), B(3, 3) and C. Now

Hence, the given points are the vertices of an equilateral triangle.

#### Question 24:

Let the given points be A(− 3, − 3), B(3, 3) and C. Now

Hence, the given points are the vertices of an equilateral triangle.

Let the given points be A(−5, 6) B(3, 0) and C(9, 8).

Also, (AB)2+(BC)2
and (AC)2 = ${\left(10\sqrt{2}\right)}^{2}=200$
Thus, (AB)2+(BC)2 = (AC)2
This show that $∆ABC$ is right- angled at B.
Therefore, the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle.
Also,

#### Question 25:

Let the given points be A(−5, 6) B(3, 0) and C(9, 8).

Also, (AB)2+(BC)2
and (AC)2 = ${\left(10\sqrt{2}\right)}^{2}=200$
Thus, (AB)2+(BC)2 = (AC)2
This show that $∆ABC$ is right- angled at B.
Therefore, the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle.
Also,

The given points are O(0, 0) A(3, $\sqrt{3}$) and B(3, − $\sqrt{3}$).

Thus, t
he points O(0, 0) A(3, $\sqrt{3}$)and B(3, − $\sqrt{3}$) are the vertices of an equilateral triangle.
Also, the area of the triangle OAB = $\frac{\sqrt{3}}{4}×{\left(\mathrm{side}\right)}^{2}$

#### Question 26:

The given points are O(0, 0) A(3, $\sqrt{3}$) and B(3, − $\sqrt{3}$).

Thus, t
he points O(0, 0) A(3, $\sqrt{3}$)and B(3, − $\sqrt{3}$) are the vertices of an equilateral triangle.
Also, the area of the triangle OAB = $\frac{\sqrt{3}}{4}×{\left(\mathrm{side}\right)}^{2}$

(i) The given points are A(3, 2), B(0, 5), C(−3, 2) and D(0, −1).

Therefore, the given points form a square.

(ii) The given points are A(6, 2), B(2, 1), C(1, 5) and D(5, 6).

Therefore, the given points form a square.

(iii) The given points are P(0, −2), Q(3, 1), R(0, 4) and S(−3, 1).

Therefore, the given points form a square.

#### Question 27:

(i) The given points are A(3, 2), B(0, 5), C(−3, 2) and D(0, −1).

Therefore, the given points form a square.

(ii) The given points are A(6, 2), B(2, 1), C(1, 5) and D(5, 6).

Therefore, the given points form a square.

(iii) The given points are P(0, −2), Q(3, 1), R(0, 4) and S(−3, 1).

Therefore, the given points form a square.

The given points are A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4).

Therefore, ABCD is a quadrilateral with equal sides and unequal diagonals.
Hence, ABCD is a rhombus.

#### Question 28:

The given points are A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4).

Therefore, ABCD is a quadrilateral with equal sides and unequal diagonals.
Hence, ABCD is a rhombus.

The given points are A(3, 0), B(4, 5), C(− 1, 4) and D(− 2, − 1).

Therefore, the given points are the vertices of a rhombus.

Hence, the area of the rhombus is 24 sq. units.

#### Question 29:

The given points are A(3, 0), B(4, 5), C(− 1, 4) and D(− 2, − 1).

Therefore, the given points are the vertices of a rhombus.

Hence, the area of the rhombus is 24 sq. units.

The given points are A(6, 1), B(8, 2), C(9, 4) and D(7, 3).

Therefore, the given points are the vertices of a rhombus. Now

Hence, the area of the rhombus is 3 sq. units.

#### Question 30:

The given points are A(6, 1), B(8, 2), C(9, 4) and D(7, 3).

Therefore, the given points are the vertices of a rhombus. Now

Hence, the area of the rhombus is 3 sq. units.

The given points are A(2, 1), B(5, 2), C(6, 4) and D(3, 3).

But diagonal AC is not equal to diagonal BD.
Hence, the given points do not form a rectangle.

#### Question 31:

The given points are A(2, 1), B(5, 2), C(6, 4) and D(3, 3).

But diagonal AC is not equal to diagonal BD.
Hence, the given points do not form a rectangle.

The given vertices are A(1, 2), B(4, 3), C(6, 6) and D(3, 5).

Therefore, ABCD is a parallelogram. Now

Thus, the diagonals AC and BD are not equal and hence ABCD is not a rectangle.

#### Question 32:

The given vertices are A(1, 2), B(4, 3), C(6, 6) and D(3, 5).

Therefore, ABCD is a parallelogram. Now

Thus, the diagonals AC and BD are not equal and hence ABCD is not a rectangle.

(i)

(ii) The given points are A(2, −2), B(14, 10) C(11, 13) and D(−1, 1).

Also, diagonal AC = diagonal BD
Hence, the given points form a rectangle.

(iii) The given points are A(0, −4), B(6, 2) C(3, 5) and D(−3, −1).

Also, diagonal AC = diagonal BD.
Hence, the given points form a rectangle.

#### Question 33:

(i)

(ii) The given points are A(2, −2), B(14, 10) C(11, 13) and D(−1, 1).

Also, diagonal AC = diagonal BD
Hence, the given points form a rectangle.

(iii) The given points are A(0, −4), B(6, 2) C(3, 5) and D(−3, −1).

Also, diagonal AC = diagonal BD.
Hence, the given points form a rectangle.

In ΔABC, the coordinates of the vertices are A(–2, 0), B(0,2), C(2,0).

$\mathrm{AB}=\sqrt{{\left(0+2\right)}^{2}+{\left(2-0\right)}^{2}}=\sqrt{8}=2\sqrt{2}\phantom{\rule{0ex}{0ex}}\mathrm{CB}=\sqrt{{\left(0-2\right)}^{2}+{\left(2-0\right)}^{2}}=\sqrt{8}=2\sqrt{2}\phantom{\rule{0ex}{0ex}}\mathrm{AC}=\sqrt{{\left(2+2\right)}^{2}+{\left(0-0\right)}^{2}}=4$

In ΔDEF, the coordinates of the vertices are D(–4, 0), E(4, 0), F(0, 4).

$\mathrm{DF}=\sqrt{{\left(4+4\right)}^{2}+{\left(0-0\right)}^{2}}=8\phantom{\rule{0ex}{0ex}}\mathrm{FE}=\sqrt{{\left(0-4\right)}^{2}+{\left(4-0\right)}^{2}}=4\sqrt{2}\phantom{\rule{0ex}{0ex}}\mathrm{DE}=\sqrt{{\left(0+4\right)}^{2}+{\left(4-0\right)}^{2}}=4\sqrt{2}$

Now, for ΔABC and ΔDEF to be similar, the corresponding sides should be proportional.

#### Question 1:

In ΔABC, the coordinates of the vertices are A(–2, 0), B(0,2), C(2,0).

$\mathrm{AB}=\sqrt{{\left(0+2\right)}^{2}+{\left(2-0\right)}^{2}}=\sqrt{8}=2\sqrt{2}\phantom{\rule{0ex}{0ex}}\mathrm{CB}=\sqrt{{\left(0-2\right)}^{2}+{\left(2-0\right)}^{2}}=\sqrt{8}=2\sqrt{2}\phantom{\rule{0ex}{0ex}}\mathrm{AC}=\sqrt{{\left(2+2\right)}^{2}+{\left(0-0\right)}^{2}}=4$

In ΔDEF, the coordinates of the vertices are D(–4, 0), E(4, 0), F(0, 4).

$\mathrm{DF}=\sqrt{{\left(4+4\right)}^{2}+{\left(0-0\right)}^{2}}=8\phantom{\rule{0ex}{0ex}}\mathrm{FE}=\sqrt{{\left(0-4\right)}^{2}+{\left(4-0\right)}^{2}}=4\sqrt{2}\phantom{\rule{0ex}{0ex}}\mathrm{DE}=\sqrt{{\left(0+4\right)}^{2}+{\left(4-0\right)}^{2}}=4\sqrt{2}$

Now, for ΔABC and ΔDEF to be similar, the corresponding sides should be proportional.

(i) The end points of AB are A(−1, 7) and B(4, −3).
Therefore, (x1 = −1, y1 = 7) and (x2 = 4, y2 = −3)
Also, m = 2 and n = 3
Let the required point be P(x, y).
By section formula, we get:

Hence, the coordinates of the required point are (1, 3).

(ii) The end points of AB are A(−5, 11) and B(4, −7).
Therefore, (x1 = −5, y1 = 11) and (x2 = 4, y2 = −7).
Also, m = 7 and n = 2
Let the required point be P(x, y).
By section formula, we have:

Hence, the required point is P(2, −3).

#### Question 2:

(i) The end points of AB are A(−1, 7) and B(4, −3).
Therefore, (x1 = −1, y1 = 7) and (x2 = 4, y2 = −3)
Also, m = 2 and n = 3
Let the required point be P(x, y).
By section formula, we get:

Hence, the coordinates of the required point are (1, 3).

(ii) The end points of AB are A(−5, 11) and B(4, −7).
Therefore, (x1 = −5, y1 = 11) and (x2 = 4, y2 = −7).
Also, m = 7 and n = 2
Let the required point be P(x, y).
By section formula, we have:

Hence, the required point is P(2, −3).

Consider the figure.
Here points P and Q trisect AB.
Therefore, P divides AB into 1 : 2 and Q divides AB into 2 : 1.
Using section formula, coordinates of P are;
$\mathrm{P}\left(x,y\right)=\left(\frac{1×1+2×7}{3},\frac{1×-5+2×-2}{3}\right)\phantom{\rule{0ex}{0ex}}\mathrm{P}\left(x,y\right)=\left(\frac{15}{3},\frac{-9}{3}\right)=\left(5,-3\right)$
Similarly, coordinates of Q are;
$\mathrm{Q}\left(a,b\right)=\left(\frac{2×1+1×7}{3},\frac{2×-5+1×-2}{3}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Q}\left(a,b\right)=\left(\frac{9}{3},\frac{-12}{3}\right)=\left(3,-4\right)$
Therefore, coordinates of points P and Q are (5, $-$3) and (3, $-$4) respectively.

#### Question 3:

Consider the figure.
Here points P and Q trisect AB.
Therefore, P divides AB into 1 : 2 and Q divides AB into 2 : 1.
Using section formula, coordinates of P are;
$\mathrm{P}\left(x,y\right)=\left(\frac{1×1+2×7}{3},\frac{1×-5+2×-2}{3}\right)\phantom{\rule{0ex}{0ex}}\mathrm{P}\left(x,y\right)=\left(\frac{15}{3},\frac{-9}{3}\right)=\left(5,-3\right)$
Similarly, coordinates of Q are;
$\mathrm{Q}\left(a,b\right)=\left(\frac{2×1+1×7}{3},\frac{2×-5+1×-2}{3}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Q}\left(a,b\right)=\left(\frac{9}{3},\frac{-12}{3}\right)=\left(3,-4\right)$
Therefore, coordinates of points P and Q are (5, $-$3) and (3, $-$4) respectively.

The coordinates of the points A and B are (−2, −2) and (2, −4) respectively, where $AP=\frac{3}{7}AB$ and P lies on the line segment AB. So

Let (x, y) be the coordinates of P which divides AB in the ratio 3 : 4 internally. Then
$x=\frac{3×2+4×\left(-2\right)}{3+4}=\frac{6-8}{7}=-\frac{2}{7}\phantom{\rule{0ex}{0ex}}y=\frac{3×\left(-4\right)+4×\left(-2\right)}{3+4}=\frac{-12-8}{7}=-\frac{20}{7}$
Hence, the coordinates of point P are .

