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#### Question 1:

Find the distance between each of the following pairs of points.
(1) A(2, 3), B(4, 1)

(2) P(–5, 7), Q(–1, 3)

(3)

(4) L(5, –8), M(–7, –3)

(5) T(–3, 6), R(9, –10)

(6)

(1) A(2, 3), B(4, 1)
$\mathrm{AB}=\sqrt{{\left(2-4\right)}^{2}+{\left(3-1\right)}^{2}}=\sqrt{{\left(-2\right)}^{2}+{2}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4+4}\phantom{\rule{0ex}{0ex}}=2\sqrt{2}$

(2) P(–5, 7), Q(–1, 3)
$\mathrm{PQ}=\sqrt{{\left(-5-\left(-1\right)\right)}^{2}+{\left(7-3\right)}^{2}}=\sqrt{{\left(-4\right)}^{2}+{4}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{16+16}\phantom{\rule{0ex}{0ex}}=4\sqrt{2}$
(3)
$\mathrm{RS}=\sqrt{{\left(0-0\right)}^{2}+{\left(-3-\frac{5}{2}\right)}^{2}}=\sqrt{{\left(\frac{-11}{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{121}{4}}\phantom{\rule{0ex}{0ex}}=\frac{11}{2}$

(4) L(5, –8), M(–7, –3)
$\mathrm{LM}=\sqrt{{\left(5-\left(-7\right)\right)}^{2}+{\left(-8-\left(-3\right)\right)}^{2}}=\sqrt{{12}^{2}+{\left(-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{144+25}=\sqrt{169}\phantom{\rule{0ex}{0ex}}=13$

(5) T(–3, 6), R(9, –10)
$\mathrm{TR}=\sqrt{{\left(-3-9\right)}^{2}+{\left(6-\left(-10\right)\right)}^{2}}=\sqrt{{\left(-12\right)}^{2}+{\left(16\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{144+256}=\sqrt{400}\phantom{\rule{0ex}{0ex}}=20$

(6)
$\mathrm{WX}=\sqrt{{\left(\frac{-7}{2}-11\right)}^{2}+{\left(4-4\right)}^{2}}=\sqrt{{\left(\frac{-29}{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{841}{4}}\phantom{\rule{0ex}{0ex}}=\frac{29}{2}$

#### Question 2:

Determine whether the points are collinear.
(1) A(1, –3), B(2, –5), C(–4, 7)
(2) L(–2, 3), M(1, –3), N(5, 4)
(3) R(0, 3), D(2, 1), S(3, –1)
(4) P(–2, 3), Q(1, 2), R(4, 1)

(1) A(1, –3), B(2, –5), C(–4, 7)
$AB=\sqrt{{\left(1-2\right)}^{2}+{\left(-3-\left(-5\right)\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(-1\right)}^{2}+{2}^{2}}=\sqrt{5}\phantom{\rule{0ex}{0ex}}BC=\sqrt{{\left(2-\left(-4\right)\right)}^{2}+{\left(-5-7\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{36+144}\phantom{\rule{0ex}{0ex}}=\sqrt{180}\phantom{\rule{0ex}{0ex}}=6\sqrt{5}\phantom{\rule{0ex}{0ex}}AC=\sqrt{{\left(-1-\left(-4\right)\right)}^{2}+{\left(-3-7\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{25+100}\phantom{\rule{0ex}{0ex}}=\sqrt{125}\phantom{\rule{0ex}{0ex}}=5\sqrt{5}$
AB + AC = BC
$\sqrt{5}+5\sqrt{5}=6\sqrt{5}$
Thus, the given points lie on the same line. Hence, they are collinear.

(2) L(–2, 3), M(1, –3), N(5, 4)
$LM=\sqrt{{\left(-2-1\right)}^{2}+{\left(3-\left(-3\right)\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(-3\right)}^{2}+{6}^{2}}=\sqrt{9+36}\phantom{\rule{0ex}{0ex}}=\sqrt{45}=3\sqrt{5}\phantom{\rule{0ex}{0ex}}MN=\sqrt{{\left(1-5\right)}^{2}+{\left(-3-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{16+49}\phantom{\rule{0ex}{0ex}}=\sqrt{65}\phantom{\rule{0ex}{0ex}}LN=\sqrt{{\left(-2-5\right)}^{2}+{\left(3-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(-7\right)}^{2}+{\left(-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{49+1}=\sqrt{50}\phantom{\rule{0ex}{0ex}}=5\sqrt{2}$
Sum of two sides is not equal to the third side.
Hence, the given points are not collinear.

(3) R(0, 3), D(2, 1), S(3, –1)
$RD=\sqrt{{\left(0-2\right)}^{2}+{\left(3-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(-2\right)}^{2}+{2}^{2}}=\sqrt{4+4}\phantom{\rule{0ex}{0ex}}=2\sqrt{2}\phantom{\rule{0ex}{0ex}}DS=\sqrt{{\left(2-3\right)}^{2}+{\left(1-\left(-1\right)\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(-1\right)}^{2}+4}=\sqrt{1+4}\phantom{\rule{0ex}{0ex}}=\sqrt{5}\phantom{\rule{0ex}{0ex}}RS=\sqrt{{\left(0-3\right)}^{2}+{\left(3-\left(-1\right)\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9+16}\phantom{\rule{0ex}{0ex}}=\sqrt{25}\phantom{\rule{0ex}{0ex}}=5$
Sum of two sides is not equal to the third side.
Hence, the given points are not collinear.

(4) P(–2, 3), Q(1, 2), R(4, 1)
$PQ=\sqrt{{\left(-2-1\right)}^{2}+{\left(3-2\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9+1}=\sqrt{10}\phantom{\rule{0ex}{0ex}}QR=\sqrt{{\left(1-4\right)}^{2}+{\left(2-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9+1}\phantom{\rule{0ex}{0ex}}=\sqrt{10}\phantom{\rule{0ex}{0ex}}PR=\sqrt{{\left(-2-4\right)}^{2}+{\left(3-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{36+4}\phantom{\rule{0ex}{0ex}}=\sqrt{40}\phantom{\rule{0ex}{0ex}}=2\sqrt{10}$
PQ + QR = PR
So, the given points lie on the same line. Hence, the given points are collinear.

#### Question 3:

Find the point on the X–axis which is equidistant from A(–3, 4) and B(1, –4).

