## Coordinate Geometry

### Class-10-Mathematics-2-Chapter-5-Maharashtra Board

### Practice Set Solutions

**Practice set 5.1**

**Question 1.1. Find the distance between each of the following pairs of points****. **

**(1) A(2, 3), B(4, 1) **

Suppose the coordinates of point A are (x_{1}, y_{1}) and of point B are (x_{2}, y_{2}).

x_{1} = 2, y_{1} = 3, x_{2} = 4 and y_{2} = 1

By distance formula,

AB = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

= \(\sqrt{(4-2)^2+(1-3)^2}\)

= \(\sqrt{(2)^2+(-2)^2}\)

= \(\sqrt{4 + 4}\)

= \(\sqrt{8}\) = \(\sqrt{4×2}\) = \(2\sqrt{2}\)

**Answer is : Distance between points A and B is \(2\sqrt{2}\)** **.**

**(2) P(−5, 7), Q(−1, 3) **

Suppose the coordinates of point P are (x_{1}, y_{1}) and of point Q are (x_{2}, y_{2}).

x_{1} = −5, y_{1} = 7, x_{2} = −1 and y_{2} = 3

By distance formula,

PQ = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

= \(\sqrt{[-1-(-5)]^2+(3-7)^2}\)

= \(\sqrt{(4)^2+(-4)^2}\)

= \(\sqrt{16 + 16}\)

= \(\sqrt{32}\)

= \(\sqrt{16×2}\)

= \(4\sqrt{2}\)

**Answer is : Distance between points P and Q is \(4\sqrt{2}\)** **.**

**(3) R(0, −3), S(0, \(\frac{5}{2}\)****)**

Suppose the coordinates of point R are (x_{1}, y_{1}) and of point S are (x_{2}, y_{2}).

x_{1} = 0, y_{1} = −3, x_{2} = 0 and y_{2} = \(\frac{5}{2}\)

By distance formula,

PQ = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

= \(\sqrt{(0-0)^2+[\frac{5}{2}-(-3)]^2}\)

= \(\sqrt{(o)^2+(\frac{5}{2}+3)^2}\)

= \(\sqrt{0+(\frac{5+6}{2})^2}\)

= \(\sqrt{(\frac{11}{2})^2}\)

= \(\frac{11}{2}\)

**Answer is : Distance between points P and Q is \(\frac{11}{2}\)** **.**

**(4) L(5, −8), M(−7, −3) **

Suppose the coordinates of point L are (x_{1}, y_{1}) and of point M are (x_{2}, y_{2}).

x_{1} = 5, y_{1} = −8, x_{2} = −7 and y_{2} = −3

By distance formula,

PQ = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

= \(\sqrt{(-7-5)^2+[-3-(-8)]^2}\)

= \(\sqrt{(-12)^2+(5)^2}\)

= \(\sqrt{144 + 25}\)

= \(\sqrt{169}\)

= 13

**Answer is : Distance between points P and Q is ****13****.**

**(5) T(−3, 6), R(9, −10) **

Suppose the coordinates of point T are (x_{1}, y_{1}) and of point R are (x_{2}, y_{2}).

x_{1} = −3, y_{1} = 6, x_{2} = 9 and y_{2} = −10

By distance formula,

PQ = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

= \(\sqrt{[9-(-3)]^2+(-10-6)^2}\)

= \(\sqrt{(12)^2+(-16)^2}\)

= \(\sqrt{144 + 256}\)

= \(\sqrt{400}\)

= 20

**Answer is : Distance between points P and Q is 20****.**

**(6) W(\(\frac{-7}{2}\) , 4), X(11, 4)**

Suppose the coordinates of point W are (x_{1}, y_{1}) and of point X are (x_{2}, y_{2}).

x_{1} = \(\frac{-7}{2}\), y_{1} = 4, x_{2} = 11 and y_{2} = 4

By distance formula,

PQ = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

= \(\sqrt{[11-(\frac{-7}{2})]^2+(4-4)^2}\)

= \(\sqrt{(11+\frac{7}{2})^2+(0)^2}\)

= \(\sqrt{(\frac{22+7}{2})^2 + 0}\)

= \(\sqrt{(\frac{29}{2})^2}\)

= \(\frac{29}{2}\)

= 14.5

**Answer is : Distance between points P and Q is 14.5****.**

**Question 1.2. Determine whether the points are collinear.**

**(1) A(1, −3), B(2, −5), C(−4, 7) **

**Given : **A(1, −3), B(2, −5), C(−4, 7)

By distance formula,

d(A, B) = \(\sqrt{(2-1)^2+[(-5-(-3)]^2}\)

= \(\sqrt{(1)^2+(-2)^2}\)

= \(\sqrt{1 + 4}\)

= \(\sqrt{5}\) ….(i)

d(B, C) = \(\sqrt{(-4-2)^2+[(7-(-5)]^2}\)

= \(\sqrt{(-6)^2+(12)^2}\)

= \(\sqrt{36+144}\)

= \(\sqrt{180}\)

= \(\sqrt{36×5}\)

= \(6\sqrt{5}\) ….(ii)

d(A, C) = \(\sqrt{(-4-1)^2+[(7-(-3)]^2}\)

= \(\sqrt{(-5)^2+(10)^2}\)

= \(\sqrt{25+100}\)

= \(\sqrt{125}\)

= \(\sqrt{25×5}\)

= \(5\sqrt{5}\) ….(iii)

Adding (i) and (iii),

d(A, B) + d(A, C)= \(\sqrt{5}\) + \(5\sqrt{5}\) = \(6\sqrt{5}\) ... (iv)

∴ d(A, B) + d(A, C) = d(B, C) ... [From (ii) and (iv)]

∴ points A, B and C are collinear.

