Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 6 Coordinate Geometry are provided here with simple step-by-step explanations. These solutions for Coordinate Geometry are extremely popular among Class 10 students for Maths Coordinate Geometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Page No 311:

(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)

#### Page No 311:

(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.

#### Page No 311:

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

Therefore, x = 2 or 8.

#### Page No 311:

The given points are .

Hence, the possible values of y are .

#### Page No 311:

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

Hence, the values of x are 1 and 17.

#### Page No 311:

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.

#### Page No 311:

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.

#### Page No 312:

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

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

#### Page No 312:

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).

#### Page No 312:

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).

#### Page No 312:

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.

#### Page No 312:

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.

#### Page No 312:

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).

#### Page No 312:

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.

#### Page No 312:

As per the question, we have

Now

Hence, x = 2 or −6 and .

#### Page No 312:

As per the question, we have

Now for $k=-1$

For $k=-3$

Hence, .

#### Page No 312:

(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.

#### Page No 312:

(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.

#### Page No 312:

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.

#### Page No 312:

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.

#### Page No 312:

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

#### Page No 312:

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.

#### Page No 312:

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.

#### Page No 312:

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,

#### Page No 313:

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 = $\frac{\sqrt{3}}{4}×{\left(\mathrm{side}\right)}^{2}$

#### Page No 313:

(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.

#### Page No 313:

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.

#### Page No 313:

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.

#### Page No 313:

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.

#### Page No 313:

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.

#### Page No 313:

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.

#### Page No 313:

(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.

#### Page No 313:

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.

#### Page No 324:

(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).

#### Page No 324:

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.

#### Page No 324:

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 .

#### Page No 325:

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.

#### Page No 325:

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.

#### Page No 325:

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.

#### Page No 325:

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

#### Page No 325:

(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.

#### Page No 325:

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.

#### Page No 325:

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

#### Page No 325:

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).

#### Page No 325:

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).

#### Page No 325:

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.

#### Page No 325:

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.

#### Page No 325:

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

#### Page No 325:

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

#### Page No 325:

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).

#### Page No 325:

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).

#### Page No 326:

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.

#### Page No 326:

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

#### Page No 326:

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

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

#### Page No 326:

Two vertices of ∆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 ∆ABC is C(−2, 7).

#### Page No 326:

Two vertices of ∆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 ∆ABC is A(3, 1).

#### Page No 326:

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.

#### Page No 326:

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.

#### Page No 326:

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).

#### Page No 326:

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.

#### Page No 326:

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 $y=-\frac{3}{2}$.

#### Page No 326:

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 .

#### Page No 326:

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 .

#### Page No 326:

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 .

#### Page No 326:

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.

#### Page No 326:

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.

#### Page No 327:

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.

#### Page No 327:

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.

#### Page No 327:

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

#### Page No 327:

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).

#### Page No 327:

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).

#### Page No 327:

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.

#### Page No 327:

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.

#### Page No 327:

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 .

#### Page No 340:

(i) A(1, 2), B(−2, 3) and C(−3, −4) are the vertices of ∆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 ∆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 ∆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 ∆ABC. Then,
(x1 = 10, y1 = −6), (x2 = 2, y2 = 5) and (x3 = −1, y3 = 3)

#### Page No 341:

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.

#### Page No 341:

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

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

#### Page No 341:

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

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

#### Page No 341:

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.

#### Page No 341:

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.

#### Page No 341:

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.

#### Page No 341:

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.

#### Page No 341:

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 .

#### Page No 341:

(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,

#### Page No 341:

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.

#### Page No 341:

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.

#### Page No 341:

(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.

#### Page No 342:

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.

#### Page No 342:

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.

#### Page No 342:

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.

#### Page No 342:

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
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(-3\right)\left(y+5\right)+2\left(-5-9\right)+4\left(9-y\right)=0\phantom{\rule{0ex}{0ex}}⇒-3y-15-28+36-4y=0\phantom{\rule{0ex}{0ex}}⇒7y=36-43$
$⇒y=-1$

#### Page No 342:

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, .

#### Page No 342:

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.

#### Page No 342:

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.

#### Page No 342:

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.

#### Page No 342:

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.

#### Page No 342:

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, .

#### Page No 342:

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 ∆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.

#### Page No 342:

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, ​the value of x is 13 and −3.

#### Page No 344:

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.

#### Page No 344:

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.

#### Page No 345:

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..

#### Page No 345:

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

Hence, k = 1 or k = 5.

#### Page No 345:

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.

#### Page No 345:

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.

#### Page No 345:

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.

#### Page No 345:

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

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

#### Page No 345:

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.

#### Page No 345:

The given points are .
Then,
Therefore,

#### Page No 345:

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

.

#### Page No 345:

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

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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:
$\frac{a+b+c}{3}=0\phantom{\rule{0ex}{0ex}}⇒a+b+c=0$

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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  .

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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,

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

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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).

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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).

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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).

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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).

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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).

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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).

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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).

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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).

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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).

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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).

#### Page No 348:

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).

#### Page No 349:

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

Hence, the correct answer is option (a).

#### Page No 349:

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).

#### Page No 349:

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).

#### Page No 349:

Here, AQ : BQ = 2 : 1. Then

Hence, the correct answer is option (c).

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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).

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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).

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

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

#### Page No 349:

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).

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

#### Page No 349:

(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.

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(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.

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

and

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

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(b) 2 : 9
Let the line​ 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.

#### Page No 350:

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

#### Page No 350:

(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).

#### Page No 350:

(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 .

#### Page No 350:

(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.

#### Page No 350:

(d) right-angled

Let P(0, 6), Q(−5, 3) and R(3, 1) be the given points. Then,

Therefore, ∆PQR is right-angled.

#### Page No 350:

(b) k = 6
The given points are A(2, 3), B(5, k) and C(6, 7).
Here,
Points A,B and C are collinear. Then,
${x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒2\left(k-7\right)+5\left(7-3\right)+6\left(3-k\right)=0\phantom{\rule{0ex}{0ex}}⇒2k-14+20+18-6k=0\phantom{\rule{0ex}{0ex}}⇒-4k=-24\phantom{\rule{0ex}{0ex}}⇒k=6$

#### Page No 350:

(c) 2a = b
The given points are A(1, 2), O(0, 0) and C(a, b).
Here, .
Points A, O and C are collinear.
$⇒{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒1\left(0-b\right)+0\left(b-2\right)+a\left(2-0\right)=0\phantom{\rule{0ex}{0ex}}⇒-b+2a=0\phantom{\rule{0ex}{0ex}}⇒2a=b$

#### Page No 350:

(c) 8 sq units
The given points are .
Here,
Therefore,

#### Page No 350:

(c) $\sqrt{34}$ units
are the three vertices of a rectangle; let C be the fourth vertex.
Then, the length of the diagonal,

Since, the diagonals of rectangle are equal .
Hence, the length of its diagonals is .

#### Page No 350:

$AB=5\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}=5\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(1-4\right)}^{2}+{\left(0-p\right)}^{2}}=5\phantom{\rule{0ex}{0ex}}⇒{\left(-3\right)}^{2}+{\left(-p\right)}^{2}=25\phantom{\rule{0ex}{0ex}}⇒9+{p}^{2}=25\phantom{\rule{0ex}{0ex}}⇒{p}^{2}=16\phantom{\rule{0ex}{0ex}}⇒p=±\sqrt{16}\phantom{\rule{0ex}{0ex}}⇒p=±4\phantom{\rule{0ex}{0ex}}$