Let the point on the x-axis be P(a, 0).
$\mathrm{PA}=\sqrt{{\left(a-\left(-3\right)\right)}^{2}+{\left(0-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(a+3\right)}^{2}+16}\phantom{\rule{0ex}{0ex}}\mathrm{PB}=\sqrt{{\left(a-1\right)}^{2}+{\left(0-\left(-4\right)\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(a-1\right)}^{2}+16}\phantom{\rule{0ex}{0ex}}{\mathrm{PA}}^{2}={\mathrm{PB}}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(a+3\right)}^{2}+16={\left(a-1\right)}^{2}+16\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+6a+9={a}^{2}-2a+1\phantom{\rule{0ex}{0ex}}⇒8a=-8\phantom{\rule{0ex}{0ex}}⇒a=-1\phantom{\rule{0ex}{0ex}}\left(a,0\right)=\left(-1,0\right)\phantom{\rule{0ex}{0ex}}$

#### Question 4:

Verify that points P(–2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle.

$\mathrm{PQ}=\sqrt{{\left(-2-2\right)}^{2}+{\left(2-2\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(-4\right)}^{2}+0}\phantom{\rule{0ex}{0ex}}=\sqrt{16}=4\phantom{\rule{0ex}{0ex}}\mathrm{QR}=\sqrt{{\left(2-2\right)}^{2}+{\left(2-7\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{0+{\left(-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{25}\phantom{\rule{0ex}{0ex}}=5\phantom{\rule{0ex}{0ex}}\mathrm{PR}=\sqrt{{\left(-2-2\right)}^{2}+{\left(2-7\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(-4\right)}^{2}+{\left(-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{16+25}\phantom{\rule{0ex}{0ex}}=\sqrt{41}$
${\mathrm{PQ}}^{2}+{\mathrm{QR}}^{2}={\mathrm{PR}}^{2}\phantom{\rule{0ex}{0ex}}⇒{4}^{2}+{5}^{2}=16+25=41=\mathrm{PR}$
Thus, the square of the third is equal to the sum of the squares of the other two sides.
Thus, they are the vertices of the right angled triangle.

#### Question 5:

Show that points P(2, –2), Q(7, 3), R(11, –1) and S (6, –6) are vertices of a parallelogram.

The given points are P(2, –2), Q(7, 3), R(11, –1) and S (6, –6).
$\mathrm{PQ}=\sqrt{{\left(3-\left(-2\right)\right)}^{2}+{\left(7-2\right)}^{2}}=\sqrt{25+25}=5\sqrt{2}\phantom{\rule{0ex}{0ex}}\mathrm{QR}=\sqrt{{\left(7-11\right)}^{2}+{\left(3-\left(-1\right)\right)}^{2}}=\sqrt{16+16}=4\sqrt{2}\phantom{\rule{0ex}{0ex}}\mathrm{RS}=\sqrt{{\left(11-6\right)}^{2}+{\left(-1-\left(-6\right)\right)}^{2}}=\sqrt{25+25}=5\sqrt{2}\phantom{\rule{0ex}{0ex}}\mathrm{PS}=\sqrt{{\left(2-6\right)}^{2}+{\left(-2-\left(-6\right)\right)}^{2}}=\sqrt{16+16}=4\sqrt{2}$
So, PQ = RS and QR = PS
Thus, opposite sides are equal.
Hence, the given points form a parallelogram.

#### Question 6:

Show that points A(–4, –7), B(–1, 2), C(8, 5) and D(5, –4) are vertices of a rhombus ABCD.

The given points are A(–4, –7), B(–1, 2), C(8, 5) and D(5, –4).
$\mathrm{AB}=\sqrt{{\left(-4-\left(-1\right)\right)}^{2}+{\left(-7-2\right)}^{2}}=\sqrt{9+81}=\sqrt{90}=3\sqrt{10}\phantom{\rule{0ex}{0ex}}\mathrm{BC}=\sqrt{{\left(-1-8\right)}^{2}+{\left(2-5\right)}^{2}}=\sqrt{81+9}=\sqrt{90}=3\sqrt{10}\phantom{\rule{0ex}{0ex}}\mathrm{CD}=\sqrt{{\left(8-5\right)}^{2}+{\left(5-\left(-4\right)\right)}^{2}}=\sqrt{9+81}=\sqrt{90}=3\sqrt{10}\phantom{\rule{0ex}{0ex}}\mathrm{DA}=\sqrt{{\left(5-\left(-4\right)\right)}^{2}+{\left(-4-\left(-7\right)\right)}^{2}}=\sqrt{81+9}=\sqrt{90}=3\sqrt{10}$
Thus, all the sides are equal.
$\mathrm{AC}=\sqrt{{\left(-4-8\right)}^{2}+{\left(-7-5\right)}^{2}}=\sqrt{144+144}=12\sqrt{2}\phantom{\rule{0ex}{0ex}}\mathrm{DB}=\sqrt{{\left(8-5\right)}^{2}+{\left(2-\left(-4\right)\right)}^{2}}=\sqrt{9+36}=3\sqrt{5}$
Thus, the diagonals are not equal.
So, the given vertices form a rhombus ABCD.

#### Question 7:

Find x if distance between points L(x, 7) and M(1, 15) is 10.

Distance between LM = 10
$\mathrm{LM}=\sqrt{{\left(x-1\right)}^{2}+{\left(7-15\right)}^{2}}=10\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(x-1\right)}^{2}+{\left(-8\right)}^{2}}=10\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(x-1\right)}^{2}+64}=100$
Squaring both sides
${\left(x-1\right)}^{2}+64=100\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2x+1+64=100\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2x-35=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-7x+5x-35=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-7\right)+5\left(x-7\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+5\right)\left(x-7\right)=0\phantom{\rule{0ex}{0ex}}⇒x=-5,7$

#### Question 8:

Show that the points A(1, 2), B(1, 6),  are vertices of an equilateral triangle.

The given points are A(1, 2), B(1, 6),  .
$\mathrm{AB}=\sqrt{{\left(1-1\right)}^{2}+{\left(2-6\right)}^{2}}=\sqrt{0+{\left(-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{16}\phantom{\rule{0ex}{0ex}}=4\phantom{\rule{0ex}{0ex}}\mathrm{BC}=\sqrt{{\left(1-1-2\sqrt{3}\right)}^{2}+{\left(6-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{12+4}=\sqrt{16}\phantom{\rule{0ex}{0ex}}=4\phantom{\rule{0ex}{0ex}}\mathrm{AC}=\sqrt{{\left(1-1-2\sqrt{3}\right)}^{2}+{\left(2-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{12+4}\phantom{\rule{0ex}{0ex}}=\sqrt{16}\phantom{\rule{0ex}{0ex}}=4$
AB = BC = AC = 4
Since, all the sides of the triangle formed by the points A, B and C are equal so, the tiangle formed is equilateral triangle.