**Answer is: Points A(1, −3), B(2, −5) and C(−4, 7) are collinear.**

**(2) L(−2, 3), M(1, −3), N(5, 4)**

**Given : **L(−2, 3), M(1, −3), N(5, 4)

By distance formula,

d(L, M) = \(\sqrt{[1-(-2)]^2+(-3-3)^2}\)

= \(\sqrt{(3)^2+(-6)^2}\)

= \(\sqrt{9 + 36}\)

= \(\sqrt{45}\)

= \(\sqrt{9×5}\)

= \(3\sqrt{5}\) ….(i)

d(M, N) = \(\sqrt{(5-1)^2+[4-(-3)]^2}\)

= \(\sqrt{(4)^2+(7)^2}\)

= \(\sqrt{16 + 49}\)

= \(\sqrt{65}\) ….(ii)

d(L, N) = \(\sqrt{[5-(-2)]^2+(4-3)^2}\)

= \(\sqrt{(7)^2+(1)^2}\)

= \(\sqrt{49 + 1}\)

= \(\sqrt{50}\)

= \(\sqrt{25×2}\)

= \(5\sqrt{2}\) ….(iii)

Adding (i) and (iii),

d(L, M) + d(L, N) = \(3\sqrt{5}\) + \(5\sqrt{2}\) ….(iv)

∴ d(L, M) + d(L, N) ≠ d(M, N) ….[From (ii) and (iv)]

∴ points L, M and N are not collinear.

**Answer is: Points L(−2, 3), M(1, −3) and N(5, 4) are not collinear.**

**(3) R(0, 3), D(2, 1), S(3, −1) **

**Given : **R(0, 3), D(2, 1), S(3, −1)

By distance formula,

d(R, D) = \(\sqrt{(2-0)^2+(1-3)^2}\)

= \(\sqrt{(2)^2+(-2)^2}\)

= \(\sqrt{4 + 4}\)

= \(\sqrt{8}\)

= \(\sqrt{4×2}\)

= \(2\sqrt{2}\) ….(i)

d(D, S) = \(\sqrt{(3-2)^2+(-1-1)^2}\)

= \(\sqrt{(1)^2+(-2)^2}\)

= \(\sqrt{1 + 4}\)

= \(\sqrt{5}\) ….(ii)

d(R, S) = \(\sqrt{(3-0)^2+(-1-3)^2}\)

= \(\sqrt{(3)^2+(-4)^2}\)

= \(\sqrt{9 + 16}\)

= \(\sqrt{25}\)

= 5 ….(iii)

Adding (ii) and (iii),

d(D, S) + d(R, S) = \(\sqrt{5}\) + 5 ….(iv)

∴ d(D, S) + d(R, S) ≠ d(R, D) ... [From (i) and (iv)]

∴ points R, D and S are not collinear.

**Answer is . Points R(0, 3), D(2, 1) and S(3, −1) are not collinear.**

**(4) P(−2, 3), Q(1, 2), R(4, 1)**

**Given : **P(−2, 3), Q(1, 2), R(4, 1)

By distance formula,

d(P, Q) = \(\sqrt{[1-(-2)]^2+(2-3)^2}\)

= \(\sqrt{(3)^2+(-1)^2}\)

= \(\sqrt{9 + 1}\)

= \(\sqrt{10}\) ….(i)

d(Q, R) = (\sqrt{(4-1)^2+(1-2)^2}\)

= \(\sqrt{(3)^2+(-1)^2}\)

= \(\sqrt{9 + 1}\)

= \(\sqrt{10}\) ….(ii)

d(P, R) = \(\sqrt{[4-(-2)]^2+(1-3)^2}\)

= \(\sqrt{(6)^2+(-2)^2}\)

= \(\sqrt{36 + 4}\)

= \(\sqrt{40}\)

= \(\sqrt{4×10}\)

= \(2\sqrt{10}\) ….(iii)

Adding (i) and (ii), we get,

d(P, Q) + d(Q, R) = \(\sqrt{10}\) + \(\sqrt{10}\)

∴ d(P, Q) + d(Q,R) = \(2\sqrt{10}\) ….(iv)

∴ d(P, Q) + d(Q, R) = d(P, R) ... [From (iii) and (iv)]

∴ points P, Q and R are collinear.

**Answer is: Points P(−2, 3), Q(1, 2) and R(4, 1) are collinear.**

**Question 1.3. Find the point on the ****X−****axis which is equidistant from ****A(−3, 4) ****and ****B(1, −4).**

A(−3, 4) and B(1, − 4).

Let P be the point on X−axis equidistant from points A and B.

PA = PB …..(i)

P lies on X−axis its y coordinate is 0.

Let P(x, 0)

Now PA = PB ….[From (i)]

By distance formula,

d(P, A) = d(P, B)

\(\sqrt{[x-(-3)]^2+(0-4)^2}\) = \(\sqrt{(x-1)^2+[0-(-4)]^2}\)

∴ \(\sqrt{(x+3)^2+(-4)^2}\) = \(\sqrt{(x-1)^2+(4)^2}\)

∴ \(\sqrt{(x+3)^2+16}\) = \(\sqrt{(x-1)^2+16}\)

Squaring both the sides,

(x + 3)^{2} + 16 = (x − 1)^{2} + 16

X^{2} + 6x + 9 = x^{2} − 2x + 1

6x + 2x = 1 — 9

8x = − 8

∴ x = −8/8

∴ x = − 1

P(−1, 0)

**Answer is: The coordinates of point on X−axis equidistant from points A and B are ( −1, 0).**

**Question 1.4. Verify that points ****P(−2, 2), Q(2, 2) ****and ****R(2, 7) ****are vertices of a right angled triangle****.**

**Given :** P( −2, 2), Q(2, 2) and R(2, 7)

By distance formula,

PQ = \(\sqrt{[2-(-2)]^2+(2-2)^2}\)

= \(\sqrt{(4)^2+(0)^2}\)

= \(\sqrt{16 + 0}\)

