## Maharashtra State Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Problem Set 5

Question 1.

Fill in the blanks using correct alternatives.

i. Seg AB is parallel to Y-axis and co-ordinates 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)

Answer: (D)

Since, seg AB || Y-axis.

∴ x co-ordinate of all points on seg AB

will be the same,

x co-ordinate of A (1, 3) = 1

x co-ordinate of B (1, – 3) = 1

∴ Option (D) is correct.

ii. 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)

Answer: (D)

iii. Distance of point (-3, 4) from the origin is _________.

(A) 7

(B) 1

(C) 5

(D) -5

Answer: (C)

Distance of (-3, 4) from origin

\(\begin{array}{l}{=\sqrt{(-3)^{2}+(4)^{2}}} \\ {=\sqrt{9+16}} \\ {=\sqrt{25}=5}\end{array}\)

iv. 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) \(\sqrt { 3 }\)

Answer: (C)

Question 2.

Determine whether the given points are collinear.

i. A (0, 2), B (1, -0.5), C (2, -3)

ii. P(1,2), Q(2,\(\frac { 8 }{ 5 } \)),R(3,\(\frac { 6 }{ 5 } \))

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

Solution:

∴ slope of line AB = slope of line BC

∴ line AB || line BC

Also, point B is common to both the lines.

∴ Both lines are the same.

∴ Points A, B and C are collinear.

∴ slope of line PQ = slope of line QR

∴ line PQ || line QR

Also, point Q is common to both the lines.

∴ Both lines are the same.

∴ Points P, Q and R are collinear.

∴ slope of line LM ≠ slope of line MN

∴ Points L, M and N are not collinear.

[Note: Students can solve the above problems by using distance formula.]

Question 3.

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

Solution:

P(x_{1},y_{1}) = P (0, 6), Q(x_{2}, y_{2}) = Q (12, 20)

Here, x_{1} = 0, y_{1} = 6, x_{2} = 12, y_{2} = 20

∴ Co-ordinates of the midpoint of seg PQ

∴ The co-ordinates of the midpoint of seg PQ are (6,13).

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.

Solution:

Let C be a point on Y-axis which divides seg AB in the ratio m : n.

Point C lies on the Y-axis

∴ its x co-ordinate is 0.

Let C = (0, y)

Here A (x_{1},y_{1}) = A(3, 8)

B (x_{2}, y_{2}) = B (-9, 3)

∴ By section formula,

∴ Y-axis divides the seg AB in the ratio 1 : 3.

Question 5.

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

Solution:

Let point R be on the X-axis which is equidistant from points P and Q.

Point R lies on X-axis.

∴ its y co-ordinate is 0.

Let R = (x, 0)

R is equidistant from points P and Q.

∴ PR = QR

∴ (x – 2)^{2} + [0 – (-5)]^{2} = [x – (- 2)]^{2} + (0 – 9)^{2} …[Squaring both sides]

∴ (x – 2)^{2} + (5)^{2} = (x + 2)^{2} + (-9)^{2}

∴ 4 – 4x + x^{2} + 25 = 4 + 4x + x^{2} + 81

∴ – 8x = 56

∴ x = -7

∴ The point on X-axis which is equidistant from points P and Q is (-7,0).

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)

Solution:

i. Let A (x_{1}, y_{1}) and B (x_{2}, y_{2}) be the given points.

∴ x_{1} = a, y_{1} = 0, x_{2} = 0, y_{2} = a

By distance formula,

∴ d(A, B) = a\(\sqrt { 2 }\) units

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

∴ x_{1} = -6, y_{1} = -3, x_{2} = -1, y_{2} = 9

By distance formula,

∴ d(P, Q) = 13 units

iii. Let R (x_{1}, y_{1}) and S (x_{2}, y_{2}) be the given points.

∴ x_{1} = -3a, y_{1} = a, x_{2} = a, y_{2} = -2a

By distance formula,

∴ d(R, S) = 5a units

Question 7.

Find the co-ordinates of the circumcentre of a triangle whose vertices are (-3,1), (0, -2) and (1,3).

Solution:

Let A (-3, 1), B (0, -2) and C (1, 3) be the vertices of the triangle.

Suppose O (h, k) is the circumcentre of ∆ABC.

∴ (h + 3)^{2} + (k – 1)^{2} = h^{2} + (k + 2)^{2}

∴ h^{2} + 6h + 9 + k^{2} – 2k + 1 = h^{2} + k^{2} + 4k + 4

∴ 6h – 2k + 10 = 4k + 4

∴ 6h – 2k – 4k = 4 – 10

∴ 6h – 6k = – 6

∴ h – k = -1 ,..(i)[Dividing both sides by 6]

OB = OC …[Radii of the same circle]

∴ h^{2} + (k + 2)^{2} = (h – 1)^{2} + (k – 3)^{2}

∴ h^{2} + k^{2} + 4k + 4 = h^{2} – 2h + 1 + k^{2} – 6k + 9

∴ 4k + 4 = -2h + 1 – 6k + 9

∴ 2h+ 10k = 6

∴ h + 5k = 3 …(ii)

Subtracting equation (ii) from (i), we get

∴ The co-ordinates of the circumcentre of the triangle are (\(\frac { -1 }{ 3 } \),\(\frac { 2 }{ 3 } \))

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.

i. L (6, 4), M (-5, -3), N (-6, 8)

ii. P (-2, -6), Q (-4, -2), R (-5, 0)

iii. A(\(\sqrt { 2 }\),\(\sqrt { 2 }\)),B(-\(\sqrt { 2 }\),-\(\sqrt { 2 }\)),C(\(\sqrt { 6 }\),\(\sqrt { 6 }\))

Solution:

i. By distance formula,

∴ d(M, N) + d (L, N) > d (L, M)

∴ Points L, M, N are non collinear points.