#### Question 1:

Find the coordinates of point P if P divides the line segment joining the points A(–1,7) and B(4,–3) in the ratio 2 : 3.

Let the coordinates of point P be (x, y).
The ratio in which P divides A(–1,7) and B(4,–3) is 2 : 3.
Using the section formula we have
$x=\frac{4×2+3×\left(-1\right)}{2+3}=\frac{8-3}{5}=1\phantom{\rule{0ex}{0ex}}y=\frac{2×\left(-3\right)+3×7}{2+3}=\frac{-6+21}{5}=3$
Thus, the coordinates of point P are (1, 3).

#### Question 2:

In each of the following examples find the co-ordinates of point A which divides segment PQ in the ratio a : b.
(1) P(–3, 7), Q(1, –4), a : b = 2 : 1
(2) P(–2, –5), Q(4, 3), a : b = 3 : 4
(3) P(2, 6), Q(–4, 1), a : b = 1 : 2

Let the coordinates of point A be (x, y).
(1) P(–3, 7), Q(1, –4), = 2 : 1
Using section formula
$x=\frac{2×1+1×\left(-3\right)}{2+1}=\frac{2-3}{3}=\frac{-1}{3}\phantom{\rule{0ex}{0ex}}y=\frac{2×\left(-4\right)+1×7}{2+1}=\frac{-8+7}{3}=\frac{-1}{3}\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(\frac{-1}{3},\frac{-1}{3}\right)$

(2) P(–2, –5), Q(4, 3), = 3 : 4
Using section formula
$x=\frac{3×4+4×\left(-2\right)}{3+4}=\frac{12-8}{7}=\frac{4}{7}\phantom{\rule{0ex}{0ex}}y=\frac{3×3+4×\left(-5\right)}{3+4}=\frac{9-20}{3}=\frac{-11}{7}\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(\frac{4}{7},\frac{-11}{7}\right)$

(3) P(2, 6), Q(–4, 1), = 1 : 2
Using section formula
$x=\frac{1×\left(-4\right)+2×2}{1+2}=\frac{-4+4}{3}=0\phantom{\rule{0ex}{0ex}}y=\frac{1×1+2×6}{1+2}=\frac{1+12}{3}=\frac{13}{3}\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(0,\frac{13}{3}\right)$

#### Question 3:

Find the ratio in which point T(–1, 6)divides the line segment joining the points P(–3, 10) and Q(6, –8).

Let the ratio be k : 1.
Using section formula we have
$-1=\frac{6k-3×1}{k+1}\phantom{\rule{0ex}{0ex}}⇒-k-1=6k-3\phantom{\rule{0ex}{0ex}}⇒-1+3=6k+k\phantom{\rule{0ex}{0ex}}⇒2=7k\phantom{\rule{0ex}{0ex}}⇒k=\frac{2}{7}$
Thus, the required ratio is 2 : 7.

#### Question 4:

Point P is the centre of the circle and AB is a diameter . Find the coordinates of point B if coordinates of point A and P are (2, –3) and (–2, 0) respectively.

Centre = P(–2, 0)
Diameter has A(2, –3) on one end and B(x, y) on the other.
The centre point P divides AB in two equal parts.
So, using the midpoint formula,
$-2=\frac{2+x}{2}\phantom{\rule{0ex}{0ex}}⇒-4=2+x\phantom{\rule{0ex}{0ex}}⇒x=-6\phantom{\rule{0ex}{0ex}}\mathrm{And}\phantom{\rule{0ex}{0ex}}0=\frac{-3+y}{2}\phantom{\rule{0ex}{0ex}}⇒0=-3+y\phantom{\rule{0ex}{0ex}}⇒y=3\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(-6,3\right)$

#### Question 5:

Find the ratio in which point P(k, 7) divides the segment joining A(8, 9) and B(1, 2). Also find k.

Let the ratio be x : 1.
Using the section formula,

So, the required ratio is 2 : 5.
Putting this value of in (1) we get
$k\left(\frac{2}{5}+1\right)=\frac{2}{5}+8\phantom{\rule{0ex}{0ex}}⇒\frac{7}{5}k=\frac{42}{5}\phantom{\rule{0ex}{0ex}}⇒k=6\phantom{\rule{0ex}{0ex}}$

#### Question 6:

Find the coordinates of midpoint of the segment joining the points (22, 20) and (0, 16).

Let the given points be A (22, 20) and B(0, 16).
Let the coordinate of the mid point be (x, y).
Using the midpoint formula
$x=\frac{22+0}{2}\phantom{\rule{0ex}{0ex}}⇒2x=22\phantom{\rule{0ex}{0ex}}⇒x=11\phantom{\rule{0ex}{0ex}}y=\frac{20+16}{2}\phantom{\rule{0ex}{0ex}}⇒2y=36\phantom{\rule{0ex}{0ex}}⇒y=18\phantom{\rule{0ex}{0ex}}$
(x, y) = (11, 18)

#### Question 7:

Find the centroids of the triangles whose vertices are given below.
(1) (–7, 6), (2, –2), (8, 5)
(2) (3, –5), (4, 3), (11, –4)
(3) (4, 7), (8, 4), (7, 11)

(1) (–7, 6), (2, –2), (8, 5)
The centroid of the triangle formed by the given coordinates is
$=\left(\frac{-7+2+8}{3},\frac{6-2+5}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(1,3\right)$

(2) (3, –5), (4, 3), (11, –4)
The centroid of the triangle formed by the given coordinates is
$=\left(\frac{3+4+11}{3},\frac{-5+3-4}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(6,-2\right)$

(3) (4, 7), (8, 4), (7, 11)
The centroid of the triangle formed by the given coordinates is

$=\left(\frac{4+8+7}{3},\frac{7+4+11}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{19}{3},\frac{22}{3}\right)$

#### Question 8:

In ∆ABC, G (–4, –7) is the centroid. If A (–14, –19) and B(3, 5) then find the co–ordinates of C.

Let the coordinates of point C be (x, y).
Centroid = G (–4, –7)

Thus, point C will be $\left(-1,-7\right)$.

#### Question 9:

A(h, –6), B(2, 3) and C(–6, k) are the co–ordinates of vertices of a triangle whose centroid is G (1, 5). Find h and k.

Centroid = G(1, 5)
The coordinates of the triangle are A(h, –6), B(2, 3) and C(–6, k).

#### Question 10:

Find the co-ordinates of the points of trisection of the line segment AB with A(2, 7) and B(–4, –8).