= \(\sqrt{16}\)

= 4 ….(i)

QR = \(\sqrt{(2-2)^2+(7-2)^2}\)

= \(\sqrt{(0)^2+(5)^2}\)

= \(\sqrt{0 + 25}\)

= \(\sqrt{25}\)

= 5 ….(ii)

PR = \(\sqrt{[2-(-2)]^2+(7-2)^2}\)

= \(\sqrt{(4)^2+(5)^2}\)

= \(\sqrt{16 + 25}\)

= \(\sqrt{41}\)

PQ^{2} + QR^{2} = 4^{2} + 5^{2} = 16 + 25 = 41 …. [From (i) and (ii)] ….(iii)

PR^{2} = \((\sqrt{41})^2\) = 41 ….(iv)

PQ^{2} + QR^{2} = PR^{2} ….[From (iii) and (iv)]

by converse of Pythagoras theorem, Δ PQR is a right angled triangle.

i.e. P( −2, 2), Q(2, 2) and R(2, 7) are the vertices of a right angled triangle.

**Question 1.5. Show that points ****P(2, −2), Q(7, 3), R(11, −1) ****and ****S (6, −6) ****are vertices of a parallelogram****.**

**Given : **P(2, −2), Q(7, 3), R(11, −1) and S (6, −6)

By distance formula,

PQ = \(\sqrt{(7-2)^2+[3-(-2)]^2}\)

= \(\sqrt{(5)^2+(5)^2}\)

= \(\sqrt{25 + 25}\)

= \(\sqrt{50}\)

= \(\sqrt{25×2}\)

= \(5\sqrt{2}\) ….(i)

QR = \(\sqrt{(11-7)^2+(-1-3)^2}\)

= \(\sqrt{(4)^2+(-4)^2}\)

= \(\sqrt{16 + 16}\)

= \(\sqrt{32}\)

= \(\sqrt{16×2}\)

= \(4\sqrt{2}\) ….(ii)

RS = \(\sqrt{(6-11)^2+[-6-(-1)]^2}\)

= \(\sqrt{(-5)^2+(-5)^2}\)

= \(\sqrt{25 + 25}\)

= \(\sqrt{50}\)

= \(\sqrt{25×2}\)

= \(5\sqrt{2}\) ….(iii)

PS = \(\sqrt{(6-2)^2+[-6-(-2)]^2}\)

= \(\sqrt{(4)^2+(-4)^2}\)

= \(\sqrt{16 + 16}\)

= \(\sqrt{32}\)

= \(\sqrt{16×2}\)

= \(4\sqrt{2}\) ….(iv)

ln □ PQRS.

PQ = RS [From (i) and (iii)]

OH = PS ' [From (ii) and (iv)]

A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent.

∴ □ PQRS is a parallelogram.

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

**Question 1.6. Show that points ****A(−4, −7), B(−1, 2), C(8, 5) ****and ****D(5, −4) ****are vertices of a rhombus ****ABCD.**

**Given : **A(−4, −7), B(−1, 2), C(8, 5) and D(5, −4)

By distance formula,

d(A, B) = \(\sqrt{[-1-(-4)]^2+[2-(-7)]^2}\)

= \(\sqrt{(3)^2+(9)^2}\)

= \(\sqrt{9+81}\)

= \(\sqrt{90}\)

= \(\sqrt{9×10}\)

= \(3\sqrt{10}\)

∴ AB = \(3\sqrt{10}\) ….(i)

d(B, C) = \(\sqrt{[8-(-1)]^2+(5-2)^2}\)

= \(\sqrt{(9)^2+(3)^2}\)

= \(\sqrt{81+9}\)

= \(\sqrt{90}\)

= \(\sqrt{9×10}\)

= \(3\sqrt{10}\)

∴ BC = \(3\sqrt{10}\ ….(ii)

d(C, D) = \(\sqrt{(5-8)^2+(-4-5)^2}\)

= \(\sqrt{(-3)^2+(-9)^2}\)

= \(\sqrt{9+81}\)

= \(\sqrt{90}\)

= \(\sqrt{9×10}\)

= \(3\sqrt{10}\)

∴ CD = \(3\sqrt{10}\) ….(iii)

d(A, D) = \(\sqrt{[5-(-4)]^2+[-4-(-7)]^2}\)

= \(\sqrt{(9)^2+(3)^2}\)

= \(\sqrt{81+9}\)

= \(\sqrt{90}\)

= \(\sqrt{9×10}\)

= \(3\sqrt{10}\)

∴ AD = \(3\sqrt{10}\) ….(iv)

from (i), (ii), (iii) and (iv)

AB = BC = CD = AD

□ ABCD is a rhombus.

∴ A(−4, −7), B(−1, 2), C(8, 5) and D(5. −4) are vertices of rhombus ABCD.

**Question 1.7. Find x if distance between points **

**L(**

*x***, 7)**

**and**

**M(1, 15)**

**is**

**10.**

**Given : **L(*x*, 7) and M(1, 15) d(L, M) = 10

By distance formula,

d(L, M) = \(\sqrt{(x-1)^2+(7-15)^2}\)

∴ 10 = \(\sqrt{(x-1)^2+(-8)^2}\)

Squaring both the sides, we get,

100 = (x − 1)^{2} + 64

∴ (x − 1)^{2} = 100 − 64

∴ (x − 1)^{2} = 36

∴ x − 1 = ± 6 …. (Taking square roots of both the sides)

Either x – 1 = 6 or x – 1 = −6

∴ x = 6 + 1 or x = −6 + 1

∴ x = 7 or x = −5

Answer is x = 7 or x = −5.