#### Question 4:

The coordinates of the points A and B are (−2, −2) and (2, −4) respectively, where $AP=\frac{3}{7}AB$ and P lies on the line segment AB. So

Let (x, y) be the coordinates of P which divides AB in the ratio 3 : 4 internally. Then
$x=\frac{3×2+4×\left(-2\right)}{3+4}=\frac{6-8}{7}=-\frac{2}{7}\phantom{\rule{0ex}{0ex}}y=\frac{3×\left(-4\right)+4×\left(-2\right)}{3+4}=\frac{-12-8}{7}=-\frac{20}{7}$
Hence, the coordinates of point P are .

Let the coordinates of A be (x, y). Here, $\frac{PA}{PQ}=\frac{2}{5}$. So,

Let (x, y) be the coordinates of A, which divides PQ in the ratio 2 : 3 internally. Then using section formula, we get
$x=\frac{2×\left(-4\right)+3×\left(6\right)}{2+3}=\frac{-8+18}{5}=\frac{10}{5}=2\phantom{\rule{0ex}{0ex}}y=\frac{2×\left(-1\right)+3×\left(-6\right)}{2+3}=\frac{-2-18}{5}=\frac{-20}{5}=-4$
Now, the point (2, −4) lies on the line 3x + k (y + 1) = 0, therefore
$3×2+k\left(-4+1\right)=0\phantom{\rule{0ex}{0ex}}⇒3k=6\phantom{\rule{0ex}{0ex}}⇒k=\frac{6}{3}=2$
Hence, k = 2.

#### Question 5:

Let the coordinates of A be (x, y). Here, $\frac{PA}{PQ}=\frac{2}{5}$. So,

Let (x, y) be the coordinates of A, which divides PQ in the ratio 2 : 3 internally. Then using section formula, we get
$x=\frac{2×\left(-4\right)+3×\left(6\right)}{2+3}=\frac{-8+18}{5}=\frac{10}{5}=2\phantom{\rule{0ex}{0ex}}y=\frac{2×\left(-1\right)+3×\left(-6\right)}{2+3}=\frac{-2-18}{5}=\frac{-20}{5}=-4$
Now, the point (2, −4) lies on the line 3x + k (y + 1) = 0, therefore
$3×2+k\left(-4+1\right)=0\phantom{\rule{0ex}{0ex}}⇒3k=6\phantom{\rule{0ex}{0ex}}⇒k=\frac{6}{3}=2$
Hence, k = 2.

Since, the points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in five equal parts, so
AP = PQ = QR = RS = SB
Here, point P divides AB in the ratio of 1 : 4 internally. So using section formula, we get

The point Q divides AB in the ratio of 2 : 3 internally. So using section formula, we get

The point R divides AB in the ratio of 3 : 2 internally. So using section formula, we get

Hence, the coordinates of the points P, Q and R are (2, 3), (3, 4) and (4, 5) respectively.

#### Question 6:

Since, the points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in five equal parts, so
AP = PQ = QR = RS = SB
Here, point P divides AB in the ratio of 1 : 4 internally. So using section formula, we get

The point Q divides AB in the ratio of 2 : 3 internally. So using section formula, we get

The point R divides AB in the ratio of 3 : 2 internally. So using section formula, we get

Hence, the coordinates of the points P, Q and R are (2, 3), (3, 4) and (4, 5) respectively.

The given points are A(1, 6) and B(5, −2).
Then, P(x, y) is a point that divides the line AB in the ratio 1:3.
By the section formula:

Therefore, the coordinates of point P are (2, 4).
Let Q be the mid point of AB.
Then, Q(x, y):

Therefore, the coordinates of Q are (3, 2).
Let R (x, y) be a point that divides AB in the ratio 3:1.
Then, by the section formula:

Therefore, the coordinates of R are (4, 0).
Hence, the coordinates of point P, Q and R are (2, 4), (3, 2) and (4, 0) respectively.

#### Question 7:

The given points are A(1, 6) and B(5, −2).
Then, P(x, y) is a point that divides the line AB in the ratio 1:3.
By the section formula:

Therefore, the coordinates of point P are (2, 4).
Let Q be the mid point of AB.
Then, Q(x, y):

Therefore, the coordinates of Q are (3, 2).
Let R (x, y) be a point that divides AB in the ratio 3:1.
Then, by the section formula:

Therefore, the coordinates of R are (4, 0).
Hence, the coordinates of point P, Q and R are (2, 4), (3, 2) and (4, 0) respectively.

Let P and Q be the points of trisection of AB.
Then, P divides AB in the ratio 1:2.
So, the coordinates of P are

Hence, the coordinates of P are ($\frac{7}{3}$, −2).
But (p, −2) are the coordinates of P.
So, $p=\frac{7}{3}$
Also, Q divides the line AB in the ratio 2:1.
So, the coordinates of Q are

But the given coordinates of
So, q = 0
Thus, $p=\frac{7}{3}$ and $q=0$53

#### Question 8:

Let P and Q be the points of trisection of AB.
Then, P divides AB in the ratio 1:2.
So, the coordinates of P are

Hence, the coordinates of P are ($\frac{7}{3}$, −2).
But (p, −2) are the coordinates of P.
So, $p=\frac{7}{3}$
Also, Q divides the line AB in the ratio 2:1.
So, the coordinates of Q are

But the given coordinates of
So, q = 0
Thus, $p=\frac{7}{3}$ and $q=0$53

(i) The given points are A(3, 0) and B(−5, 4).
Let (x, y) be the mid point of AB. Then:

Therefore, (−1, 2) are the coordinates of mid point of AB.

(ii) The given points are P(−11, −8) and Q(8, −2).
Let (x, y) be the mid point of PQ. Then:

Therefore, are the coordinates of midpoint of PQ.

#### Question 9:

(i) The given points are A(3, 0) and B(−5, 4).
Let (x, y) be the mid point of AB. Then:

Therefore, (−1, 2) are the coordinates of mid point of AB.

(ii) The given points are P(−11, −8) and Q(8, −2).
Let (x, y) be the mid point of PQ. Then:

Therefore, are the coordinates of midpoint of PQ.

The given points are A(6, −5) and B(−2, 11).
Let (x, y) be the mid point of AB. Then:

So, the midpoint of AB is (2, 3).
But it is given that the midpoint of AB is (2, p).
Therefore, the value of p = 3.

#### Question 10:

The given points are A(6, −5) and B(−2, 11).
Let (x, y) be the mid point of AB. Then:

So, the midpoint of AB is (2, 3).
But it is given that the midpoint of AB is (2, p).
Therefore, the value of p = 3.

The points are A(2a, 4) and B(−2, 3b).
Let C(1, 2a + 1) be the mid point of AB. Then:

#### Question 11:

The points are A(2a, 4) and B(−2, 3b).
Let C(1, 2a + 1) be the mid point of AB. Then:

The given points are A(−2, 9) and B(6, 3).
Then, C(x, y) is the midpoint of AB.

Therefore, the coordinates of point C are (2, 6).

#### Question 12:

The given points are A(−2, 9) and B(6, 3).
Then, C(x, y) is the midpoint of AB.

Therefore, the coordinates of point C are (2, 6).

C(2, −3) is the centre of the given circle. Let A(a, b) and B(1, 4) be the two end-points of the given diameter AB. Then, the coordinates of C are

Therefore, the coordinates of point A are (3, -10).

#### Question 13:

C(2, −3) is the centre of the given circle. Let A(a, b) and B(1, 4) be the two end-points of the given diameter AB. Then, the coordinates of C are

Therefore, the coordinates of point A are (3, -10).

Let the point P(2, 5) divide AB in the ratio k : 1.
Then, by section formula, the coordinates of P are

Therefore, the point P(2, 5) divides AB in the ratio 3 : 4.

#### Question 14:

Let the point P(2, 5) divide AB in the ratio k : 1.
Then, by section formula, the coordinates of P are

Therefore, the point P(2, 5) divides AB in the ratio 3 : 4.

Let k : 1 be the ratio in which the point divides the line segment joining the points and $\left(2,-5\right)$. Then

Hence, the required ratio is 1 : 5.

#### Question 15:

Let k : 1 be the ratio in which the point divides the line segment joining the points and $\left(2,-5\right)$. Then

Hence, the required ratio is 1 : 5.

Let the point P(m, 6) divide the line AB in the ratio k : 1.
Then, by the section formula:

#### Question 16:

Let the point P(m, 6) divide the line AB in the ratio k : 1.
Then, by the section formula:

Let the point P(−3, k) divide the line AB in the ratio s : 1.
Then, by the section formula:

#### Question 17:

Let the point P(−3, k) divide the line AB in the ratio s : 1.
Then, by the section formula:

Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are

But P lies on the x-axis; so, its ordinate is 0.

Therefore, the required ratio is $\frac{1}{2}$ : 1, which is same as 1 : 2.
Thus, the x-axis divides the line AB in the ratio 1 : 2 at the point P.
Applying k = $\frac{1}{2}$, we get the coordinates of point :

Hence, the point of intersection of AB and the x-axis is P(3, 0).

#### Question 18:

Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are

But P lies on the x-axis; so, its ordinate is 0.

Therefore, the required ratio is $\frac{1}{2}$ : 1, which is same as 1 : 2.
Thus, the x-axis divides the line AB in the ratio 1 : 2 at the point P.
Applying k = $\frac{1}{2}$, we get the coordinates of point :

Hence, the point of intersection of AB and the x-axis is P(3, 0).

Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are
$P\left(\frac{3k-2}{k+1},\frac{7k-3}{k+1}\right)$
But P lies on the y-axis; so, its abscissa is 0.

Therefore, the required ratio is $\frac{2}{3}$ : 1, which is same as 2 : 3.
Thus, the x-axis divides the line AB in the ratio 2:3 at the point P.
Applying k=$\frac{2}{3}$, we get the coordinates of point P:

Hence, the point of intersection of AB and the x-axis is P(0, 1).

#### Question 19:

Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are
$P\left(\frac{3k-2}{k+1},\frac{7k-3}{k+1}\right)$
But P lies on the y-axis; so, its abscissa is 0.

Therefore, the required ratio is $\frac{2}{3}$ : 1, which is same as 2 : 3.
Thus, the x-axis divides the line AB in the ratio 2:3 at the point P.
Applying k=$\frac{2}{3}$, we get the coordinates of point P:

Hence, the point of intersection of AB and the x-axis is P(0, 1).

Let the line xy − 2 = 0 divide the line segment joining the points A(3, −1) and B(8, 9) in the ratio k : 1 at P.
Then, the coordinates of P are

Since, P lies on the line xy − 2 = 0, we have:

$\left(\frac{8k+3}{k+1}\right)-\left(\frac{9k-1}{k+1}\right)-2=0\phantom{\rule{0ex}{0ex}}⇒8k+3-9k+1-2k-2=0\phantom{\rule{0ex}{0ex}}⇒8k-9k-2k+3+1-2=0\phantom{\rule{0ex}{0ex}}⇒-3k+2=0\phantom{\rule{0ex}{0ex}}⇒-3k=-2\phantom{\rule{0ex}{0ex}}⇒k=\frac{2}{3}$
So, the required ratio is $\frac{2}{3}$ : 1, which is equal to 2 : 3.

#### Question 20:

Let the line xy − 2 = 0 divide the line segment joining the points A(3, −1) and B(8, 9) in the ratio k : 1 at P.
Then, the coordinates of P are

Since, P lies on the line xy − 2 = 0, we have:

$\left(\frac{8k+3}{k+1}\right)-\left(\frac{9k-1}{k+1}\right)-2=0\phantom{\rule{0ex}{0ex}}⇒8k+3-9k+1-2k-2=0\phantom{\rule{0ex}{0ex}}⇒8k-9k-2k+3+1-2=0\phantom{\rule{0ex}{0ex}}⇒-3k+2=0\phantom{\rule{0ex}{0ex}}⇒-3k=-2\phantom{\rule{0ex}{0ex}}⇒k=\frac{2}{3}$
So, the required ratio is $\frac{2}{3}$ : 1, which is equal to 2 : 3.