Let P$\left({x}_{1},{y}_{1}\right)$ and Q$\left({x}_{2},{y}_{2}\right)$ divide the line AB into 3 equal parts.
AP = PQ = QB
$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AP}}{\mathrm{PQ}+\mathrm{QB}}=\frac{\mathrm{AP}}{\mathrm{AP}+\mathrm{AP}}=\frac{\mathrm{AP}}{2\mathrm{AP}}=\frac{1}{2}$
So, P divides AB in the ratio 1 : 2.
${x}_{1}=\frac{1×\left(-4\right)+2×2}{2+1}=0\phantom{\rule{0ex}{0ex}}{y}_{1}=\frac{1×\left(-8\right)+2×7}{2+1}=2\phantom{\rule{0ex}{0ex}}\left({x}_{1},{y}_{1}\right)=\left(0,2\right)$
Also, $\frac{\mathrm{AQ}}{\mathrm{QB}}=\frac{\mathrm{AP}+\mathrm{PB}}{\mathrm{QB}}=\frac{\mathrm{QB}+\mathrm{QB}}{\mathrm{QB}}=\frac{2}{1}$
${x}_{2}=\frac{2×\left(-4\right)+1×2}{2+1}=-2\phantom{\rule{0ex}{0ex}}{y}_{2}=\frac{2×\left(-8\right)+1×7}{2+1}=-3\phantom{\rule{0ex}{0ex}}\left({x}_{2},{y}_{2}\right)=\left(-2,-3\right)$
Thus, the coordinates of the point of intersection are
P$\left({x}_{1},{y}_{1}\right)=\left(0,2\right)$
Q$\left({x}_{2},{y}_{2}\right)$ = $\left(-2,-3\right)$

#### Question 11:

If A (–14, –10), B(6, –2) is given, find the coordinates of the points which divide segment AB into four equal parts.

Let  be the points which divide the line segment AB into 4 equal parts.
AP = PQ = QR = RB
$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AP}}{\mathrm{PQ}+\mathrm{QR}+\mathrm{RB}}=\frac{\mathrm{AP}}{\mathrm{AP}+\mathrm{AP}+\mathrm{AP}}=\frac{\mathrm{AP}}{3\mathrm{AP}}=\frac{1}{3}$
${x}_{1}=\left(\frac{6×1+3×\left(-14\right)}{1+3}\right)=\frac{6-42}{4}=-9\phantom{\rule{0ex}{0ex}}{y}_{1}=\left(\frac{1×\left(-2\right)+3×\left(-10\right)}{1+3}\right)=\frac{-2-30}{4}=-8\phantom{\rule{0ex}{0ex}}\mathrm{P}\left({x}_{1},{y}_{1}\right)=\left(-9,-8\right)$
Now, $\frac{\mathrm{PQ}}{\mathrm{QB}}=\frac{\mathrm{PQ}}{\mathrm{QR}+\mathrm{RB}}=\frac{\mathrm{PQ}}{\mathrm{PQ}+\mathrm{PQ}}=\frac{1}{2}$
${x}_{2}=\left(\frac{1×6+2×\left(-9\right)}{1+2}\right)=-4\phantom{\rule{0ex}{0ex}}{y}_{2}=\left(\frac{1×\left(-2\right)+2×\left(-8\right)}{1+2}\right)=-6\phantom{\rule{0ex}{0ex}}\mathrm{Q}\left({x}_{2},{y}_{2}\right)=\left(-4,-6\right)$
Now R divides QB into 2 equal parts so, using the midpoint formula we have
${x}_{3}=\frac{-4+6}{2}=1\phantom{\rule{0ex}{0ex}}{y}_{3}=\frac{-6+\left(-2\right)}{2}=-4\phantom{\rule{0ex}{0ex}}\mathrm{R}\left({x}_{3},{y}_{3}\right)=\left(1,-4\right)$
Thus,  are .

#### Question 12:

If A (20, 10), B(0, 20) are given, find the coordinates of the points which divide segment AB into five congruent parts.

Let the points  be the points which divide the line segment AB into 5 equal parts.
$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AP}}{\mathrm{PQ}+\mathrm{QR}+\mathrm{RS}}=\frac{\mathrm{AP}}{4\mathrm{AP}}=\frac{1}{4}$
${x}_{1}=\left(\frac{1×0+4×20}{1+4}\right)=16\phantom{\rule{0ex}{0ex}}{y}_{1}=\left(\frac{1×20+4×10}{1+4}\right)=12\phantom{\rule{0ex}{0ex}}\mathrm{P}\left({x}_{1},{y}_{1}\right)=\left(16,12\right)$
$\frac{\mathrm{PQ}}{\mathrm{QB}}=\frac{\mathrm{PQ}}{\mathrm{QR}+\mathrm{RS}+\mathrm{SB}}=\frac{\mathrm{PQ}}{\mathrm{PQ}+\mathrm{PQ}+\mathrm{PQ}}=\frac{\mathrm{PQ}}{3\mathrm{PQ}}=\frac{1}{3}$
${x}_{2}=\left(\frac{1×0+3×16}{1+3}\right)=12\phantom{\rule{0ex}{0ex}}{y}_{2}=\left(\frac{1×20+3×12}{1+3}\right)=14\phantom{\rule{0ex}{0ex}}\mathrm{Q}\left({x}_{2},{y}_{2}\right)=\left(12,14\right)$
$\frac{\mathrm{QR}}{\mathrm{RB}}=\frac{\mathrm{QR}}{\mathrm{RS}+\mathrm{SB}}=\frac{\mathrm{QR}}{\mathrm{QR}+\mathrm{QR}}=\frac{\mathrm{QR}}{2\mathrm{QR}}=\frac{1}{2}$
${x}_{3}=\left(\frac{1×0+2×12}{1+2}\right)=8\phantom{\rule{0ex}{0ex}}{y}_{3}=\left(\frac{1×20+2×14}{1+3}\right)=16\phantom{\rule{0ex}{0ex}}\mathrm{R}\left({x}_{3},{y}_{3}\right)=\left(8,16\right)$
S is the midpoint of RB so, using the midpoint formula
${x}_{4}=\frac{8+0}{2}=4\phantom{\rule{0ex}{0ex}}{y}_{4}=\frac{16+20}{2}=18\phantom{\rule{0ex}{0ex}}\mathrm{S}\left({x}_{4},{y}_{4}\right)=\left(4,18\right)$
So, the points

#### Question 1:

Angles made by the line with the positive direction of X–axis are given. Find the slope of these lines.
(1) 45° (2) 60° (3) 90°

(1) 45°
$m=\mathrm{tan}45°=1$
Thus, slope = 1

(2) 60°
$m=\mathrm{tan}60°=\sqrt{3}$
Thus, slope = $\sqrt{3}$

(3) 90°
$m=\mathrm{tan}90°=\mathrm{undefined}$
Thus, the slope cannot be defined.