**Question 1.8. Show that the points ****A(1, 2), B(1, 6), C(1+\(2\sqrt{3}\)** **, 4) ****are vertices of an equilateral triangle****.**

**Given : **A(1, 2), B(1, 6), C(1+\(2\sqrt{3}\), 4)

By distance formula,

AB = \(\sqrt{(1-1)^2+(6-2)^2}\)

= \(\sqrt{(0)^2+(4)^2}\)

= \(\sqrt{0+16}\)

= \(\sqrt{16}\)

= 4

BC = \(\sqrt{(1+2\sqrt{3}-1)^2+(4-6)^2}\)

= \(\sqrt{(2\sqrt{3})^2+(-4)^2}\)

= \(\sqrt{12+4}\)

= \(\sqrt{16}\)

= 4 ….(ii)

AC = \(\sqrt{(1+2\sqrt{3}-1)^2+(4-2)^2}\)

= \(\sqrt{(2\sqrt{3})^2+(2)^2}\)

= \(\sqrt{12+4}\)

= \(\sqrt{16}\)

= 4

∴ AB = BC = AC

∴ Δ ABC is an equilateral triangle.

∴ All, A(1, 2), B(1, 6), C(1+ , 4) are vertices of an equilateral triangle,

**Practice set 5.2**

**Question 2.1. 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.**

A(−1, 7) and B(4, −3).

Let the coordinates of point P be (x, y).

Suppose coordinates of point A are (x_{1}, y_{1}) and B are (x_{2}, y_{2}) then

x_{1} = −1, y_{1} = 7, x_{2} = 4 and y_{2} = −3.

P divides AB in the ratio 2 : 3.

∴ m : n = 2 : 3

By section formula,

x = \(\frac{mx_2+nx_1}{m+n}\)

∴ x = \(\frac{2(4)+3(-1)}{2+3}\) = \(\frac{8-3}{5}\) = \(\frac{5}{5}\) = 1

y = \(\frac{my_2+ny_1}{m+n}\)

∴ y = \(\frac{2(-3)+3(7)}{2+3}\) = \(\frac{-6+21}{5}\) = \(\frac{15}{5}\) = 3

P(1, 3) divides the line segment joining the points A(−1, 7) and B(4, −3) in the ratio 2 : 3.

**Question 2.2. 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**

P(−3, 7) and Q(1, − 4).

Let coordinates of point A be (x, y).

Suppose coordinates of point P are (x_{1}, y_{1}) and Q are (x_{2}, y_{2})

then x_{1} = −3, y_{1} = 7, x_{2} = 1 and y_{2} = −4.

A divides PQ in the ratio a : b.

a : b = 2 : 1.

By section formula,

x = \(\frac{ax_2+bx_1}{a+b}\)

∴ x = \(\frac{2(1)+1(-3)}{2+1}\) = \(\frac{2-3}{3}\) = \(-\frac{1}{3}\)

y = \(\frac{ay_2+by_1}{a+b}\)

∴ x = \(\frac{2(-4)+1(7)}{2+1}\) = \(\frac{-8+7}{3}\) = \(-\frac{1}{3}\)

**∴ ****the co−ordinates of point ****A are \((-\frac{1}{3},\,-\frac{1}{3})\)**

**(2) P(−2, −5), Q(4, 3), ***a ***:** *b ***= 3 ****:**** 4**

P(−2, −5), Q(4, 3), *a *: *b *= 3 : 4

Let coordinates of point A be (x, y).

Suppose coordinates of point P are (x_{1}, y_{1}) and Q are (x_{2}, y_{2})

then x_{1} = −2, y_{1} = −5, x_{2} = 4 and y_{2} = 3.

A divides PQ in the ratio a : b.

a : b = 3 : 4.

By section formula,

x = \(\frac{ax_2+bx_1}{a+b}\)

∴ x = \(\frac{3(4)+4(-2)}{3+4}\) = \(\frac{12-8}{7}\) = \(\frac{4}{7}\)

y = \(\frac{ay_2+by_1}{a+b}\)

∴ y = \(\frac{3(3)+4(-5)}{3+4}\) = \(\frac{9-20}{7}\) = \(-\frac{11}{7}\)

**∴ ****the co−ordinates of point ****A are \((\frac{4}{7},\,-\frac{11}{7})\)**

**(3) P(2, 6), Q(−4, 1), ***a ***:** *b ***= 1 ****:**** 2**

P(2, 6), Q(−4, 1), *a *: *b *= 1 : 2

Let coordinates of point A be (x, y).

Suppose coordinates of point P are (x_{1}, y_{1}) and Q are (x_{2}, y_{2})

then x_{1} = 2, y_{1} = 6, x_{2} = −4 and y_{2} = 1.

A divides PQ in the ratio a : b.

a : b = 1 : 2.

By section formula,

x = \(\frac{ax_2+bx_1}{a+b}\)

∴ x = \(\frac{1(-4)+2(2)}{1+2}\) = \(\frac{-4+4}{3}\) = \(\frac{0}{3}\) = 0

y = \(\frac{ay_2+by_1}{a+b}\)

∴ y = \(\frac{1(1)+2(6)}{1+2}\) = \(\frac{1+12}{3}\) = \(\frac{13}{3}\)

**∴ ****the co−ordinates of point ****A are (0, \(\frac{13}{3}\))**

**Question 2.3. Find the ratio in which point ****T(−1, 6)****divides the line segment joining the points ****P(−3, 10) ****and ****Q(6, −8)****.**

P(−3, 10), Q(6, −8) and T(−1, 6).

Suppose the coordinates of point P are (x_{1}, y_{1}), the coordinates of point Q are (x_{2}, y_{2}) and the coordinates of point T are (x, y).

Here, x_{1} = −3, y_{1} = 10, x_{2} = 6, y_{2} = −8, x = −1 and y = 6

By section formula,

x = \(\frac{mx_2+nx_1}{m+n}\)

∴ −1 = \(\frac{m(6)+n(-3)}{m+n}\)

∴ −1(m + n) = 6m – 3n

∴ −m − n = 6m – 3n

∴ 7m = 2n

∴ \(\frac{m}{n}=\frac{2}{7}\)

∴ m : n = 2 : 7

**Answer is : T divides line segment PQ in the ratio 2 : 7.**

**Question 2.4. 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.**

A(2, −3) and P(−2, 0). Let B(x, y).