The vertices of âˆ†ABC are A(0, −1), B(2, 1) and C(0, 3).
Let AD, BE and CF be the medians of âˆ†ABC.

Let D be the midpoint of BC. So, the coordinates of D are

Let E be the midpoint of AC. So, the coordinates of E are

Let F be the midpoint of AB. So, the coordinates of F are

Therefore, the lengths of the medians: AD$\sqrt{10}$ units, BE = 2 units and CF = $\sqrt{10}$ units

#### Question 21:

The vertices of âˆ†ABC are A(0, −1), B(2, 1) and C(0, 3).
Let AD, BE and CF be the medians of âˆ†ABC.

Let D be the midpoint of BC. So, the coordinates of D are

Let E be the midpoint of AC. So, the coordinates of E are

Let F be the midpoint of AB. So, the coordinates of F are

Therefore, the lengths of the medians: AD$\sqrt{10}$ units, BE = 2 units and CF = $\sqrt{10}$ units

Here, (x1 = −1, y1 = 0), (x2 = 5, y2 = −2) and (x3 = 8, y3 = 2).
Let G(x, y) be the centroid of the âˆ†ABC. Then,

Hence, the centroid of âˆ†ABC is G(4, 0).

#### Question 22:

Here, (x1 = −1, y1 = 0), (x2 = 5, y2 = −2) and (x3 = 8, y3 = 2).
Let G(x, y) be the centroid of the âˆ†ABC. Then,

Hence, the centroid of âˆ†ABC is G(4, 0).

Two vertices of âˆ†ABC are A(1, −6) and B(−5, 2). Let the third vertex be C(a, b).
Then the coordinates of its centroid are

But it is given that G(−2, 1) is the centroid. Therefore,

Therefore, the third vertex of âˆ†ABC is C(−2, 7).

#### Question 23:

Two vertices of âˆ†ABC are A(1, −6) and B(−5, 2). Let the third vertex be C(a, b).
Then the coordinates of its centroid are

But it is given that G(−2, 1) is the centroid. Therefore,

Therefore, the third vertex of âˆ†ABC is C(−2, 7).

Two vertices of âˆ†ABC are B(−3,1) and C(0, −2). Let the third vertex be A(a, b).
Then, the coordinates of its centroid are

But it is given that the centroid is at the origin, that is G(0, 0). Therefore,

Therefore, the third vertex of âˆ†ABC is A(3, 1).

#### Question 24:

Two vertices of âˆ†ABC are B(−3,1) and C(0, −2). Let the third vertex be A(a, b).
Then, the coordinates of its centroid are

But it is given that the centroid is at the origin, that is G(0, 0). Therefore,

Therefore, the third vertex of âˆ†ABC is A(3, 1).

The points are A(3, 1), B(0, −2), C(1, 1) and D(4, 4).
Join AC and BD, intersecting at O.

We know that the diagonals of a parallelogram bisect each other.

Thus, the diagonals AC and BD have the same midpoint.
Therefore, ABCD is a parallelogram.

#### Question 25:

The points are A(3, 1), B(0, −2), C(1, 1) and D(4, 4).
Join AC and BD, intersecting at O.

We know that the diagonals of a parallelogram bisect each other.

Thus, the diagonals AC and BD have the same midpoint.
Therefore, ABCD is a parallelogram.

The points are P(a, −11), Q(5, b), R(2, 15) and S(1, 1).

Join PR and QS, intersecting at O.
We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of PR as well as QS.

#### Question 26:

The points are P(a, −11), Q(5, b), R(2, 15) and S(1, 1).

Join PR and QS, intersecting at O.
We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of PR as well as QS.

Let A(1, −2), B(3, 6) and C(5, 10) be the three vertices of a parallelogram ABCD and the fourth vertex be D(a, b).
Join AC and BD intersecting at O.

We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of AC as well as BD.

Therefore, the fourth vertex is D(3, 2).

#### Question 27:

Let A(1, −2), B(3, 6) and C(5, 10) be the three vertices of a parallelogram ABCD and the fourth vertex be D(a, b).
Join AC and BD intersecting at O.

We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of AC as well as BD.

Therefore, the fourth vertex is D(3, 2).

Let y-axis divides the line segment joining the points (−4, 7) and (3, −7) in the ratio k : 1. Then
$0=\frac{3k\mathit{-}4}{k+1}\phantom{\rule{0ex}{0ex}}⇒3k=4\phantom{\rule{0ex}{0ex}}⇒k=\frac{4}{3}$
Hence, the required ratio is 4 : 3.

#### Question 28:

Let y-axis divides the line segment joining the points (−4, 7) and (3, −7) in the ratio k : 1. Then
$0=\frac{3k\mathit{-}4}{k+1}\phantom{\rule{0ex}{0ex}}⇒3k=4\phantom{\rule{0ex}{0ex}}⇒k=\frac{4}{3}$
Hence, the required ratio is 4 : 3.

Let the point  divides the line segment joining the points A(3, −5) and B(−7, 9) in the ratio k : 1. Then

Now, substituting $k=\frac{1}{3}$ in $\frac{9k-5}{k+1}=y$, we get
$\frac{\frac{9}{3}-5}{\frac{1}{3}+1}=y⇒y=\frac{9-15}{1+3}=-\frac{3}{2}$
Hence, required ratio is 1 : 3 and $y=-\frac{3}{2}$.

#### Question 29:

Let the point  divides the line segment joining the points A(3, −5) and B(−7, 9) in the ratio k : 1. Then

Now, substituting $k=\frac{1}{3}$ in $\frac{9k-5}{k+1}=y$, we get
$\frac{\frac{9}{3}-5}{\frac{1}{3}+1}=y⇒y=\frac{9-15}{1+3}=-\frac{3}{2}$
Hence, required ratio is 1 : 3 and $y=-\frac{3}{2}$.

The line segment joining the points A(3, − 3) and B(− 2, 7) is divided by x-axis. Let the required ratio be k : 1. So,
$0=\frac{k\left(7\right)-3}{k+1}⇒k=\frac{3}{7}$
Now

Hence, the required ratio is 3 : 7 and the point of division is .

#### Question 30:

The line segment joining the points A(3, − 3) and B(− 2, 7) is divided by x-axis. Let the required ratio be k : 1. So,
$0=\frac{k\left(7\right)-3}{k+1}⇒k=\frac{3}{7}$
Now

Hence, the required ratio is 3 : 7 and the point of division is .

Let (x, 0) be the coordinates of R. Then
$0=\frac{-4+x}{2}⇒x=4$
Thus, the coordinates of R are (4, 0).
Here, PQ = QR = PR and the coordinates of P lies on y-axis. Let the coordinates of P be (0, y). Then
$PQ=QR⇒P{Q}^{2}=Q{R}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(0+4\right)}^{2}+{\left(y-0\right)}^{2}={8}^{2}\phantom{\rule{0ex}{0ex}}⇒{y}^{2}=64-16=48\phantom{\rule{0ex}{0ex}}⇒y=±4\sqrt{3}$
Hence, the required coordinates are .

#### Question 31:

Let (x, 0) be the coordinates of R. Then
$0=\frac{-4+x}{2}⇒x=4$
Thus, the coordinates of R are (4, 0).
Here, PQ = QR = PR and the coordinates of P lies on y-axis. Let the coordinates of P be (0, y). Then
$PQ=QR⇒P{Q}^{2}=Q{R}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(0+4\right)}^{2}+{\left(y-0\right)}^{2}={8}^{2}\phantom{\rule{0ex}{0ex}}⇒{y}^{2}=64-16=48\phantom{\rule{0ex}{0ex}}⇒y=±4\sqrt{3}$
Hence, the required coordinates are .

Let (0, y) be the coordinates of B. Then
$0=\frac{-3+y}{2}⇒y=3$
Thus, the coordinates of B are (0, 3).
Here, AB = BC = AC and by symmetry the coordinates of A lies on x-axis. Let the coordinates of A be (x, 0). Then
$AB=BC⇒A{B}^{2}=B{C}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(0-3\right)}^{2}={6}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=36-9=27\phantom{\rule{0ex}{0ex}}⇒x=±3\sqrt{3}$
If the coordinates of point A are , then the coordinates of D are .
If the coordinates of point A are , then the coordinates of D are .
Hence, the required coordinates are or .

#### Question 32:

Let (0, y) be the coordinates of B. Then
$0=\frac{-3+y}{2}⇒y=3$
Thus, the coordinates of B are (0, 3).
Here, AB = BC = AC and by symmetry the coordinates of A lies on x-axis. Let the coordinates of A be (x, 0). Then
$AB=BC⇒A{B}^{2}=B{C}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(0-3\right)}^{2}={6}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=36-9=27\phantom{\rule{0ex}{0ex}}⇒x=±3\sqrt{3}$
If the coordinates of point A are , then the coordinates of D are .
If the coordinates of point A are , then the coordinates of D are .
Hence, the required coordinates are or .

Let k be the ratio in which P( −1, y) divides the line segment joining the points A(−3, 10) and B(6, −8). Then

Substituting $k=\frac{2}{7}$ in , we get
$y=\frac{\frac{-8×2}{7}+10}{\frac{2}{7}+1}=\frac{-16+70}{9}=6$
Hence, the required ratio is 2 : 7 and y = 6.

#### Question 33:

Let k be the ratio in which P( −1, y) divides the line segment joining the points A(−3, 10) and B(6, −8). Then

Substituting $k=\frac{2}{7}$ in , we get
$y=\frac{\frac{-8×2}{7}+10}{\frac{2}{7}+1}=\frac{-16+70}{9}=6$
Hence, the required ratio is 2 : 7 and y = 6.

Here, the points P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Then

Now
$PQ=\sqrt{{\left(2+1\right)}^{2}+{\left(4-\frac{3}{2}\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\phantom{\rule{0ex}{0ex}}QR=\sqrt{{\left(5-2\right)}^{2}+{\left(\frac{3}{2}-4\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\phantom{\rule{0ex}{0ex}}RS=\sqrt{{\left(5-2\right)}^{2}+{\left(\frac{3}{2}+1\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\phantom{\rule{0ex}{0ex}}SP=\sqrt{{\left(2+1\right)}^{2}+{\left(-1-\frac{3}{2}\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\phantom{\rule{0ex}{0ex}}PR=\sqrt{{\left(5+1\right)}^{2}+{\left(\frac{3}{2}-\frac{3}{2}\right)}^{2}}=\sqrt{36}=6\phantom{\rule{0ex}{0ex}}QS=\sqrt{{\left(2-2\right)}^{2}+{\left(-1-4\right)}^{2}}=\sqrt{25}=5\phantom{\rule{0ex}{0ex}}$
Thus, PQ = QR = RS = SP and $PR\ne QS$ therefore PQRS is a rhombus.

#### Question 34:

Here, the points P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Then

Now
$PQ=\sqrt{{\left(2+1\right)}^{2}+{\left(4-\frac{3}{2}\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\phantom{\rule{0ex}{0ex}}QR=\sqrt{{\left(5-2\right)}^{2}+{\left(\frac{3}{2}-4\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\phantom{\rule{0ex}{0ex}}RS=\sqrt{{\left(5-2\right)}^{2}+{\left(\frac{3}{2}+1\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\phantom{\rule{0ex}{0ex}}SP=\sqrt{{\left(2+1\right)}^{2}+{\left(-1-\frac{3}{2}\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\phantom{\rule{0ex}{0ex}}PR=\sqrt{{\left(5+1\right)}^{2}+{\left(\frac{3}{2}-\frac{3}{2}\right)}^{2}}=\sqrt{36}=6\phantom{\rule{0ex}{0ex}}QS=\sqrt{{\left(2-2\right)}^{2}+{\left(-1-4\right)}^{2}}=\sqrt{25}=5\phantom{\rule{0ex}{0ex}}$
Thus, PQ = QR = RS = SP and $PR\ne QS$ therefore PQRS is a rhombus.