#### Question 2:

Find the slopes of the lines passing through the given points.
(1) A (2, 3) , B (4, 7)
(2) P (–3, 1) , Q (5, –2)
(3) C (5, –2) , ∆ (7, 3)
(4) L (–2, –3) , M (–6, –8)
(5) E(–4, –2) , F (6, 3)
(6) T (0, –3) , S (0, 4)

(1) A (2, 3) , B (4, 7)
Slope = $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{7-3}{4-2}=\frac{4}{2}=2$

(2) P (–3, 1) , Q (5, –2)
Slope = $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{-2-1}{5-\left(-3\right)}=\frac{-3}{8}$

(3) C (5, –2) , ∆ (7, 3)
Slope = $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{3-\left(-2\right)}{7-5}=\frac{5}{2}$

(4) L (–2, –3) , M (–6, –8)
Slope = $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{-8-\left(-3\right)}{-6-\left(-2\right)}=\frac{-5}{-4}=\frac{5}{4}$

(5) E(–4, –2) , F (6, 3)
Slope = $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{3-\left(-2\right)}{6-\left(-4\right)}=\frac{5}{10}=\frac{1}{2}$

(6) T (0, –3) , S (0, 4)
Slope =

#### Question 3:

Determine whether the following points are collinear.
(1) A(–1, –1), B(0, 1), C(1, 3)

(2) D(–2, –3), E(1, 0), F(2, 1)

(3) L(2, 5), M(3, 3), N(5, 1)

(4) P(2, –5), Q(1, –3), R(–2, 3)

(5) R(1, –4), S(–2, 2), T(–3, 4)

(6) A(–4, 4),  N (4, –2)

(1) A(–1, –1), B(0, 1), C(1, 3)
Slope of AB = $\frac{1-\left(-1\right)}{0-\left(-1\right)}=\frac{2}{1}=2$
Slope of BC = $\frac{3-1}{1-0}=\frac{2}{1}=2$
Slope of AB = Slope of BC = 2
Thus, the given points are collinear.

(2) D(–2, –3), E(1, 0), F(2, 1)

Slope of DE = Slope of EF = 1
So, the given points are collinear.

(3) L(2, 5), M(3, 3), N(5, 1)

Slope of LM not equal to slope of MN. Thus, the given points are not collinear.

(4) P(2, –5), Q(1, –3), R(–2, 3)

Slope of PQ = Slope of QR
So, the given points are collinear.

(5) R(1, –4), S(–2, 2), T(–3, 4)

Slope of RS = Slope of ST
So, the given points are collinear.

(6) A(–4, 4),  N (4, –2)

Slope of AK=Slope of KN
Thus, the given points are collinear.

#### Question 4:

If A (1, –1), B (0, 4), C (–5, 3) are vertices of a triangle then find the slope of each side.

A (1, –1), B (0, 4), C (–5, 3) form a triangle.
Slope of AB = $\frac{4-\left(-1\right)}{0-1}=\frac{5}{-1}=-5$
Slope of BC = $\frac{3-4}{-5-0}=\frac{-1}{-5}=\frac{1}{5}$
Slope of AC = $\frac{3-\left(-1\right)}{-5-1}=\frac{4}{-6}=\frac{-2}{3}$

#### Question 5:

Show that A(–4, –7), B (–1, 2), C (8, 5) and D (5, –4) are the vertices of a parallelogram.

The given points are A(–4, –7), B(–1, 2), C(8, 5) and D(5, –4).
Slope of AB = $\frac{2-\left(-7\right)}{-1-\left(-4\right)}=\frac{9}{3}=3$
Slope of BC = $\frac{5-2}{8-\left(-1\right)}=\frac{3}{9}=\frac{1}{3}$
Slope of CD = $\frac{-4-5}{5-8}=\frac{-9}{-3}=3$
Slope of AD = $\frac{-4-\left(-7\right)}{5-\left(-4\right)}=\frac{3}{9}=\frac{1}{3}$
Slope of AB = Slope of CD
Slope of BC = Slope of AD
So, AB || CD and BC || AD
Hence, ABCD is a parallelogram.

#### Question 6:

Find k, if R(1, –1), S (–2, k) and slope of line RS is –2.

Slope of line RS is –2
Slope of RS will be
$\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{k-\left(-1\right)}{-2-1}=\frac{k+1}{-3}=-2\phantom{\rule{0ex}{0ex}}⇒k+1=6\phantom{\rule{0ex}{0ex}}⇒k=5$

#### Question 7:

Find k, if B(k, –5), C (1, 2) and slope of the line is 7.

Slope of the line BC is 7.
Slope of BC =
$\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{2-\left(-5\right)}{1-k}=7\phantom{\rule{0ex}{0ex}}⇒\frac{7}{1-k}=7\phantom{\rule{0ex}{0ex}}⇒1-k=1\phantom{\rule{0ex}{0ex}}⇒k=0$

#### Question 8:

Find k, if PQ || RS and P(2, 4), Q (3, 6), R(3, 1), S(5, k).

Given P(2, 4), Q (3, 6), R(3, 1), S(5, k)
PQ || RS so, slope of PQ = slope of RS
$\frac{6-4}{3-2}=\frac{k-1}{5-3}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{1}=\frac{k-1}{2}\phantom{\rule{0ex}{0ex}}⇒k-1=4\phantom{\rule{0ex}{0ex}}⇒k=5$

#### Question 1:

Fill in the blanks using correct alternatives.
(1) Seg AB is parallel to Y-axis and coordinates of point A are (1,3) then co–ordinates of point B can be ........ .
(A) (3,1)
(B) (5,3)
(C) (3,0)
(D) (1,–3)

(2) Out of the following, point ........ lies to the right of the origin on X– axis.
(A) (–2,0)
(B) (0,2)
(C) (2,3)
(D) (2,0)

(3) Distance of point (–3,4) from the origin is ...... .
(A) 7
(B) 1
(C) 5
(D) –5

(4) A line makes an angle of 30° with the positive direction of X– axis. So the slope of the line is .......... .

(A) $\frac{1}{2}$

(B) $\frac{\sqrt{3}}{2}$

(C) $\frac{1}{\sqrt{3}}$

(D) $\sqrt{3}$

(1) Slope of y-axis will be not defined because the denominator will be 0.
Since AB is parallel to y-axis, so the slope will be same.
Let the coordinate of point B be (x, y).
Slope of AB = $\frac{y-3}{x-1}$ = slope of y-axis
Slope of y-axis is not defined as the denominator is 0 so, the denominator of $\frac{y-3}{x-1}$ will also be 0.
So,
$x-1=0\phantom{\rule{0ex}{0ex}}⇒x=1$
with $x=1$ there is only 1 option that (D).
Hence, the correct answer is option D.