The centre of the circle is the midpoint of the diameter.

∴ by midpoint formula,

−2 = \(\frac{2+x}{2}\)

∴ −4 = 2 + x

∴ x = −6

and 0 = \(\frac{-3+y}{2}\)

∴ −3 + y = 0

∴ y = 3

**Answer is: The coordinates of point B are (−6, 3).**

**Question 2.5. Find the ratio in which point ****P(***k***, 7) ****divides the segment joining ****A(8, 9) ****and ****B(1, 2). ****Also find k **

**.**

A(8, 9), B(1, 2) and P(k, 7).

Let P divide the seg AB in the ratio m : n.

Here, x_{1} = 8, y_{1} = 9, x_{2} = 1, y_{2} = 2, x = k and y = 7

By section formula,

y = \(\frac{m(2)+n(9)}{m+n}\)

∴ 7 = \(\frac{2m+9n}{m+n}\)

∴ 7(m + n) = 2m + 9n

∴ 7m + 7n = 2m + 9n

∴ 7m − 2m = 9n − 7n

∴ 5m = 2n

∴ \(\frac{m}{n}=\frac{2}{5}\)

∴ m : n = 2 : 5

P divides segment AB in the ratio 2 : 5.

By section formula,

k = \(\frac{2(1)+5(8)}{2+5}\) = \(\frac{2+40}{7}\) = 6

**Answer is: The ratio in which point P divides seg AB is 2 : 5 and value of k is 6.**

**Question 2.6. Find the coordinates of midpoint of the segment joining the points ****(22, 20) ****and ****(0, 16).**

Let A(22, 20) and B(0, 16).

Let P(x, y) be its midpoint of seg AB.

If A(x_{1}, y_{1}) and B(x_{2}, y_{2})

then x_{1} = 22, y_{1} = 20, X_{2} = 0 and y_{2} = 16

By midpoint formula,

x = \(\frac{x_1+x_2}{2}\) and y **=** \(\frac{y_1+y_2}{2}\)

∴ x = \(\frac{22+0}{2}\)** **= **11** and y **=** \(\frac{20+16}{2}\) = **18**

**Answer is: Coordinates of midpoint of the segment joining the points (22, 20) and (0, 16) are (11, 18).**

**Question 2.7. Find the centroids of the triangles whose vertices are given below****.**

**(1)(−7, 6), (2, −2), (8, 5)**

Let A(−7, 6), B(2, −2) and C(8, 5).

Suppose the coordinates of point A are (x_{1}, y_{1}), coordinates of point B are (x_{2}, y_{2}) and coordinates of point C are (x_{3}, y_{3}).

∴ x_{1} = −7, y_{1} = 6, x_{2} = 2, y_{2} = −2, x_{3} = 8 and y_{3} = 5.

Let O(x, y) be the centroid.

By centroid formula,

x** = **\(\frac{x_1+x_2+x_3}{3}\) ** **and y **=** \(\frac{y_1+y_2+y_3}{3}\)

∴ x = \(\frac{-7+2+8}{3}\) = \(\frac{3}{3}\)** **= **1** and y **= **\(\frac{6+(-2)+5}{3}\) = \(\frac{9}{3}\)** **= **3**

**Answer is : The coordinates of the centroid are (1, 3).**

**(2) (3, −5), (4, 3), (11, −4)**

Let A(3, −5), B(4, 3) and C(11, −4).

Suppose the coordinates of point A are (x_{1}, y_{1}), coordinates of point B are (x_{2}, y_{2}) and coordinates of point C are (x_{3}, y_{3}).

∴ x_{1} = 3, y_{1} = −5, x_{2} = 4, y_{2} = 3, x_{3} = 11 and y_{3} = −4.

Let O(x, y) be the centroid.

By centroid formula,

x** = **\(\frac{x_1+x_2+x_3}{3}\) ** **and y **=** \(\frac{y_1+y_2+y_3}{3}\)

∴ x = \(\frac{3+4+11}{3}\) = \(\frac{18}{3}\)** **= **6** and y **=** \(\frac{-5+3+(-4)}{3}\) = \(\frac{-6}{3}\) ** **= **−2**

**Answer is : The coordinates of the centroid are (6, −2).**

**(3) (4, 7), (8, 4), (7, 11)**

Let A(4, 7), B(8, 4) and C(7, 11).

Suppose the coordinates of point A are (x_{1}, y_{1}), coordinates of point B are (x_{2}, y_{2}) and coordinates of point C are (x_{3}, y_{3}).

∴ x_{1} = 4, y_{1} = 7, x_{2} = 8, y_{2} = 4, x_{3} = 7 and y_{3} = 11.

Let O(x, y) be the centroid.

By centroid formula,

x** = **\(\frac{x_1+x_2+x_3}{3}\) ** **and y **=** \(\frac{y_1+y_2+y_3}{3}\)

∴ x = \(\frac{4+8+7}{3}\) = \(\frac{19}{3}\) and y **= ** \(\frac{7+4+11}{3}\) = \(\frac{22}{3}\)

**Answer is : The coordinates of the centroid are ( \(\frac{19}{3}\), \(\frac{22}{3}\)).**

**Question 2.8. In **Δ **ABC, G (−4, −7) ****is the centroid. If ****A (−14, −19) ****and ****B(3, 5) ****then find the co−ordinates of ****C.**

A(−14, −19), B(3,5) and G(−4, −7)

Suppose the coordinates of point A are (x_{1}, y_{1}), coordinates of point B are (x_{2}, y_{2}) coordinates of point C are (x_{3}, y_{3}) and coordinates of point G are (x, y).