The midpoint of AB is .
Let k be the ratio in which P divides CD. So

Now, substituting $k=\frac{3}{2}$ in $\frac{k\left(\mathrm{y}\right)-4}{k+1}=2$, we get
$\frac{y×\frac{3}{2}-4}{\frac{3}{2}+1}=2\phantom{\rule{0ex}{0ex}}⇒\frac{3y-8}{5}=2\phantom{\rule{0ex}{0ex}}⇒y=\frac{10+8}{3}=6$
Hence, the required ratio is 3 : 2 and y = 6.

#### Question 35:

The midpoint of AB is .
Let k be the ratio in which P divides CD. So

Now, substituting $k=\frac{3}{2}$ in $\frac{k\left(\mathrm{y}\right)-4}{k+1}=2$, we get
$\frac{y×\frac{3}{2}-4}{\frac{3}{2}+1}=2\phantom{\rule{0ex}{0ex}}⇒\frac{3y-8}{5}=2\phantom{\rule{0ex}{0ex}}⇒y=\frac{10+8}{3}=6$
Hence, the required ratio is 3 : 2 and y = 6.

Suppose the line intersects the y-axis at P(0, y) and the x-axis at Q(x, 0).
It is given that (2, –5) is the mid-point of PQ.

Using mid-point formula, we have

Thus, the coordinates of P and Q are (0, −10) and (4, 0), respectively.

#### Question 36:

Suppose the line intersects the y-axis at P(0, y) and the x-axis at Q(x, 0).
It is given that (2, –5) is the mid-point of PQ.

Using mid-point formula, we have

Thus, the coordinates of P and Q are (0, −10) and (4, 0), respectively.

Let the point P divides the line PQ in the ratio k : 1.
Then, by the section formula:

#### Question 37:

Let the point P divides the line PQ in the ratio k : 1.
Then, by the section formula:

Let the coordinates of A, B, C be
Because D is the mid-point of BC, using mid-point formula, we have

Similarly, E is the mid point of AC. Using mid-point formula, we have;

Again, F is the mid point of AB. Using mid point formula, we have

Adding (i), (ii) and (iii), we get

On solving equation (iv)using equations (i), (ii) and (iii), we get
${x}_{1}=11\phantom{\rule{0ex}{0ex}}{x}_{2}=1\phantom{\rule{0ex}{0ex}}{x}_{3}=5$
Similarly,
${y}_{1}=12\phantom{\rule{0ex}{0ex}}{y}_{2}=2\phantom{\rule{0ex}{0ex}}{y}_{3}=6$
Hence, the points are: A(11,12), B(1,2) and C(5,6).

#### Question 38:

Let the coordinates of A, B, C be
Because D is the mid-point of BC, using mid-point formula, we have

Similarly, E is the mid point of AC. Using mid-point formula, we have;

Again, F is the mid point of AB. Using mid point formula, we have

Adding (i), (ii) and (iii), we get

On solving equation (iv)using equations (i), (ii) and (iii), we get
${x}_{1}=11\phantom{\rule{0ex}{0ex}}{x}_{2}=1\phantom{\rule{0ex}{0ex}}{x}_{3}=5$
Similarly,
${y}_{1}=12\phantom{\rule{0ex}{0ex}}{y}_{2}=2\phantom{\rule{0ex}{0ex}}{y}_{3}=6$
Hence, the points are: A(11,12), B(1,2) and C(5,6).

Let ABCD be the parallelogram with two adjacent vertices A(3, 2) and B(−1, 0). Suppose O(2, −5) be the point of intersection of the diagonals AC and BD.
Let C(x1y1) and D(x2y2) be the coordinates of the other vertices of the parallelogram.

We know that the diagonals of the parallelogram bisect each other. Therefore, O is the mid-point of AC and BD.
Using the mid-point formula, we have

So, the coordinates of C are (1, −12).
Also,

So, the coordinates of D are (5, −10).

#### Question 39:

Let ABCD be the parallelogram with two adjacent vertices A(3, 2) and B(−1, 0). Suppose O(2, −5) be the point of intersection of the diagonals AC and BD.
Let C(x1y1) and D(x2y2) be the coordinates of the other vertices of the parallelogram.

We know that the diagonals of the parallelogram bisect each other. Therefore, O is the mid-point of AC and BD.
Using the mid-point formula, we have

So, the coordinates of C are (1, −12).
Also,

So, the coordinates of D are (5, −10).

Given: Point A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD.

Diagonals of a parallelogram bisect each other.

∴ Mid point of AC = Mid point of BD

Hence,  the values of a and is 6 and 3, respectively.

#### Question 40:

Given: Point A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD.

Diagonals of a parallelogram bisect each other.

∴ Mid point of AC = Mid point of BD

Hence,  the values of a and is 6 and 3, respectively.

Let the points A(2, 1) and B(5, –8) is trisected at the points P(x, y) and Q(a, b).

Thus, AP = PQ = QB

Therefore, P divides AB internally in the ratio 1 : 2.

Section formula: if the point (xy) divides the line segment joining the points (x1y1) and (x2y2) internally in the ratio : n, then the coordinates (xy) =

Therefore, using section formula, the coordinates of P are:

Hence, the coordinates of P are (3, –2).

Since, P also lies on the line given by 2x – y + k = 0,
Therefore, (3, –2) satisfies the equation 2x – y + k = 0
$2\left(3\right)-\left(-2\right)+k=0\phantom{\rule{0ex}{0ex}}⇒6+2+k=0\phantom{\rule{0ex}{0ex}}⇒k=-8$

Hence,  the values of k is –8.

#### Question 41:

Let the points A(2, 1) and B(5, –8) is trisected at the points P(x, y) and Q(a, b).

Thus, AP = PQ = QB

Therefore, P divides AB internally in the ratio 1 : 2.

Section formula: if the point (xy) divides the line segment joining the points (x1y1) and (x2y2) internally in the ratio : n, then the coordinates (xy) =

Therefore, using section formula, the coordinates of P are:

Hence, the coordinates of P are (3, –2).

Since, P also lies on the line given by 2x – y + k = 0,
Therefore, (3, –2) satisfies the equation 2x – y + k = 0
$2\left(3\right)-\left(-2\right)+k=0\phantom{\rule{0ex}{0ex}}⇒6+2+k=0\phantom{\rule{0ex}{0ex}}⇒k=-8$

Hence,  the values of k is –8.

Section formula: if the point (xy) divides the line segment joining the points (x1y1) and (x2y2) internally in the ratio : 1, then the coordinates (xy) =

Let the point P(0, y) divides the line segment joining the points A(5, –6) and B(–1, 4) in the ratio : 1.

Therefore, using section formula, the coordinates of P are:

Hence, the y-axis divides the line segment joining the points A(5, –6) and B(–1, 4) in the ratio 5 : 1.
and the coordinates of the point of division are .

#### Question 1:

Section formula: if the point (xy) divides the line segment joining the points (x1y1) and (x2y2) internally in the ratio : 1, then the coordinates (xy) =

Let the point P(0, y) divides the line segment joining the points A(5, –6) and B(–1, 4) in the ratio : 1.

Therefore, using section formula, the coordinates of P are:

Hence, the y-axis divides the line segment joining the points A(5, –6) and B(–1, 4) in the ratio 5 : 1.
and the coordinates of the point of division are .

(i) A(1, 2), B(−2, 3) and C(−3, −4) are the vertices of âˆ†ABC. Then,
(x1 = 1, y1 = 2), (x2 = −2, y2 = 3) and (x3 = −3, y3 = -4)

(ii) A(−5, 7), B(−4, −5) and C(4, 5) are the vertices of âˆ†ABC. Then,
(x1 = −5, y1 = 7), (x2 = −4, y2 = −5) and (x3 = 4, y3 = 5)

(iii) A(3, 8), B(−4, 2) and C(5, −1) are the vertices of âˆ†ABC. Then,
(x1 = 3, y1 = 8), (x2 = −4, y2 = 2) and (x3 = 5, y3 = −1)

(iv) A(10, −6), B(2, 5) and C(−1, −3) are the vertices of âˆ†ABC. Then,
(x1 = 10, y1 = −6), (x2 = 2, y2 = 5) and (x3 = −1, y3 = 3)

#### Question 2:

(i) A(1, 2), B(−2, 3) and C(−3, −4) are the vertices of âˆ†ABC. Then,
(x1 = 1, y1 = 2), (x2 = −2, y2 = 3) and (x3 = −3, y3 = -4)

(ii) A(−5, 7), B(−4, −5) and C(4, 5) are the vertices of âˆ†ABC. Then,
(x1 = −5, y1 = 7), (x2 = −4, y2 = −5) and (x3 = 4, y3 = 5)

(iii) A(3, 8), B(−4, 2) and C(5, −1) are the vertices of âˆ†ABC. Then,
(x1 = 3, y1 = 8), (x2 = −4, y2 = 2) and (x3 = 5, y3 = −1)

(iv) A(10, −6), B(2, 5) and C(−1, −3) are the vertices of âˆ†ABC. Then,
(x1 = 10, y1 = −6), (x2 = 2, y2 = 5) and (x3 = −1, y3 = 3)

By joining A and C, we get two triangles ABC and ACD.
Let . Then

So, the area of the quadrilateral is 25 + 107 = 132 sq. units.

#### Question 3:

By joining A and C, we get two triangles ABC and ACD.
Let . Then

So, the area of the quadrilateral is 25 + 107 = 132 sq. units.

By joining P and R, we get two triangles PQR and PRS.
Let . Then

So, the area of the quadrilateral PQRS is  sq. units.

#### Question 4:

By joining P and R, we get two triangles PQR and PRS.
Let . Then

So, the area of the quadrilateral PQRS is  sq. units.

By joining A and C, we get two triangles ABC and ACD.
Let . Then

So, the area of the quadrilateral ABCD is  sq. units.

#### Question 5:

By joining A and C, we get two triangles ABC and ACD.
Let . Then

So, the area of the quadrilateral ABCD is  sq. units.

Consider the figure.
Construction: Produce AC by joining points A to C to form two triangles,
In $∆\mathrm{ABC},$

We know that,

Similarly, in $∆\mathrm{ADC},$

Now, ar(quad. ABCD) = $ar\left(∆\mathrm{ABC}\right)+ar\left(∆\mathrm{ADC}\right)$

ar(quad. ABCD) = $\frac{35}{2}+\frac{109}{2}=\frac{144}{2}=72$
Therefore, area of quadrilateral ABCD is 72 sq. units

Disclaimer: The answer thus calculated does not match with the answer given in the book.
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#### Question 6:

Consider the figure.
Construction: Produce AC by joining points A to C to form two triangles,
In $∆\mathrm{ABC},$

We know that,

Similarly, in $∆\mathrm{ADC},$

Now, ar(quad. ABCD) = $ar\left(∆\mathrm{ABC}\right)+ar\left(∆\mathrm{ADC}\right)$

ar(quad. ABCD) = $\frac{35}{2}+\frac{109}{2}=\frac{144}{2}=72$
Therefore, area of quadrilateral ABCD is 72 sq. units

Disclaimer: The answer thus calculated does not match with the answer given in the book.
â€‹

The vertices of the triangle are A(2, 1), B(4, 3) and C(2, 5).

Now

Hence, the area of the required triangle is 1 sq. unit.

#### Question 7:

The vertices of the triangle are A(2, 1), B(4, 3) and C(2, 5).

Now

Hence, the area of the required triangle is 1 sq. unit.

The vertices of the triangle are A(7, −3), B(5, 3), C(3, −1).

For the area of the triangle ADC, let . Then

Now, for the area of triangle ABD, let . Then

Thus, .
Hence, AD divides $∆ABC$ into two triangles of equal areas.

#### Question 8:

The vertices of the triangle are A(7, −3), B(5, 3), C(3, −1).