(2) A point lying on the right of the origin on X – axis will have positive x-coordinate and non-zero y-coordinate.
So, among the given points (2, 3) lies on the right of the origin on X – axis.
Hence, the correct answer is option (C).

(3) Let the given point be P(–3,4).
Distance of P(–3,4) from the origin O(0, 0) is
$\sqrt{{\left(0-\left(-3\right)\right)}^{2}+{\left(0-4\right)}^{2}}=\sqrt{9+16}=\sqrt{25}=5$
Hence, the correct answer is option (C).

(4) Slope of line will be $\mathrm{tan}30°=\frac{1}{\sqrt{3}}$
Hence, the correct answer is option (C).

#### Question 2:

Determine whether the given points are collinear.
(1) A(0,2), B(1,–0.5), C(2,–3)

(2)

(3) L(1,2), M(5,3) , N(8,6)

(1) A(0,2), B(1,–0.5), C(2,–3)
Slope of AB = $\frac{-0.5-2}{1-0}=\frac{-2.5}{1}=-2.5$
Slope of BC = $\frac{-3-\left(-0.5\right)}{2-1}=\frac{-2.5}{1}=-2.5$
So, the slope of AB = slope of BC.
Point B lies on both the lines.
Hence, the given points are collinear.

(2)
Slope of PQ = $\frac{\frac{8}{5}-2}{2-1}=\frac{\frac{-2}{5}}{1}=\frac{-2}{5}$
Slope of QR = $\frac{\frac{6}{5}-\frac{8}{5}}{3-2}=\frac{\frac{-2}{5}}{1}=\frac{-2}{5}$
So, the slope of PQ = slope of QR.
Point Q lies on both the lines.
Hence, the given points are collinear.

(3) L(1,2), M(5,3) , N(8,6)
Slope of LM$=\frac{3-2}{5-1}=\frac{1}{4}$
Slope of MN = $\frac{6-3}{8-5}=\frac{3}{3}=1$
Thus, the slope of LM not equal to slope MN.
So, the given points are not collinear.

#### Question 3:

Find the coordinates of the midpoint of the line segment joining P(0,6) and Q(12,20).

The given points are P(0,6) and Q(12,20).
Midpoint of PQ = $=\left(\frac{0+12}{2},\frac{6+20}{2}\right)=\left(6,13\right)$.

#### Question 4:

Find the ratio in which the line segment joining the points A(3,8) and B(–9, 3) is divided by the Y– axis.

Let the ratio in which the line AB is divided by the y-axis be k : 1.
The point on the - axis be (0, y).

Thus, the ratio is 1 : 3.

#### Question 5:

Find the point on X–axis which is equidistant from P(2,–5) and Q(–2,9).

Let the point on x-axis equidistant from P(2,–5) and Q(–2,9) be $\mathrm{A}\left(x,0\right)$.
$\mathrm{AP}=\sqrt{{\left(x-2\right)}^{2}+{\left(0-\left(-5\right)\right)}^{2}}=\sqrt{{\left(x-2\right)}^{2}+25}\phantom{\rule{0ex}{0ex}}\mathrm{QA}=\sqrt{{\left(x-\left(-2\right)\right)}^{2}+{\left(0-9\right)}^{2}}=\sqrt{{\left(x+2\right)}^{2}+81}\phantom{\rule{0ex}{0ex}}\mathrm{AP}=\mathrm{QA}\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(x-2\right)}^{2}+25}=\sqrt{{\left(x+2\right)}^{2}+81}\phantom{\rule{0ex}{0ex}}$
Squaring both sides
${\left(x-2\right)}^{2}+25={\left(x+2\right)}^{2}+81\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4-4x+25={x}^{2}+4+4x+81\phantom{\rule{0ex}{0ex}}⇒-8x=56\phantom{\rule{0ex}{0ex}}⇒x=-7$
Thus, the required point is $\left(-7,0\right)$.

#### Question 6:

Find the distances between the following points.
(i) A(a, 0), B(0, a) (ii) P(–6, –3), Q(–1, 9) (iii) R(–3a, a), S(a, –2a)

(i) A(a, 0), B(0, a)
$\mathrm{AB}=\sqrt{{\left(0-a\right)}^{2}+{\left(a-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{a}^{2}+{a}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{2{a}^{2}}\phantom{\rule{0ex}{0ex}}=a\sqrt{2}$

(ii) P(–6, –3), Q(–1, 9)
$\mathrm{PQ}=\sqrt{{\left(-6-\left(-1\right)\right)}^{2}+{\left(-3-9\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{25+144}\phantom{\rule{0ex}{0ex}}=\sqrt{169}\phantom{\rule{0ex}{0ex}}=13$

(iii) R(–3aa), S(a, –2a)
$\mathrm{RS}=\sqrt{{\left(-3a-a\right)}^{2}+{\left(a-\left(-2a\right)\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{16{a}^{2}+9{a}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{25{a}^{2}}\phantom{\rule{0ex}{0ex}}=5a$

#### Question 7:

Find the coordinates of the circumcentre of a triangle whose vertices are (–3,1), (0,–2) and (1,3)

Let the given vertices A(–3, 1), B(0, –2) and C(1, 3).
Circumcircle passes through the vertices of a triangle.
Let the circumcentre be P(a, b)
The distance of point P will be equal from A, B and C.
PA = PB = PC
$⇒$${\mathrm{PA}}^{2}={\mathrm{PB}}^{2}={\mathrm{PC}}^{2}$

solving (1) and (2) we get
$a=\frac{-1}{3},b=\frac{2}{3}$
Thus, the circumcentre of the triangle is $\mathrm{P}\left(\frac{-1}{3},\frac{2}{3}\right)$

#### Question 8:

In the following examples, can the segment joining the given points form a triangle ? If triangle is formed, state the type of the triangle considering sides of the triangle.
(1) L(6,4) , M(–5,–3) , N(–6,8)

(2) P(–2,–6) , Q(–4,–2), R(–5,0)

(3)

(1) L(6,4) , M(–5,–3) , N(–6,8)
$\mathrm{LM}=\sqrt{{\left(6+5\right)}^{2}+{\left(4+3\right)}^{2}}=\sqrt{121+49}=\sqrt{170}=2\sqrt{35}\phantom{\rule{0ex}{0ex}}\mathrm{MN}=\sqrt{{\left(-5+6\right)}^{2}+{\left(-3-8\right)}^{2}}=\sqrt{1+121}=\sqrt{122}=2\sqrt{61}\phantom{\rule{0ex}{0ex}}\mathrm{LN}=\sqrt{{\left(6+6\right)}^{2}+{\left(4-8\right)}^{2}}=\sqrt{144+16}=\sqrt{160}=4\sqrt{10}$
LM, MN and LN form a scalene triangle.