Here, x_{1} = −14, y_{1} = −19, x_{2} = 3, y_{2} = 5, x = − 4, y = −7.

By centroid formula,

x** = ** \(\frac{x_1+x_2+x_3}{3}\)

∴ −4 = \(\frac{-14+3+x_3}{3}\)** **

∴ −4 × 3 = −11 + x_{3}

∴ −12 = −11 + x_{3}

∴ x_{3 }= −1

y **= **\(\frac{y_1+y_2+y_3}{3}\)

∴ −7 **= **\(\frac{-19+5+y_3}{3}\)

∴ −7 × 3 = −14 + y_{3}

∴ −21 + 14 = y_{3}

∴ y_{3} = −7

**Answer is : The coordinates of C are (−1, −7).**

**Question 2.9. ****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**

**.**

A(h, −6), B(2,3), C(−6, k) and G(1, 5).

Suppose the coordinates of point A are (x_{1}, y_{1}), coordinates of point B are (x_{2}, y_{2}) coordinates of point C are (x_{3}, y_{3}) and coordinates of point G are (x, y).

Here, x_{1} = h, y_{1} =− 6, x_{2} = 2, y_{2} = 3, x_{3} = −6, y_{3} = k, x = 1 and y = 5.

By centroid formula,

x** = **\(\frac{x_1+x_2+x_3}{3}\)

∴ 1 = \(\frac{h+2+(-6)}{3}\)** **** **

∴ 1 × 3 = h − 4

∴ 3 + 4 = h

∴ h = 7

y **=** \(\frac{y_1+y_2+y_3}{3}\)

∴ 5 **= **\(\frac{-6+3+k}{3}\)

∴ 5 × 3 = −3 + k

∴ 15 + 3 = k

∴ k = 18

**Answer is : h = 7 and k = 18**

**Question 2.10. Find the co−ordinates of the points of trisection of the line segment ****AB ****with ****A(2, 7) ****and ****B(−4, −8).**

A(2,7) and B(−4, −8)

Let P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) be the points of trisection.

∴ AP = PQ = QB ….(i)

∴ \(\frac{AP}{PB}\) = \(\frac{AP}{PQ+QB}\) ….(P—Q—B)

∴ \(\frac{AP}{PB}\) = \(\frac{AP}{AP+AP}\) ….[from (i)]

∴ \(\frac{AP}{PB}\) = \(\frac{AP}{2AP}\)

∴ AP : PB = 1 : 2

∴ P divides segment AB in the ratio 1 : 2.

By section formula for ratio m : n

x_{1 }= \(\frac{1(-4)+2(2)}{1+2}\) = \(\frac{-4+4}{3}\) = 0

and y_{1 }= \(\frac{1(-8)+2(7)}{1+2}\) = \(\frac{-8+14}{3}\) = \(\frac{6}{3}\) = 2

∴ x_{1} = 0 and y_{1} = 2

∴ the coordinates of point P are (0, 2).

AP = PQ ….[from (i)]

∴ P(x_{1}, y_{1}) is the midpoint of AQ.

By midpoint formula,

x_{1 }= \(\frac{x_2+2}{2}\) and y_{1 }= \(\frac{y_2+7}{2}\)

∴ 0 = \(\frac{x_2+2}{2}\) and 2 = \(\frac{y_2+7}{2}\)

∴ x_{2 }= −2 and y_{2} = −3

∴ the coordinates of point Q are (−2, −3).

**Answer is: The coordinates of the points of trisection of the line segment AB are (0, 2) and (−2. −3).**

**Question 2.11. If ****A(−14, −10), B(6, −2) ****is given, find the coordinates of the points which divide segment ****AB ****into four equal parts****.**

A(−14, −10) and B(6, −2)

Let P, Q and R be the points that divide segment AB into four equal parts.

Let P(x_{1}, y_{1}), Q(x_{2}, y_{2}) and R(x_{3}, y_{3}).

Q is the midpoint of seg AB.

∴ by midpoint formula,

x_{2 }= \(\frac{-14+6}{2}\) = \(\frac{-8}{2}\) = −4

y_{2 }= \(\frac{-10+(-6)}{2}\) = \(\frac{-12}{2}\) = −6

The coordinates of point Q are (−4, −6).

P is the midpoint of seg AQ.

∴ by midpoint formula,

x_{1 }= \(\frac{-14+(-4)}{2}\) = \(\frac{-18}{2}\) = −9

y_{1 }= \(\frac{-10+(-6)}{2}\) = \(\frac{-16}{2}\) = −8

∴ the coordinates of point P are (−9, −8).

R is the midpoint of seg QB.

∴ by midpoint formula,

x_{3 }= \(\frac{-4+6)}{2}\) = \(\frac{2}{2}\) = 1

y_{3 }= \(\frac{-6+(-2)}{2}\) = \(\frac{-8}{2}\) = −4

∴ the coordinates of point R are (1, −4).

**Answer is: The coordinates of the points which divide segment AB into four equal parts are (−9, −8), (−4, −6) and (1, − 4).**

**Question 2.12. If ****A(20, 10), B(0, 20) ****are given, find the coordinates of the points which divide segment ****AB ****into five congruent parts****.**

A(20, 10) and B(0, 20).

Let points P(x1, y1), Q(x2, y2), R(X3, y3) and S(x4, y4)

divide seg AB into five congruent parts.

∴ AP = PQ = QR = RS = SB ... (1)

∴ \(\frac{AP}{PB}\) = \(\frac{AP}{PQ+QR+RS+SB}\) ….(P—Q—R, Q—R—S, R—S—B,)

∴ \(\frac{AP}{PB}\) = \(\frac{AP}{4AP}\)

∴ AP : PB = 1 : 4

∴ P divides AB in the ratio 1 : 4.