For the area of the triangle ADC, let . Then

Now, for the area of triangle ABD, let . Then

Thus, .
Hence, AD divides $∆ABC$ into two triangles of equal areas.

Let be the coordinates of B and C respectively. Since, the coordinates of A are (1, −4), therefore
$\frac{1+{x}_{2}}{2}=2⇒{x}_{2}=3\phantom{\rule{0ex}{0ex}}\frac{-4+{y}_{2}}{2}=-1⇒{y}_{2}=2\phantom{\rule{0ex}{0ex}}\frac{1+{x}_{3}}{2}=0⇒{x}_{3}=-1\phantom{\rule{0ex}{0ex}}\frac{-4+{y}_{3}}{2}=-1⇒{y}_{3}=2$
Let . Now

Hence, the area of the triangle $∆ABC$ is 12 sq. units.

#### Question 9:

Let be the coordinates of B and C respectively. Since, the coordinates of A are (1, −4), therefore
$\frac{1+{x}_{2}}{2}=2⇒{x}_{2}=3\phantom{\rule{0ex}{0ex}}\frac{-4+{y}_{2}}{2}=-1⇒{y}_{2}=2\phantom{\rule{0ex}{0ex}}\frac{1+{x}_{3}}{2}=0⇒{x}_{3}=-1\phantom{\rule{0ex}{0ex}}\frac{-4+{y}_{3}}{2}=-1⇒{y}_{3}=2$
Let . Now

Hence, the area of the triangle $∆ABC$ is 12 sq. units.

Let (x, y) be the coordinates of D and be the coordinates of E. Since, the diagonals of a parallelogram bisect
each other at the same point, therefore
$\frac{x+8}{2}=\frac{6+9}{2}⇒x=7\phantom{\rule{0ex}{0ex}}\frac{y+2}{2}=\frac{1+4}{2}⇒y=3$
Thus, the coordinates of D are (7, 3).
E is the midpoint of DC, therefore
$x\text{'}=\frac{7+9}{2}⇒x\text{'}=8\phantom{\rule{0ex}{0ex}}y\text{'}=\frac{3+4}{2}⇒y\text{'}=\frac{7}{2}$
Thus, the coordinates of E are .
Let . Now

Hence, the area of the triangle $∆ADE$ is .

#### Question 10:

Let (x, y) be the coordinates of D and be the coordinates of E. Since, the diagonals of a parallelogram bisect
each other at the same point, therefore
$\frac{x+8}{2}=\frac{6+9}{2}⇒x=7\phantom{\rule{0ex}{0ex}}\frac{y+2}{2}=\frac{1+4}{2}⇒y=3$
Thus, the coordinates of D are (7, 3).
E is the midpoint of DC, therefore
$x\text{'}=\frac{7+9}{2}⇒x\text{'}=8\phantom{\rule{0ex}{0ex}}y\text{'}=\frac{3+4}{2}⇒y\text{'}=\frac{7}{2}$
Thus, the coordinates of E are .
Let . Now

Hence, the area of the triangle $∆ADE$ is .

(i) Let . Now
$\mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{\mathit{1}}\mathit{-}{y}_{\mathit{2}}\right)\right]\phantom{\rule{0ex}{0ex}}⇒15=\frac{1}{2}\left[1\left(p-7\right)+4\left(7+3\right)-9\left(-3-p\right)\right]\phantom{\rule{0ex}{0ex}}⇒15=\frac{1}{2}\left[10p+60\right]\phantom{\rule{0ex}{0ex}}⇒\left|10p+60\right|=30$
Therefore

Hence, .

(ii)
Let
Now

Hence,

#### Question 11:

(i) Let . Now
$\mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{\mathit{1}}\mathit{-}{y}_{\mathit{2}}\right)\right]\phantom{\rule{0ex}{0ex}}⇒15=\frac{1}{2}\left[1\left(p-7\right)+4\left(7+3\right)-9\left(-3-p\right)\right]\phantom{\rule{0ex}{0ex}}⇒15=\frac{1}{2}\left[10p+60\right]\phantom{\rule{0ex}{0ex}}⇒\left|10p+60\right|=30$
Therefore

Hence, .

(ii)
Let
Now

Hence,

Let . Now
$\mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}\mathit{-}{y}_{2}\right)\right]\phantom{\rule{0ex}{0ex}}⇒6=\frac{1}{2}\left[\left(k+1\right)\left(-3+k\right)+4\left(-k-1\right)+7\left(1+3\right)\right]\phantom{\rule{0ex}{0ex}}⇒6=\frac{1}{2}\left[{k}^{2}-2k-3-4k-4+28\right]\phantom{\rule{0ex}{0ex}}⇒{k}^{2}-6k+9=0$
$⇒{\left(k-3\right)}^{2}=0⇒k=3$
Hence, k = 3.

#### Question 12:

Let . Now
$\mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}\mathit{-}{y}_{2}\right)\right]\phantom{\rule{0ex}{0ex}}⇒6=\frac{1}{2}\left[\left(k+1\right)\left(-3+k\right)+4\left(-k-1\right)+7\left(1+3\right)\right]\phantom{\rule{0ex}{0ex}}⇒6=\frac{1}{2}\left[{k}^{2}-2k-3-4k-4+28\right]\phantom{\rule{0ex}{0ex}}⇒{k}^{2}-6k+9=0$
$⇒{\left(k-3\right)}^{2}=0⇒k=3$
Hence, k = 3.

Let be the vertices of
the triangle. So
$\mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]\phantom{\rule{0ex}{0ex}}⇒53=\frac{1}{2}\left[\left(-2\right)\left(-4-10\right)+k\left(10-5\right)+\left(2k+1\right)\left(5+4\right)\right]\phantom{\rule{0ex}{0ex}}⇒53=\frac{1}{2}\left[28+5k+9\left(2k+1\right)\right]\phantom{\rule{0ex}{0ex}}⇒28+5k+18k+9=106$

$⇒37+23k=106\phantom{\rule{0ex}{0ex}}⇒23k=106-37=69\phantom{\rule{0ex}{0ex}}⇒k=\frac{69}{23}=3$
Hence, k = 3.

#### Question 13:

Let be the vertices of
the triangle. So
$\mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]\phantom{\rule{0ex}{0ex}}⇒53=\frac{1}{2}\left[\left(-2\right)\left(-4-10\right)+k\left(10-5\right)+\left(2k+1\right)\left(5+4\right)\right]\phantom{\rule{0ex}{0ex}}⇒53=\frac{1}{2}\left[28+5k+9\left(2k+1\right)\right]\phantom{\rule{0ex}{0ex}}⇒28+5k+18k+9=106$

$⇒37+23k=106\phantom{\rule{0ex}{0ex}}⇒23k=106-37=69\phantom{\rule{0ex}{0ex}}⇒k=\frac{69}{23}=3$
Hence, k = 3.

(i)
Let A(x1 = 2, y1 = −2), B(x2 = −3, y2 = 8) and C(x3 = −1, y3 = 4) be the given points. Now
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\phantom{\rule{0ex}{0ex}}=2\left(8-4\right)+\left(-3\right)\left(4+2\right)+\left(-1\right)\left(-2-8\right)\phantom{\rule{0ex}{0ex}}=8-18+10\phantom{\rule{0ex}{0ex}}=0$
Hence, the given points are collinear.

(ii)
Let A(x1 = −5, y1 = 1), B(x2 = 5, y2 = 5) and C(x3 = 10, y3 = 7) be the given points. Now
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\phantom{\rule{0ex}{0ex}}=\left(-5\right)\left(5-7\right)+5\left(7-1\right)+10\left(1-5\right)\phantom{\rule{0ex}{0ex}}=-5\left(-2\right)+5\left(6\right)+10\left(-4\right)\phantom{\rule{0ex}{0ex}}=10+30-40\phantom{\rule{0ex}{0ex}}=0$
Hence, the given points are collinear.

(iii)
Let A(x1 = 5, y1 = 1), B(x2 = 1, y2 = −1) and C(x3 = 11, y3 = 4) be the given points. Now
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\phantom{\rule{0ex}{0ex}}=5\left(-1-4\right)+1\left(4-1\right)+11\left(1+1\right)\phantom{\rule{0ex}{0ex}}=-25+3+22\phantom{\rule{0ex}{0ex}}=0$
Hence, the given points are collinear.

(iv)
Let A(x1 = 8, y1 = 1), B(x2 = 3, y2 = −4) and C(x3 = 2, y3 = −5) be the given points. Now
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\phantom{\rule{0ex}{0ex}}=8\left(-4+5\right)+3\left(-5-1\right)+2\left(1+4\right)\phantom{\rule{0ex}{0ex}}=8-18+10\phantom{\rule{0ex}{0ex}}=0$
Hence, the given points are collinear.

#### Question 14:

(i)
Let A(x1 = 2, y1 = −2), B(x2 = −3, y2 = 8) and C(x3 = −1, y3 = 4) be the given points. Now
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\phantom{\rule{0ex}{0ex}}=2\left(8-4\right)+\left(-3\right)\left(4+2\right)+\left(-1\right)\left(-2-8\right)\phantom{\rule{0ex}{0ex}}=8-18+10\phantom{\rule{0ex}{0ex}}=0$
Hence, the given points are collinear.

(ii)
Let A(x1 = −5, y1 = 1), B(x2 = 5, y2 = 5) and C(x3 = 10, y3 = 7) be the given points. Now
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\phantom{\rule{0ex}{0ex}}=\left(-5\right)\left(5-7\right)+5\left(7-1\right)+10\left(1-5\right)\phantom{\rule{0ex}{0ex}}=-5\left(-2\right)+5\left(6\right)+10\left(-4\right)\phantom{\rule{0ex}{0ex}}=10+30-40\phantom{\rule{0ex}{0ex}}=0$
Hence, the given points are collinear.

(iii)
Let A(x1 = 5, y1 = 1), B(x2 = 1, y2 = −1) and C(x3 = 11, y3 = 4) be the given points. Now
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\phantom{\rule{0ex}{0ex}}=5\left(-1-4\right)+1\left(4-1\right)+11\left(1+1\right)\phantom{\rule{0ex}{0ex}}=-25+3+22\phantom{\rule{0ex}{0ex}}=0$
Hence, the given points are collinear.

(iv)
Let A(x1 = 8, y1 = 1), B(x2 = 3, y2 = −4) and C(x3 = 2, y3 = −5) be the given points. Now
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\phantom{\rule{0ex}{0ex}}=8\left(-4+5\right)+3\left(-5-1\right)+2\left(1+4\right)\phantom{\rule{0ex}{0ex}}=8-18+10\phantom{\rule{0ex}{0ex}}=0$
Hence, the given points are collinear.

Let . So, the condition for three collinear points is
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}\mathit{-}{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒x\left(-4+5\right)-3\left(-5-2\right)+7\left(2+4\right)=0\phantom{\rule{0ex}{0ex}}⇒x+21+42=0\phantom{\rule{0ex}{0ex}}⇒x=-63$
Hence, x = − 63.

#### Question 15:

Let . So, the condition for three collinear points is
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}\mathit{-}{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒x\left(-4+5\right)-3\left(-5-2\right)+7\left(2+4\right)=0\phantom{\rule{0ex}{0ex}}⇒x+21+42=0\phantom{\rule{0ex}{0ex}}⇒x=-63$
Hence, x = − 63.

A(−3, 12), B(7, 6) and C(x, 9) are the given points. Then:
(x1 = −3, y1 = 12), (x2 = 7, y2 = 6) and (x3 = x, y3 = 9)
It is given that the points A, B and C are collinear. Therefore,

Therefore, when x= 2, the given points are collinear.

#### Question 16:

A(−3, 12), B(7, 6) and C(x, 9) are the given points. Then:
(x1 = −3, y1 = 12), (x2 = 7, y2 = 6) and (x3 = x, y3 = 9)
It is given that the points A, B and C are collinear. Therefore,

Therefore, when x= 2, the given points are collinear.