(2) P(–2,–6) , Q(–4,–2), R(–5,0)
$\mathrm{PQ}=\sqrt{{\left(-2+4\right)}^{2}+{\left(-6+2\right)}^{2}}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}\phantom{\rule{0ex}{0ex}}\mathrm{QR}=\sqrt{{\left(-4+5\right)}^{2}+{\left(-2+0\right)}^{2}}=\sqrt{1+4}=\sqrt{5}\phantom{\rule{0ex}{0ex}}\mathrm{PR}=\sqrt{{\left(-2+5\right)}^{2}+{\left(-6+0\right)}^{2}}=\sqrt{9+36}=\sqrt{45}=3\sqrt{5}\phantom{\rule{0ex}{0ex}}\mathrm{PQ}+\mathrm{QR}=\mathrm{PR}\phantom{\rule{0ex}{0ex}}$
Since the sum of the two sides is not greater than the third side so, the given vertices do not form a triangle.

(3)
$\mathrm{AB}=\sqrt{{\left(\sqrt{2}+\sqrt{2}\right)}^{2}+{\left(\sqrt{2}+\sqrt{2}\right)}^{2}}=\sqrt{8+8}=\sqrt{16}=4\phantom{\rule{0ex}{0ex}}\mathrm{BC}=\sqrt{{\left(-\sqrt{2}+\sqrt{6}\right)}^{2}+{\left(-\sqrt{2}-\sqrt{6}\right)}^{2}}=\sqrt{16}=4\phantom{\rule{0ex}{0ex}}\mathrm{AC}=\sqrt{{\left(\sqrt{2}+\sqrt{6}\right)}^{2}+{\left(\sqrt{2}-\sqrt{6}\right)}^{2}}=\sqrt{16}=4\phantom{\rule{0ex}{0ex}}\mathrm{AB}=\mathrm{BC}=\mathrm{AC}$
So, these vertices form an equilateral triangle.

#### Question 9:

Find k if the line passing through points P(–12, –3) and Q(4, k) has slope $\frac{1}{2}$.

Slope = $\frac{1}{2}$
Given points are P(–12, –3) and Q(4, k
Slope of PQ =
$\frac{k+3}{4+12}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒2k+6=16\phantom{\rule{0ex}{0ex}}⇒2k=10\phantom{\rule{0ex}{0ex}}⇒k=5$

#### Question 10:

Show that the line joining the points A(4, 8) and B(5, 5) is parallel to the line joining the points C(2, 4) and D(1, 7).

Slope of the line joining the points A(4, 8) and B(5, 5) will be
$=\frac{5-8}{5-4}=\frac{-3}{1}=-3$
Slope of the line joining the points C(2, 4) and D(1, 7) will be
$=\frac{7-4}{1-2}=\frac{3}{-1}=-3$
Since the slope of AB = slope of CD
so, the given lines are parallel.

#### Question 11:

Show that points P(1, –2), Q(5, 2), R(3, –1), S(–1, –5) are the vertices of a parallelogram

The given points are P(1, –2), Q(5, 2), R(3, –1), S(–1, –5).

Slope of PQ = Slope of RS and Slope of QR = Slope of PS
So, PQ || RS and QR || PS
Thus, PQRS is a parallelogram.

#### Question 12:

Show that the ▢PQRS formed by P(2, 1), Q(–1, 3), R(–5, –3) and S(–2, –5) is a rectangle

The given points are P(2, 1), Q(–1, 3), R(–5, –3) and S(–2, –5).
PQ = $\sqrt{{\left(-1-2\right)}^{2}+{\left(3-1\right)}^{2}}=\sqrt{9+4}=\sqrt{13}$
QR = $\sqrt{{\left(-5+1\right)}^{2}+{\left(-3-3\right)}^{2}}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}$
RS = $\sqrt{{\left(-2+5\right)}^{2}+{\left(-5+3\right)}^{2}}=\sqrt{9+4}=\sqrt{13}$
PS = $\sqrt{{\left(-5-1\right)}^{2}+{\left(-2-2\right)}^{2}}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}$
PQ = RS and QR = PS
Opposite sides are equal.
$\mathrm{PR}=\sqrt{{\left(-3-1\right)}^{2}+{\left(-5-2\right)}^{2}}=\sqrt{16+49}=\sqrt{65}\phantom{\rule{0ex}{0ex}}\mathrm{QS}=\sqrt{{\left(-2+1\right)}^{2}+{\left(-5-3\right)}^{2}}=\sqrt{1+64}=\sqrt{65}$
Diagonals are equal.
Thus, the given vertices form a rectangle.

#### Question 13:

Find the lengths of the medians of a triangle whose vertices are A(–1, 1), B(5, –3) and C(3, 5).

Let the medians meet the lines BC, AC and AB at points be $\mathrm{P}\left({x}_{1},{y}_{1}\right)$$\mathrm{Q}\left({x}_{2},{y}_{2}\right)$ and $\mathrm{R}\left({x}_{3},{y}_{3}\right)$ respectively.
P is thus the mid point of line BC
$\mathrm{P}\left({x}_{1},{y}_{1}\right)=\left(\frac{5+3}{2},\frac{5-3}{2}\right)=\left(4,2\right)\phantom{\rule{0ex}{0ex}}\mathrm{AP}=\sqrt{{\left(4+1\right)}^{2}+{\left(1-1\right)}^{2}}=\sqrt{25}=5$
Q is the mid point of line AC.
$Q\left({x}_{2},{y}_{2}\right)=\left(\frac{-1+3}{2},\frac{1+5}{2}\right)=\left(1,3\right)\phantom{\rule{0ex}{0ex}}\mathrm{BQ}=\sqrt{{\left(5-1\right)}^{2}+{\left(-3-3\right)}^{2}}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}$
R is the mid point of AB.
$\mathrm{R}\left({x}_{3},{y}_{3}\right)=\left(\frac{-1+5}{2},\frac{1-3}{2}\right)=\left(2,-1\right)\phantom{\rule{0ex}{0ex}}\mathrm{RC}=\sqrt{{\left(3-2\right)}^{2}+{\left(5+1\right)}^{2}}=\sqrt{1+36}=\sqrt{37}$

#### Question 14:

Find the coordinates of centroid of the triangles if points D(–7, 6), E(8, 5) and F(2, –2) are the mid points of the sides of that triangle.