By section formula,

∴ x_{1} = \(\frac{(1)(0)+4(20)}{1+4}\) = \(\frac{80}{5}\) = 16

and y_{1} = \(\frac{(1)(20)+4(10)}{1+4}\) = \(\frac{60}{5}\) = 12

∴ the coordinates of point P are (16, 12).

P is the midpoint of seg AQ,

∴ by midpoint formula,

16 = \(\frac{20+x_2}{2}\)
∴ 16 × 2 = 20 + x ∴ 32 – 20 = x ∴ x |
12 = \(\frac{10+y_2}{2}\)
∴ 12 × 2 = 10 + y ∴ 24 – 10 = y ∴ y |

∴ the coordinates of point Q are (12, 14).

R is the midpoint of seg PB.

∴ by midpoint formula,

x_{3} = \(\frac{16+0}{2}\)
∴ x |
y_{3} = \(\frac{12+20}{2}\)
∴ y |

∴ the coordinates of point R are (8, 16).

S is the midpoint of seg RB.

∴ by midpoint formula,

x_{4} = \(\frac{8+0}{2}\) = 4 |
y_{4} = \(\frac{16+20}{2}\) = 18 |

∴ the coordinates of point S are (4, 18).

**Answer is: (16, 12), (12, 14), (8, 16) and (4, 18) are the coordinates of the points which divide segment AB into five congruent parts.**

**Practice set 5.3**

**Question 3.1. Angles made by the line with the positive direction of ****X−****axis are given. Find the slope of these lines.**

**(1) 45****° **

Angle made by the line with the positive direction of X−axis (θ) = 45°

Slope of the line (m) = tan θ = tan 45° = 1

Answer is: Slope of the given line is 1.

**(2) 60****° **

Angle made by the line with the positive direction of X−axis (θ) = 60°

Slope of the line (m) = tan θ = tan 60° =

Answer is: Slope of the given line is .

**(3) 90****°**

Angle made by the line with the positive direction of X−axis (θ) = 90°

Slope of the line (m) = tan θ = tan 90°

tan 90° is not defined

Answer is: Slope of the given line is not defined

**Question 3.2. Find the slopes of the lines passing through the given points.**

**(1) A(2, 3) ****, ****B(4, 7) **

A(2, 3) and B(4, 7)

Let A (x_{1}, y_{1}) and B(x_{2}, y_{2})

then x_{1} = 2, y_{1} = 3, x_{2} = 4, y_{2} = 7

Slope of line AB = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{7-3}{4-2}\) = \(\frac{4}{2}\) = 2

**Answer is: Slope of line AB is 2.**

**(2) P(−3, 1) , Q(5, −2)**

P(−3, 1) and Q(5, −2)

Let P(x_{1}, y_{1}) and Q(x_{2}, y_{2})

then x_{1} = −3, y_{1} = 1, x_{2} = 5, y_{2} = −2

Slope of line PQ = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{-2-(-1)}{5-(-3)}\) = \(\frac{-3}{8}\)

**Answer is: Slope of line PQ is \(\frac{-3}{8}\)** **.**

**(3) C(5, −2) , D(7, 3)**

C(5, −2) , D(7, 3)

Let C(x_{1}, y_{1}) and D(x_{2}, y_{2})

then x_{1} = 5, y_{1} = −2, x_{2} = 7, y_{2} = 3

Slope of line CD = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{-3-(-2)}{7-5}\) = \(\frac{5}{2}\)

**Answer is: Slope of line CD is \(\frac{5}{2}\)** **.**

**(4) L(−2, −3), M(−6, −8)**

L(−2, −3), M(−6, −8)

Let L(x_{1}, y_{1}) and M(x_{2}, y_{2})

then x_{1} = −2, y_{1} = −3, x_{2} = −6, y_{2} = −8

Slope of line LM = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{-8-(-3)}{-6-(-2)}\) = \(\frac{-5}{-4}\) = \(\frac{5}{4}\)

**Answer is: Slope of line LM is \(\frac{5}{4}\)** **.**

**(5) E(−4, −2) , F(6, 3)**

E(−4, −2) , F(6, 3)

Let E(x_{1}, y_{1}) and F(x_{2}, y_{2})

then x_{1} = −4, y_{1} = −2, x_{2} = 6, y_{2} = 3

Slope of line EF = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{3-(-2)}{6-(-4)}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\)

**Answer is: Slope of line EF is \(\frac{1}{2}\)** **.**

**(6) T(0, −3), S(0, 4)**

T(0, −3), S(0, 4)

Let T(x_{1}, y_{1}) and S(x_{2}, y_{2})

then x_{1} = 0, y_{1} = −3, x_{2} = 0, y_{2} = 4

Slope of line TS = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{4-(-3)}{0-0}\) = \(\frac{7}{0}\) which is not defined

**Answer is: Slope of line TS cannot**** determined.**

**Question 3.3. Determine whether the following points are collinear.**

**(1) A(−1, −1), B(0, 1), C(1, 3) **

A(−1, −1), B(0, 1), C(1, 3)

Slope of line AB = \(\frac{1-(-1)}{0-(-1)}\) = \(\frac{2}{1}\) = 2

Slope of line BC = \(\frac{3-1}{1-0}\) = \(\frac{2}{1}\) = 2

Slope of line AB = slope of line BC and B is the common point.

∴ points A, B and C are collinear.

**Answer is: Points A(−1, −1), B(0, 1) and C(1, 3) are collinear.**

**(2) D(−2, −3), E(1, 0), F(2, 1)**

D(−2, −3), E(1, 0), F(2, 1)

Slope of line DE = \(\frac{0-(-3)}{1-(-2)}\) = \(\frac{3}{3}\) = 1

Slope of line EF = \(\frac{1-0}{2-1}\) = \(\frac{1}{1}\) = 1

Slope of line DE = slope of line EF and E is the common point.

∴ points D, E and F are collinear.