If the area of the triangle formed by three points is equal to zero, then the points are collinear.

Area of the triangle formed by the vertices  $\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|$.

Now, the given points A(–5, 1), B(1, p) and C(4, –2) are collinear.

Therefore, Area of triangle formed by them is equal to zero.

Hence, the value of p is –1.

#### Question 17:

If the area of the triangle formed by three points is equal to zero, then the points are collinear.

Area of the triangle formed by the vertices  $\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|$.

Now, the given points A(–5, 1), B(1, p) and C(4, –2) are collinear.

Therefore, Area of triangle formed by them is equal to zero.

Hence, the value of p is –1.

Let A(x1 = −3, y1 = 9), B(x2 = 2, y2 = y) and C(x3 = 4, y3 = −5) be the given points.
The given points are collinear if
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(-3\right)\left(y+5\right)+2\left(-5-9\right)+4\left(9-y\right)=0\phantom{\rule{0ex}{0ex}}⇒-3y-15-28+36-4y=0\phantom{\rule{0ex}{0ex}}⇒7y=36-43$
$⇒y=-1$

#### Question 18:

Let A(x1 = −3, y1 = 9), B(x2 = 2, y2 = y) and C(x3 = 4, y3 = −5) be the given points.
The given points are collinear if
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(-3\right)\left(y+5\right)+2\left(-5-9\right)+4\left(9-y\right)=0\phantom{\rule{0ex}{0ex}}⇒-3y-15-28+36-4y=0\phantom{\rule{0ex}{0ex}}⇒7y=36-43$
$⇒y=-1$

Let A(x1 = 8, y1 = 1), B(x2 = 3, y2 = −2k) and C(x3 = ky3 = −5) be the given points.
The given points are collinear if
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒8\left(-2k+5\right)+3\left(-5-1\right)+k\left(1+2k\right)=0\phantom{\rule{0ex}{0ex}}⇒-16k+40-18+k+2{k}^{2}=0\phantom{\rule{0ex}{0ex}}⇒2{k}^{2}-15k+22=0$

Hence, .

#### Question 19:

Let A(x1 = 8, y1 = 1), B(x2 = 3, y2 = −2k) and C(x3 = ky3 = −5) be the given points.
The given points are collinear if
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒8\left(-2k+5\right)+3\left(-5-1\right)+k\left(1+2k\right)=0\phantom{\rule{0ex}{0ex}}⇒-16k+40-18+k+2{k}^{2}=0\phantom{\rule{0ex}{0ex}}⇒2{k}^{2}-15k+22=0$

Hence, .

Let A(x1 = 2, y1 = 1), B(x2 = xy2 = y) and C(x3 = 7, y3 = 5) be the given points.
The given points are collinear if
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒2\left(y-5\right)+x\left(5-1\right)+7\left(1-y\right)=0\phantom{\rule{0ex}{0ex}}⇒2y-10+4x+7-7y=0\phantom{\rule{0ex}{0ex}}⇒4x-5y-3=0$
Hence, the required relation is 4x − 5y − 3 = 0.

#### Question 20:

Let A(x1 = 2, y1 = 1), B(x2 = xy2 = y) and C(x3 = 7, y3 = 5) be the given points.
The given points are collinear if
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒2\left(y-5\right)+x\left(5-1\right)+7\left(1-y\right)=0\phantom{\rule{0ex}{0ex}}⇒2y-10+4x+7-7y=0\phantom{\rule{0ex}{0ex}}⇒4x-5y-3=0$
Hence, the required relation is 4x − 5y − 3 = 0.

Let A(x1 = xy1 = y), B(x2 = −5, y2 = 7) and C(x3 = −4, y3 = 5) be the given points.
The given points are collinear if
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒x\left(7-5\right)+\left(-5\right)\left(5-y\right)+\left(-4\right)\left(y-7\right)=0\phantom{\rule{0ex}{0ex}}⇒7x-5x-25+5y-4y+28=0\phantom{\rule{0ex}{0ex}}⇒2x+y+3=0$
Hence, the required relation is 2x + y + 3 = 0.

#### Question 21:

Let A(x1 = xy1 = y), B(x2 = −5, y2 = 7) and C(x3 = −4, y3 = 5) be the given points.
The given points are collinear if
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒x\left(7-5\right)+\left(-5\right)\left(5-y\right)+\left(-4\right)\left(y-7\right)=0\phantom{\rule{0ex}{0ex}}⇒7x-5x-25+5y-4y+28=0\phantom{\rule{0ex}{0ex}}⇒2x+y+3=0$
Hence, the required relation is 2x + y + 3 = 0.

Consider the points A(a, 0), B(0, b) and C(1, 1).
Here, (x1 = a, y1 = 0), (x2 = 0, y2 = b) and (x3 = 1, y3 = 1).
It is given that the points are collinear. So,

Therefore, the given points are collinear if $\left(\frac{1}{a}+\frac{1}{b}\right)$ = 1.

#### Question 22:

Consider the points A(a, 0), B(0, b) and C(1, 1).
Here, (x1 = a, y1 = 0), (x2 = 0, y2 = b) and (x3 = 1, y3 = 1).
It is given that the points are collinear. So,

Therefore, the given points are collinear if $\left(\frac{1}{a}+\frac{1}{b}\right)$ = 1.

Let A(x1 = −3, y1 = 9), B(x2 = ay2 = b) and C(x3 = 4, y3 = 5) be the given points.
The given points are collinear if
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(-3\right)\left(b+5\right)+a\left(-5-9\right)+4\left(9-b\right)=0\phantom{\rule{0ex}{0ex}}⇒-3b-15-14a+36-4b=0\phantom{\rule{0ex}{0ex}}⇒2a+b=3$
Now, solving a + b = 1 and 2a + b = 3, we get a = 2 and b = −1.
Hence, a = 2 and b = −1.

#### Question 23:

Let A(x1 = −3, y1 = 9), B(x2 = ay2 = b) and C(x3 = 4, y3 = 5) be the given points.
The given points are collinear if
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(-3\right)\left(b+5\right)+a\left(-5-9\right)+4\left(9-b\right)=0\phantom{\rule{0ex}{0ex}}⇒-3b-15-14a+36-4b=0\phantom{\rule{0ex}{0ex}}⇒2a+b=3$
Now, solving a + b = 1 and 2a + b = 3, we get a = 2 and b = −1.
Hence, a = 2 and b = −1.

Let A(x1 = 0, y1 = −1), B(x2 = 2, y2 = 1) and C(x3 = 0, y3 = 3) be the given points. Then

So, the area of the triangle $∆ABC$ is 4 sq. units.
Let D(a1, b1), E(a2, b2) and F(a3, b3) be the midpoints of AB, BC and AC respectively. Then

Thus, the coordinates of D, E and F are D(a1 = 1, b1 = 0), E(a2 = 1, b2 = 2) and F(a3 = 0, b3 = 1). Now

So, the area of the triangle $∆DEF$ is 1 sq. unit.
Hence, .

#### Question 24:

Let A(x1 = 0, y1 = −1), B(x2 = 2, y2 = 1) and C(x3 = 0, y3 = 3) be the given points. Then

So, the area of the triangle $∆ABC$ is 4 sq. units.
Let D(a1, b1), E(a2, b2) and F(a3, b3) be the midpoints of AB, BC and AC respectively. Then

Thus, the coordinates of D, E and F are D(a1 = 1, b1 = 0), E(a2 = 1, b2 = 2) and F(a3 = 0, b3 = 1). Now

So, the area of the triangle $∆DEF$ is 1 sq. unit.
Hence, .

Let A(aa2), B(bb2) and C(0, 0) be the coordinates of the given points.
We know that the area of triangle having vertices  is $\left|\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]\right|$ square units.
So,
Area of âˆ†ABC

Since the area of the triangle formed by the points (aa2), (bb2) and (0, 0) is not zero, so the given points are not collinear.

#### Question 25:

Let A(aa2), B(bb2) and C(0, 0) be the coordinates of the given points.
We know that the area of triangle having vertices  is $\left|\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]\right|$ square units.
So,
Area of âˆ†ABC

Since the area of the triangle formed by the points (aa2), (bb2) and (0, 0) is not zero, so the given points are not collinear.

Area of the triangle formed by the vertices  $\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|$.

Now, the given vertices are (x, 3), (4, 4) and  (3, 5)
and the given area is 4 square units.

Therefore,

Hence, â€‹the value of x is 13 and −3.

#### Question 1:

Area of the triangle formed by the vertices  $\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|$.

Now, the given vertices are (x, 3), (4, 4) and  (3, 5)
and the given area is 4 square units.

Therefore,

Hence, â€‹the value of x is 13 and −3.

The given points are A(−1, y), B(5, 7) and O(2, −3y).
Here, AO and BO are the radii of the circle. So
$AO=BO⇒A{O}^{2}=B{O}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(2+1\right)}^{2}+{\left(-3y-y\right)}^{2}={\left(2-5\right)}^{2}+{\left(-3y-7\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒9+{\left(4y\right)}^{2}={\left(-3\right)}^{2}+{\left(3y+7\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒9+16{y}^{2}=9+9{y}^{2}+49+42y$
$⇒7{y}^{2}-42y-49=0\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-6y-7=0\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-7y+y-7=0\phantom{\rule{0ex}{0ex}}⇒y\left(y-7\right)+1\left(y-7\right)=0$

Hence, y = 7 or y = −1.

#### Question 2:

The given points are A(−1, y), B(5, 7) and O(2, −3y).
Here, AO and BO are the radii of the circle. So
$AO=BO⇒A{O}^{2}=B{O}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(2+1\right)}^{2}+{\left(-3y-y\right)}^{2}={\left(2-5\right)}^{2}+{\left(-3y-7\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒9+{\left(4y\right)}^{2}={\left(-3\right)}^{2}+{\left(3y+7\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒9+16{y}^{2}=9+9{y}^{2}+49+42y$
$⇒7{y}^{2}-42y-49=0\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-6y-7=0\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-7y+y-7=0\phantom{\rule{0ex}{0ex}}⇒y\left(y-7\right)+1\left(y-7\right)=0$

Hence, y = 7 or y = −1.

The given points are A(0, 2), B(3, p) and C(p, 5).
$AB=AC⇒A{B}^{2}=A{C}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(3-0\right)}^{2}+{\left(p-2\right)}^{2}={\left(p-0\right)}^{2}+{\left(5-2\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒9+{p}^{2}-4p+4={p}^{2}+9\phantom{\rule{0ex}{0ex}}⇒4p=4⇒p=1$
Hence, p = 1.

#### Question 3:

The given points are A(0, 2), B(3, p) and C(p, 5).
$AB=AC⇒A{B}^{2}=A{C}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(3-0\right)}^{2}+{\left(p-2\right)}^{2}={\left(p-0\right)}^{2}+{\left(5-2\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒9+{p}^{2}-4p+4={p}^{2}+9\phantom{\rule{0ex}{0ex}}⇒4p=4⇒p=1$
Hence, p = 1.

The given vertices are B(4, 0), C(4, 3) and D(0, 3). Here, BD is one of the diagonals. So

Hence, the length of the diagonal is 5 units..

#### Question 4:

The given vertices are B(4, 0), C(4, 3) and D(0, 3). Here, BD is one of the diagonals. So

Hence, the length of the diagonal is 5 units..

The given points are P(k − 1, 2), A(3, k) and B(k, 5).

Hence, k = 1 or k = 5.

#### Question 5:

The given points are P(k − 1, 2), A(3, k) and B(k, 5).

Hence, k = 1 or k = 5.

Let k be the ratio in which the point P(x, 2) divides the line joining the points A(x1 = 12, y1 = 5) and B(x2 = 4, y2 = −3). Then

Now
$2=\frac{k×\left(-3\right)+5}{k+1}⇒2k+2=-3k+5⇒k=\frac{3}{5}\phantom{\rule{0ex}{0ex}}$
Hence, the required ratio is 3 : 5.