The given points are D(–7, 6), E(8, 5) and F(2, –2).
Centroid is the meeting point of the medians.
Let the centroid be O(a, b)
$\left(a,b\right)=\left(\frac{2+8-7}{3},\frac{-2+5+6}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(1,3\right)$

#### Question 15:

Show that A(4, –1), B(6, 0), C(7, –2) and D(5, –3) are vertices of a square.

The given points are A(4, –1), B(6, 0), C(7, –2) and D(5, –3).
AB = $\sqrt{{\left(6-4\right)}^{2}+{\left(0+1\right)}^{2}}=\sqrt{4+1}=\sqrt{5}$
BC = $\sqrt{{\left(6-7\right)}^{2}+{\left(0+2\right)}^{2}}=\sqrt{1+4}=\sqrt{5}$
CD = $\sqrt{{\left(7-5\right)}^{2}+{\left(-2+3\right)}^{2}}=\sqrt{4+1}=\sqrt{5}$
AD = $\sqrt{{\left(5-4\right)}^{2}+{\left(-3+1\right)}^{2}}=\sqrt{1+4}=\sqrt{5}$
AB = BC = CD = DA
Slope of AB = $\frac{0+1}{6-4}=\frac{1}{2}$
Slope of BC = $\frac{-2-0}{7-6}=-2$
Slope of CD = $\frac{-3+2}{5-7}=\frac{1}{2}$
Slope of AD = $\frac{-3+1}{5-4}=-2$
Thus, AB perpendicular to BC and AD. Also, CD perpendicular to AD and AB.
So, all the sides are equal to each other and are perpendicular.
Thus, they form a square.

#### Question 16:

Find the coordinates of circumcentre and radius of circumcircle of ∆ABC if A(7, 1), B(3, 5) and C(2, 0) are given.

Let the circumcentre be $\mathrm{P}\left(a,b\right)$.
The given points are A(7, 1), B(3, 5) and C(2, 0).
The circumcircle passes through the points A, B and C and the thus,
PA = PB = PC
$⇒{\mathrm{PA}}^{2}={\mathrm{PB}}^{2}={\mathrm{PC}}^{2}$

$=\sqrt{{\left(\frac{25}{6}-2\right)}^{2}+{\left(\frac{13}{6}-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{13}{6}\right)}^{2}+{\left(\frac{13}{6}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{13}{6}\sqrt{2}$

#### Question 17:

Given A(4, –3), B(8, 5). Find the coordinates of the point that divides segment AB in the ratio 3 : 1.

Let he coordinate of the point which divide the line AB in the ratio 3 : 1 be P(a, b)
$a=\frac{3×8+1×4}{3+1}=\frac{24+4}{4}=7\phantom{\rule{0ex}{0ex}}b=\frac{3×5+1×\left(-3\right)}{3+1}=\frac{15-3}{4}=3$
P(a, b) = (7, 3)

#### Question 18:

Find the type of the quadrilateral if points A(–4, –2), B(–3, –7) C(3, –2) and D(2, 3) are joined serially.

The given points are A(–4, –2), B(–3, –7) C(3, –2) and D(2, 3).
If they are joined serially so,
Slope of AB = $\frac{-7+2}{-3+4}=-5$
Slope of BC = $\frac{-2+7}{3+3}=\frac{5}{6}$
Slope of CD = $\frac{3+2}{2-3}=-5$
Slope of AD = $\frac{3+2}{2+4}=\frac{5}{6}$
Opposite sides are parallel.
AC = $\sqrt{{\left(3+4\right)}^{2}+{\left(-2+2\right)}^{2}}=\sqrt{49}=7$
BD = $\sqrt{{\left(3+7\right)}^{2}+{\left(2+3\right)}^{2}}=\sqrt{125}=5\sqrt{5}$
Diagonals are not equal.
Hence, the given points form a parallelogram.

#### Question 19:

The line segment AB is divided into five congruent parts at P, Q, R and S such that A–P–Q–R–S–B. If point Q(12, 14) and S(4, 18) are given find the coordinates of A, P, R, B.

Let the coordinates be $\mathrm{A}\left({x}_{1},{y}_{1}\right)$ $\mathrm{P}\left({x}_{2},{y}_{2}\right)$$R\left({x}_{3},{y}_{3}\right)$ and B$\left({x}_{4},{y}_{4}\right)$
QR = RS
$\mathrm{R}\left({x}_{3},{y}_{3}\right)=\left(\frac{12+4}{2},\frac{14+18}{2}\right)=\left(8,16\right)$
RS = SB

$\frac{\mathrm{AQ}}{\mathrm{QB}}=\frac{\mathrm{AP}+\mathrm{PQ}}{\mathrm{QR}+\mathrm{RS}+\mathrm{SB}}=\frac{\mathrm{AP}+\mathrm{AP}}{\mathrm{AP}+\mathrm{AP}+\mathrm{AP}}=\frac{2}{3}$

$\mathrm{A}\left({x}_{1},{y}_{1}\right)$ = (20, 10)
AP = PQ

#### Question 20:

Find the coordinates of the centre of the circle passing through the points P(6, –6), Q(3, –7) and R (3, 3).

Let the centre be P(a, b)
RO = PO = QO
${\mathrm{RO}}^{2}={\mathrm{PO}}^{2}$

${\mathrm{PO}}^{2}={\mathrm{OQ}}^{2}$

Multiplying (2) with 3 we get

#### Question 21:

Find the possible pairs of coordinates of the fourth vertex D of the parallelogram, if three of its vertices are A(5, 6), B(1, –2) and C(3, –2).

ABCD is a parallelogram. So, AD = CB and CD = AB.

Point D lies on the line passing through the point A. So, the ordinate of the point D will also be same as that of point A which is 6.
So, = 6
Putting the value of in (1) we get
${a}^{2}+{\left(6\right)}^{2}-10a-12×6+57=0\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+36-10a-72+57=0\phantom{\rule{0ex}{0ex}}⇒{a}^{2}-10a+21=0\phantom{\rule{0ex}{0ex}}⇒{a}^{2}-7a-3a+21=0\phantom{\rule{0ex}{0ex}}⇒a\left(a-7\right)-3\left(a-7\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(a-3\right)\left(a-7\right)=0\phantom{\rule{0ex}{0ex}}⇒a=3,7$
Thus, the possible values of point D are (3, 6) and (7, 6).

#### Question 22:

Find the slope of the diagonals of a quadrilateral with vertices A(1, 7), B(6, 3), C(0, –3) and D(–3, 3).

Slope of BD = $\frac{3-3}{6+3}=0$
Slope of AC = $\frac{7+3}{1-0}=\frac{10}{1}=10$