**Answer is: Points D(−2, −3), E(1, 0) and F(2, 1) are collinear.**

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

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

Slope of line LM = \(\frac{3-5}{3-2}\) = \(\frac{-2}{1}\) = −2

Slope of line MN = \(\frac{1-3}{5-3}\) = \(\frac{-2}{2}\) = −1

Slope of line LM ≠ slope of line MN

∴ points L, M and N are not collinear.

**Answer is: Points L(2, 5), M(3, 3) and N(5, 1) are not collinear.**

**(4) P(2, −5), Q(1, −3), R(−2, 3)**

P(2, −5), Q(1, −3), R(−2, 3)

Slope of line PQ = \(\frac{-3-(-5)}{1-2}\) = \(\frac{2}{-1}\) = −2

Slope of line QR = \(\frac{3-(-3)}{-2-1}\) = \(\frac{6}{-3}\) = −2

Slope of line PQ = slope of line QR and Q is the common point.

∴ points P, Q and R are collinear.

**Answer is: Points P(2, −5), Q(1, −3) and R(−2, 3) are collinear.**

**(5) R(1, −4), S(−2, 2), T(−3, 4)**

R(1, −4), S(−2, 2), T(−3, 4)

Slope of line RS = \(\frac{2-(-4)}{-2-1}\) = \(\frac{6}{-3}\) = −2

Slope of line ST = \(\frac{4-2}{-3-(-2)}\) = \(\frac{2}{-1}\) = −2

Slope of line RS = slope of line ST and S is the common point.

∴ points R, S and T are collinear.

**Answer is: Points R(1, −4), S(−2, 2) and T( −3, 4) are collinear.**

**(6) A(−4, 4), K(−2, \(\frac{5}{2}\)****), N(4, −2)**

A(−4, 4), K(−2, \(\frac{5}{2}\)), N(4, −2)

Slope of line AK = \(\frac{\frac{5}{2}-4}{-2-(-4)}\) = \(\frac{-3/2}{2}\) = \(-\frac{3}{4}\)

Slope of line KN = \(\frac{-2-\frac{5}{2}}{4-(-2)}\) = \(\frac{-9/2}{6}\) = \(-\frac{-9}{12}\) = \(-\frac{3}{4}\)

Slope of line AK = slope of line KN and K is the common point.

∴ points A, K and N are collinear.

**Answer is: Points A(−4, 4), K(−2, ** ** ****)**** and N(4, −2) are collinear.**

**Question 3.4. 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)

Slope of side AB = \(\frac{4-(-1)}{0-1}\) = \(\frac{5}{-1}\) = −5

Slope of side BC = \(\frac{3-4}{-5-0}\) = \(\frac{-1}{-5}\) = \(\frac{1}{5}\)

Slope of side AC = \(\frac{3-(-1)}{-5-1}\) = \(\frac{4}{-6}\) = \(-\frac{1}{3}\)

**Answer is : Slope of side AB is −5, slope of side BC is \(\frac{1}{5}\) ** **, ****slope of side AC is \(-\frac{1}{3}\) **

**Question 3.5. Show that ****A (−4, −7), B (−1, 2), C (8, 5) ****and ****D (5, −4) ****are the vertices of a ****parallelogram.**

A(−4, −7), B(−1, 2), C(8, 5) and D(5, −4)

Slope of side AB = \(\frac{2-(-7)}{-1-(-4)}\) = \(\frac{9}{3}\) = 3 ……(i)

Slope of side BC = \(\frac{5-2}{8-(-1)}\) = \(\frac{3}{9}\) = \(\frac{1}{3}\) ……(ii)

Slope of side CD = \(\frac{-4-5}{5-8}\) = \(\frac{-9}{-3}\) = 3 ……(iii)

Slope of side AD = \(\frac{-4-(-7)}{5-(-4)}\) = \(\frac{3}{9}\) = \(\frac{1}{3}\) ……(iv)

From (i) and (iii),

Slope of line AB = slope of line CD

∴ line AB || line CD ….(If two lines have equal slopes then they are parallel)

From (ii) and (iv),

Slope of line BC = slope of line AD

∴ line BC || line AD ….(If two lines have equal slopes then they are parallel)

∴ □ ABCD is a parallelogram. A(−4, −7), B(−1, 2), C(8, 5) and D(5, −4) are the vertices of a parallelogram.

**Question 3.6. Find k, if **

**R(1, −1), S (−2,**

*k***)**

**and slope of line**

**RS**

**is**

**−2.**

R(1, −1) and S(−2, k)

Slope of line RS = −2

Slope of side RS = \(\frac{k-(-1)}{-2-1}\)

∴ -2 = \(\frac{k+1}{-3}\)

∴ -2 × -3 = k + 1

∴ 6 = k + 1

∴ k = 6 – 1 = 5

**Answer is: The value of k is 5**

**Question 3.7. Find k**

**,**

**if**

**B(**

*k***, −5), C (1, 2)**

**and slope of the line is**

**7.**

B(*k*, −5), C (1, 2)

Slope of the line BC = 7

Slope of the line BC = \(\frac{2-(-5)}{1-k}\)

∴ 7 = \(\frac{7}{1-k}\)

∴ 7(1 – k) = 7

∴ 1 − k = 1

∴ k = 1 – 1 = 0

**Answer is: The value of k is 0**

**Question 3.8. Find k, if PQ || RS and P(2, 4), Q (3, 6), R(3, 1), S(5, k) .**

**Given : **Line PQ || line RS.

P(2, 4), Q (3, 6), R(3, 1), S(5, *k*)

Parallel lines have equal slopes,

∴ slope of line PQ = slope of line RS

∴ \(\frac{6-4}{3-2}\) = \(\frac{k-1}{5-3}\)

∴ \(\frac{2}{1}\) = \(\frac{k-1}{2}\)

∴ 4 = k – 1

∴ k = 4 + 1 = 5

**Answer is: The value of k is 5**

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