#### Question 6:

Let k be the ratio in which the point P(x, 2) divides the line joining the points A(x1 = 12, y1 = 5) and B(x2 = 4, y2 = −3). Then

Now
$2=\frac{k×\left(-3\right)+5}{k+1}⇒2k+2=-3k+5⇒k=\frac{3}{5}\phantom{\rule{0ex}{0ex}}$
Hence, the required ratio is 3 : 5.

The vertices of the rectangle ABCD are A(2, −1), B(5, −1), C(5, 6) and D(2, 6). Now

Since, the midpoints of AC and BD coincide, therefore the diagonals of rectangle ABCD bisect each other.

#### Question 7:

The vertices of the rectangle ABCD are A(2, −1), B(5, −1), C(5, 6) and D(2, 6). Now

Since, the midpoints of AC and BD coincide, therefore the diagonals of rectangle ABCD bisect each other.

The given vertices are A(7, −3), B(5, 3) and C(3, −1).
Since D and E are the midpoints of BC and AC respectively, therefore

Now
$AD=\sqrt{{\left(7-4\right)}^{2}+{\left(-3-1\right)}^{2}}=\sqrt{9+16}=5\phantom{\rule{0ex}{0ex}}BE=\sqrt{{\left(5-5\right)}^{2}+{\left(3+2\right)}^{2}}=\sqrt{0+25}=5$
Hence, AD = BE = 5 units.

#### Question 8:

The given vertices are A(7, −3), B(5, 3) and C(3, −1).
Since D and E are the midpoints of BC and AC respectively, therefore

Now
$AD=\sqrt{{\left(7-4\right)}^{2}+{\left(-3-1\right)}^{2}}=\sqrt{9+16}=5\phantom{\rule{0ex}{0ex}}BE=\sqrt{{\left(5-5\right)}^{2}+{\left(3+2\right)}^{2}}=\sqrt{0+25}=5$
Hence, AD = BE = 5 units.

Here, the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2 : 3. So

Hence, $k=\frac{16}{5}$.

#### Question 9:

Here, the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2 : 3. So

Hence, $k=\frac{16}{5}$.

Let P(x, 0) be the point on x-axis. Then
$AP=BP⇒A{P}^{2}=B{P}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x+1\right)}^{2}+{\left(0-0\right)}^{2}={\left(x-5\right)}^{2}+{\left(0-0\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x+1={x}^{2}-10x+25\phantom{\rule{0ex}{0ex}}⇒12x=24⇒x=2$
Hence, x = 2.

#### Question 10:

Let P(x, 0) be the point on x-axis. Then
$AP=BP⇒A{P}^{2}=B{P}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x+1\right)}^{2}+{\left(0-0\right)}^{2}={\left(x-5\right)}^{2}+{\left(0-0\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x+1={x}^{2}-10x+25\phantom{\rule{0ex}{0ex}}⇒12x=24⇒x=2$
Hence, x = 2.

The given points are .
Then,
Therefore,

#### Question 11:

The given points are .
Then,
Therefore,

$⇒6-3a=5\phantom{\rule{0ex}{0ex}}⇒3a=1\phantom{\rule{0ex}{0ex}}⇒a=\frac{1}{3}$
.

#### Question 12:

$⇒6-3a=5\phantom{\rule{0ex}{0ex}}⇒3a=1\phantom{\rule{0ex}{0ex}}⇒a=\frac{1}{3}$
.

.

#### Question 13:

.

Let the point  be equidistant from the points A(7, 1) and B(3, 5).
Then,
$PA=PB\phantom{\rule{0ex}{0ex}}⇒P{A}^{2}=P{B}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-7\right)}^{2}+{\left(y-1\right)}^{2}={\left(x-3\right)}^{2}+{\left(y-5\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-14x-2y+50={x}^{2}+{y}^{2}-6x-10y+34\phantom{\rule{0ex}{0ex}}⇒8x-8y=16\phantom{\rule{0ex}{0ex}}⇒x-y=2$

#### Question 14:

Let the point  be equidistant from the points A(7, 1) and B(3, 5).
Then,
$PA=PB\phantom{\rule{0ex}{0ex}}⇒P{A}^{2}=P{B}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-7\right)}^{2}+{\left(y-1\right)}^{2}={\left(x-3\right)}^{2}+{\left(y-5\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-14x-2y+50={x}^{2}+{y}^{2}-6x-10y+34\phantom{\rule{0ex}{0ex}}⇒8x-8y=16\phantom{\rule{0ex}{0ex}}⇒x-y=2$

The given points are A(a, b), B(b, c) and C(c, a).
Here,

Let the centroid be (x, y).
Then,

But it is given that the centroid of the triangle is the origin.
Then, we have:
$\frac{a+b+c}{3}=0\phantom{\rule{0ex}{0ex}}⇒a+b+c=0$

#### Question 15:

The given points are A(a, b), B(b, c) and C(c, a).
Here,

Let the centroid be (x, y).
Then,

But it is given that the centroid of the triangle is the origin.
Then, we have:
$\frac{a+b+c}{3}=0\phantom{\rule{0ex}{0ex}}⇒a+b+c=0$

The given points are A(2, 2), B(−4, −4) and C(5, −8).
Here,
Let G(x, y) be the centroid of $∆ABC$. Then,

Hence, the centroid of  .

#### Question 16:

The given points are A(2, 2), B(−4, −4) and C(5, −8).
Here,
Let G(x, y) be the centroid of $∆ABC$. Then,

Hence, the centroid of  .

Let the required ratio be .
Then, by section formula, the coordinates of C are
$C\left(\frac{7k+2}{k+1},\frac{8k+3}{k+1}\right)$
Therefore,

#### Question 17:

Let the required ratio be .
Then, by section formula, the coordinates of C are
$C\left(\frac{7k+2}{k+1},\frac{8k+3}{k+1}\right)$
Therefore,

The given points are .
Here,
It is given that the points A, B and Câ€‹ are collinear. Then,
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒2\left(k+3\right)+4\left(-3-3\right)+6\left(3-k\right)=0\phantom{\rule{0ex}{0ex}}⇒2k+6-24+18-6k=0\phantom{\rule{0ex}{0ex}}⇒-4k=0\phantom{\rule{0ex}{0ex}}⇒k=0$

#### Question 1:

The given points are .
Here,
It is given that the points A, B and Câ€‹ are collinear. Then,
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒2\left(k+3\right)+4\left(-3-3\right)+6\left(3-k\right)=0\phantom{\rule{0ex}{0ex}}⇒2k+6-24+18-6k=0\phantom{\rule{0ex}{0ex}}⇒-4k=0\phantom{\rule{0ex}{0ex}}⇒k=0$

The distance of a point (x, y) from the origin O(0, 0) is $\sqrt{{x}^{2}+{y}^{2}}$.
Let P(x = −6, y = 8) be the given point. Then

Hence, the correct answer is option (d).

#### Question 2:

The distance of a point (x, y) from the origin O(0, 0) is $\sqrt{{x}^{2}+{y}^{2}}$.
Let P(x = −6, y = 8) be the given point. Then

Hence, the correct answer is option (d).

The distance of a point (x, y) from x-axis is $\left|y\right|$.
Here, the point is (−3, 4). So, its distance from x-axis is $\left|4\right|=4$.
Hence, the correct answer is option (c).

#### Question 3:

The distance of a point (x, y) from x-axis is $\left|y\right|$.
Here, the point is (−3, 4). So, its distance from x-axis is $\left|4\right|=4$.
Hence, the correct answer is option (c).

Let P(x, 0) the point on x-axis, then
$AP=BP⇒A{P}^{2}=B{P}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x+1\right)}^{2}+{\left(0-0\right)}^{2}={\left(x-5\right)}^{2}+{\left(0-0\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x+1={x}^{2}-10x+25\phantom{\rule{0ex}{0ex}}⇒12x=24⇒x=2$
Thus, the required point is (2, 0).
Hence, the correct answer is option (b).

#### Question 4:

Let P(x, 0) the point on x-axis, then
$AP=BP⇒A{P}^{2}=B{P}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x+1\right)}^{2}+{\left(0-0\right)}^{2}={\left(x-5\right)}^{2}+{\left(0-0\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x+1={x}^{2}-10x+25\phantom{\rule{0ex}{0ex}}⇒12x=24⇒x=2$
Thus, the required point is (2, 0).
Hence, the correct answer is option (b).

Since R(5, 6) is the midpoint of the line segment AB joining the points A(6, 5) and B(4, y), therefore
$\frac{5+y}{2}=6\phantom{\rule{0ex}{0ex}}⇒5+y=12\phantom{\rule{0ex}{0ex}}⇒y=12-5=7$
Hence, the correct option is (b).

#### Question 5:

Since R(5, 6) is the midpoint of the line segment AB joining the points A(6, 5) and B(4, y), therefore
$\frac{5+y}{2}=6\phantom{\rule{0ex}{0ex}}⇒5+y=12\phantom{\rule{0ex}{0ex}}⇒y=12-5=7$
Hence, the correct option is (b).

The point C(k, 4) divides the join of the points A(2, 6) and B(5, 1) in the ratio 2 : 3. So
$k=\frac{2×5+3×2}{2+3}=\frac{10+6}{5}=\frac{16}{5}$
Hence, the correct answer is option (c).

#### Question 6:

The point C(k, 4) divides the join of the points A(2, 6) and B(5, 1) in the ratio 2 : 3. So
$k=\frac{2×5+3×2}{2+3}=\frac{10+6}{5}=\frac{16}{5}$
Hence, the correct answer is option (c).

Let A(0, 4), B(0, 0) and C(3, 0) be the given vertices. So
$AB=\sqrt{{\left(0-0\right)}^{2}+{\left(4-0\right)}^{2}}=\sqrt{16}=4\phantom{\rule{0ex}{0ex}}BC=\sqrt{{\left(0-3\right)}^{2}+{\left(0-0\right)}^{2}}=\sqrt{9}=3\phantom{\rule{0ex}{0ex}}AC=\sqrt{{\left(0-3\right)}^{2}+{\left(4-0\right)}^{2}}=\sqrt{9+16}=5$
Therefore
AB + BC + AC = 4 + 3 + 5 = 12
Hence, the correct answer is option (d).

#### Question 7:

Let A(0, 4), B(0, 0) and C(3, 0) be the given vertices. So
$AB=\sqrt{{\left(0-0\right)}^{2}+{\left(4-0\right)}^{2}}=\sqrt{16}=4\phantom{\rule{0ex}{0ex}}BC=\sqrt{{\left(0-3\right)}^{2}+{\left(0-0\right)}^{2}}=\sqrt{9}=3\phantom{\rule{0ex}{0ex}}AC=\sqrt{{\left(0-3\right)}^{2}+{\left(4-0\right)}^{2}}=\sqrt{9+16}=5$
Therefore
AB + BC + AC = 4 + 3 + 5 = 12
Hence, the correct answer is option (d).

The diagonals of a parallelogram bisect each other. The vertices of the ||gm ABCD are A(1, 3), B(−1, 2) and C(2, 5) and D(x, 4).
Here, AC and BD are the diagonals. So
$\frac{1+2}{2}=\frac{-1+x}{2}\phantom{\rule{0ex}{0ex}}⇒x-1=3\phantom{\rule{0ex}{0ex}}⇒x=1+3=4$
Hence, the correct answer is option (b).

#### Question 8:

The diagonals of a parallelogram bisect each other. The vertices of the ||gm ABCD are A(1, 3), B(−1, 2) and C(2, 5) and D(x, 4).
Here, AC and BD are the diagonals. So
$\frac{1+2}{2}=\frac{-1+x}{2}\phantom{\rule{0ex}{0ex}}⇒x-1=3\phantom{\rule{0ex}{0ex}}⇒x=1+3=4$
Hence, the correct answer is option (b).

Let A(x1 = x, y1 = 2), B(x2 = −3, y2 = −4) and C(x3 = 7, y3 = −5) be collinear points. Then
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒x\left(-4+5\right)+\left(-3\right)\left(-5-2\right)+7\left(2+4\right)=0\phantom{\rule{0ex}{0ex}}⇒x+21+42=0\phantom{\rule{0ex}{0ex}}⇒x=-63$
Hence, the correct answer is option (a).

#### Question 9:

Let A(x1 = x, y1 = 2), B(x2 = −3, y2 = −4) and C(x3 = 7, y3 = −5) be collinear points. Then
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒x\left(-4+5\right)+\left(-3\right)\left(-5-2\right)+7\left(2+4\right)=0\phantom{\rule{0ex}{0ex}}⇒x+21+42=0\phantom{\rule{0ex}{0ex}}⇒x=-63$
Hence, the correct answer is option (a).

Let A(x1 = 5, y1 = 0), B(x2 = 8, y2 = 0) and C(x3 = 8, y3 = 4) be the vertices of the triangle. Then

Hence, the correct answer is option (c).

#### Question 10:

Let A(x1 = 5, y1 = 0), B(x2 = 8, y2 = 0) and C(x3 = 8, y3 = 4) be the vertices of the triangle. Then

Hence, the correct answer is option (c).

Let A(x1 = a, y1 = 0), O(x2 = 0, y2 = 0) and B(x3 = 0, y3 = b) be the given vertices. So

Hence, the correct answer is (b).

#### Question 11:

Let A(x1 = a, y1 = 0), O(x2 = 0, y2 = 0) and B(x3 = 0, y3 = b) be the given vertices. So

Hence, the correct answer is (b).

The point is the midpoint of the line segment joining the points A(−6, 5) and B(−2, 3). So
$\frac{a}{2}=\frac{-6-2}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{a}{2}=-4\phantom{\rule{0ex}{0ex}}⇒a=-8$
Hence, the correct answer is option (a).

#### Question 12:

The point is the midpoint of the line segment joining the points A(−6, 5) and B(−2, 3). So
$\frac{a}{2}=\frac{-6-2}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{a}{2}=-4\phantom{\rule{0ex}{0ex}}⇒a=-8$
Hence, the correct answer is option (a).

Here, AC and BD are two diagonals of the rectangle ABCD. So

Hence, the correct answer is option (a).

#### Question 13:

Here, AC and BD are two diagonals of the rectangle ABCD. So

Hence, the correct answer is option (a).

Here, the point P divides the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1. Then

Hence, the correct answer is option (b).

#### Question 14:

Here, the point P divides the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1. Then

Hence, the correct answer is option (b).

Let (x, y) be the coordinates of the other end of the diameter. Then
$-2=\frac{2+x}{2}⇒x=-6\phantom{\rule{0ex}{0ex}}5=\frac{3+y}{2}⇒y=7$
Hence, the correct answer is option (a).

#### Question 15:

Let (x, y) be the coordinates of the other end of the diameter. Then
$-2=\frac{2+x}{2}⇒x=-6\phantom{\rule{0ex}{0ex}}5=\frac{3+y}{2}⇒y=7$
Hence, the correct answer is option (a).

Here, AQ : BQ = 2 : 1. Then

Hence, the correct answer is option (c).

#### Question 16:

Here, AQ : BQ = 2 : 1. Then

Hence, the correct answer is option (c).

Let (x, y) be the coordinates of A. Then
$0=\frac{-2+x}{2}⇒x=2\phantom{\rule{0ex}{0ex}}4=\frac{3+y}{2}⇒y=8-3=5$
Thus, the coordinates of A are (2, 5).
Hence, the correct answer is option (a).

#### Question 17:

Let (x, y) be the coordinates of A. Then
$0=\frac{-2+x}{2}⇒x=2\phantom{\rule{0ex}{0ex}}4=\frac{3+y}{2}⇒y=8-3=5$
Thus, the coordinates of A are (2, 5).
Hence, the correct answer is option (a).

Let (x, y) be the coordinates of P. Then
$x=\frac{2×5+3×2}{2+3}=\frac{10+6}{5}=\frac{16}{5}\phantom{\rule{0ex}{0ex}}y=\frac{2×2+3×\left(-5\right)}{2+3}=\frac{4-15}{5}=\frac{-11}{5}$
Thus, the coordinates of point P are and so it lies in the fourth quadrant.
Hence, the correct answer is option (d).

#### Question 18:

Let (x, y) be the coordinates of P. Then
$x=\frac{2×5+3×2}{2+3}=\frac{10+6}{5}=\frac{16}{5}\phantom{\rule{0ex}{0ex}}y=\frac{2×2+3×\left(-5\right)}{2+3}=\frac{4-15}{5}=\frac{-11}{5}$
Thus, the coordinates of point P are and so it lies in the fourth quadrant.
Hence, the correct answer is option (d).

The given points are A(−6, 7) and B(−1, −5). So

Thus, 2AB = 26.
Hence, the correct answer is option (b).

#### Question 19:

The given points are A(−6, 7) and B(−1, −5). So

Thus, 2AB = 26.
Hence, the correct answer is option (b).

Let P(x, 0) be the point on x-axis. Then as per the question
$AP=BP⇒A{P}^{2}=B{P}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-7\right)}^{2}+{\left(0-6\right)}^{2}={\left(x+3\right)}^{2}+{\left(0-4\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-14x+49+36={x}^{2}+6x+9+16\phantom{\rule{0ex}{0ex}}⇒60=20x\phantom{\rule{0ex}{0ex}}⇒x=\frac{60}{20}=3$
Thus, the required point is (3, 0).
Hence, the correct answer is option (c).

#### Question 20:

Let P(x, 0) be the point on x-axis. Then as per the question
$AP=BP⇒A{P}^{2}=B{P}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-7\right)}^{2}+{\left(0-6\right)}^{2}={\left(x+3\right)}^{2}+{\left(0-4\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-14x+49+36={x}^{2}+6x+9+16\phantom{\rule{0ex}{0ex}}⇒60=20x\phantom{\rule{0ex}{0ex}}⇒x=\frac{60}{20}=3$
Thus, the required point is (3, 0).
Hence, the correct answer is option (c).

(b) 4 units
The y-coordinate is the distance of the point from the x-axis.
Here, the y-coordinate is 4.

#### Question 21:

(b) 4 units
The y-coordinate is the distance of the point from the x-axis.
Here, the y-coordinate is 4.

(c) 1 : 2
Let AB be divided by the x axis in the ratio  at the point P.
Then, by section formula, the coordinates of are
$P\left(\frac{5k+2}{k+1},\frac{6k-3}{k+1}\right)$
Butlies on the x axis: so, its ordinate is 0.
$\frac{6k-3}{k+1}=0$
$⇒6k-3=0$
$⇒6k=3$
$⇒k=\frac{1}{2}$
Hence, the required ratio is , which is same as 1 : 2.

#### Question 22:

(c) 1 : 2
Let AB be divided by the x axis in the ratio  at the point P.
Then, by section formula, the coordinates of are
$P\left(\frac{5k+2}{k+1},\frac{6k-3}{k+1}\right)$
Butlies on the x axis: so, its ordinate is 0.
$\frac{6k-3}{k+1}=0$
$⇒6k-3=0$
$⇒6k=3$
$⇒k=\frac{1}{2}$
Hence, the required ratio is , which is same as 1 : 2.

(d) 1 : 2
Let AB be divided by the y axis in the ratio  at the point P.
Then, by section formula, the coordinates of are
$P\left(\frac{8k-4}{k+1},\frac{3k+2}{k+1}\right)$
But, P lies on the y axis; so, its abscissa is 0.
$⇒\frac{8k-4}{k+1}=0$
$⇒8k-4=0$
$⇒8k=4$
$⇒k=\frac{1}{2}$
Hence, the required ratio is , which is same as 1 : 2.

#### Question 23:

(d) 1 : 2
Let AB be divided by the y axis in the ratio  at the point P.
Then, by section formula, the coordinates of are
$P\left(\frac{8k-4}{k+1},\frac{3k+2}{k+1}\right)$
But, P lies on the y axis; so, its abscissa is 0.
$⇒\frac{8k-4}{k+1}=0$
$⇒8k-4=0$
$⇒8k=4$
$⇒k=\frac{1}{2}$
Hence, the required ratio is , which is same as 1 : 2.

(b) −1
The given points are A(−3, b) and B(1, b+4).
Then,
Therefore,

and

But the midpoint is .
Therefore,
$b+2=1\phantom{\rule{0ex}{0ex}}⇒b=-1$

#### Question 24:

(b) −1
The given points are A(−3, b) and B(1, b+4).
Then,
Therefore,

and

But the midpoint is .
Therefore,
$b+2=1\phantom{\rule{0ex}{0ex}}⇒b=-1$

(b) 2 : 9
Let the lineâ€‹ divide the line segment in the ratio k : 1 at the point P.
Then, by section formula, the coordinates of P are
$P\left(\frac{3k+2}{k+1},\frac{7k-2}{k+1}\right)$
Since P lies on the line , we have:
$\frac{2\left(3k+2\right)}{k+1}+\frac{7k-2}{k+1}-4=0\phantom{\rule{0ex}{0ex}}⇒\left(6k+4\right)+\left(7k-2\right)-\left(4k+4\right)=0\phantom{\rule{0ex}{0ex}}⇒9k=2\phantom{\rule{0ex}{0ex}}⇒k=\frac{2}{9}$
Hence, the required ratio is , which is same as 2 : 9.

#### Question 25:

(b) 2 : 9
Let the lineâ€‹ divide the line segment in the ratio k : 1 at the point P.
Then, by section formula, the coordinates of P are
$P\left(\frac{3k+2}{k+1},\frac{7k-2}{k+1}\right)$
Since P lies on the line , we have:
$\frac{2\left(3k+2\right)}{k+1}+\frac{7k-2}{k+1}-4=0\phantom{\rule{0ex}{0ex}}⇒\left(6k+4\right)+\left(7k-2\right)-\left(4k+4\right)=0\phantom{\rule{0ex}{0ex}}⇒9k=2\phantom{\rule{0ex}{0ex}}⇒k=\frac{2}{9}$
Hence, the required ratio is , which is same as 2 : 9.

(c) $\left(\frac{7}{2},\frac{9}{2}\right)$
D is the midpoint of BC.
So, the coordinates of D are

#### Question 26:

(c) $\left(\frac{7}{2},\frac{9}{2}\right)$
D is the midpoint of BC.
So, the coordinates of D are

(d) (4, 0)
The given points are .
Here,
Let G(x, y) be the centroid of $∆ABC$. Then,

and

Hence, the centroid of $∆ABC$ is G(4, 0).

#### Question 27:

(d) (4, 0)
The given points are .
Here,
Let G(x, y) be the centroid of $∆ABC$. Then,

and

Hence, the centroid of $∆ABC$ is G(4, 0).

(c) (−4, −15)
Two vertices of .
Let the third vertex be C(a, b).
Then, the coordinates of its centroid are

But it is given that the centroid is $G\left(0,-3\right)$.
Therefore,

Hence, the third vertex of .

#### Question 28:

(c) (−4, −15)
Two vertices of .
Let the third vertex be C(a, b).
Then, the coordinates of its centroid are

But it is given that the centroid is $G\left(0,-3\right)$.
Therefore,

Hence, the third vertex of .

(a) isosceles
Let A(−4, 0), B(4, 0) and C(0, 3) be the given points. Then,

BC = AC = 5 units
Therefore, $∆ABC$ is isosceles.

#### Question 29:

(a) isosceles
Let A(−4, 0), B(4, 0) and C(0, 3) be the given points. Then,

BC = AC = 5 units
Therefore, $∆ABC$ is isosceles.