We can construct a triangle through 3 non collinear points.

∴ The segment joining the given points form a triangle.

Since MN ≠ LN ≠ LM

∴ ∆LMN is a scalene triangle.

∴ The segments joining the points L, M and N will form a scalene triangle.

ii. By distance formula,

∴ d(P, Q) + d(Q, R) = d (P, R) …[From (iii)]

∴ Points P, Q, R are collinear points.

We cannot construct a triangle through 3 collinear points.

∴ The segments joining the points P, Q and R will not form a triangle.

iii. By distance formula,

∴ d(A, B) + d(B, C) + d(A, C) … [From (iii)]

∴ Points A, B, C are non collinear points.

We can construct a triangle through 3 non collinear points.

∴ The segment joining the given points form a triangle.

Since, AB = BC = AC

∴ ∆ABC is an equilateral triangle.

∴ The segments joining the points A, B and C will form an equilateral triangle.

Question 9.

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

Solution:

P(x_{1},y_{1}) = P(-12,-3),

Q(X_{2},T_{2}) = Q(4, k)

Here, x_{1} = -12, x_{2} = 4, y_{1} = -3, y_{2} = k

But, slope of line PQ (m) is \(\frac { 1 }{ 2 } \) ….[Given]

∴ \(\frac { 1 }{ 2 } \) = \(\frac { k+3 }{ 16 } \)

∴ \(\frac { 16 }{ 2 } \) = k + 3

∴ 8 = k + 3

∴ k = 5

The value of k is 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).

Proof:

∴ Slope of line AB = Slope of line CD

Parallel lines have equal slope.

∴ line AB || line CD

Question 11.

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

Proof:

By distance formula,

In ꠸PQRS,

PQ = RS … [From (i) and (iii)]

QR = PS … [From (ii) and (iv)]

∴ ꠸ PQRS is a parallelogram.

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

∴ Points P, Q, R and S are the vertices of 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.

Proof:

By distance formula,

In ꠸PQRS,

PQ = RS …[From (i) and (iii)]

QR = PS …[From (ii) and (iv)]

꠸PQRS is a parallelogram.

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

In parallelogram PQRS,

PR = QS … [From (v) and (vi)]

∴ ꠸PQRS is a rectangle.

[A parallelogram is a rectangle if its diagonals are equal]

Question 13.

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

Solution:

Suppose AD, BE and CF are the medians.

∴ Points D, E and F are the midpoints of sides BC, AC and AB respectively.

∴ By midpoint formula,

∴ The lengths of the medians of the triangle 5 units, 2\(\sqrt { 13 }\) units and \(\sqrt { 37 }\) units.

Question 14.

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

Solution:

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

D (-7, 6), E (8, 5) and F (2, -2) are the midpoints of sides BC, AC and AB respectively.

Let G be the centroid of ∆ABC.

D is the midpoint of seg BC.

By midpoint formula,

E is the midpoint of seg AC.

By midpoint formula,

Adding (i), (iii) and (v),

x_{2} + x_{3} + x_{1} + x_{3} + x_{1} + x_{2} = -14 + 16 + 4

∴ 2x_{1} + 2x_{2} + 2x_{3} = 6

∴ x_{1} + x_{2} + x_{3} = 3 …(vii)

Adding (ii), (iv) and (vi),

y_{2} + y_{3} + y_{1} + y_{3} + y_{1} +y_{2} = 12 + 10 – 4

∴ 2y_{1} + 2y_{2} + 2y_{3} = 18

∴ y_{1} + y_{2} + y_{3} = 9 …(viii)

G is the centroid of ∆ABC.

By centroid formula,

∴ The co-ordinates of the centroid of the triangle are (1,3).

Question 15.

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

Proof:

By distance formula,

∴ □ABCD is a square.

[A rhombus is a square if its diagonals are equal]

Question 16.

Find the co-ordinates of circumcentre and radius of circumcircle of AABC if A (7, 1), B (3,5) and C (2,0) are given.

Solution:

Suppose, O (h, k) is the circumcentre of ∆ABC

∴ h^{2} – 6h + 9 + k^{2} – 10k + 25 = h^{2} – 4h + 4 + k^{2}

∴ 2h + 10k = 30

∴ h + 5k = 15 … (ii)[Dividing both sides by 2]

Multiplying equation (i) by 5, we get

25h + 5k = 115 …(iii)

Subtracting equation (ii) from (iii), we get

Substituting the value of h in equation (i), we get

∴ The co-ordinates of the circumcentre of the triangle are (\(\frac { 25 }{ 6 } \),\(\frac { 13 }{ 6 } \)) and radius of circumcircle is \(\frac{13 \sqrt{2}}{6}\) units.

Question 17.

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

Solution:

Suppose point C divides seg AB in the ratio 3:1.

Here; A(x_{1}, y_{1}) = A (4, -3)

B (x_{2}, y_{2}) = B (8, 5)

By section formula,

∴ The co-ordinates of point dividing seg AB in ratio 3 : 1 are (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.

Solution:

Slope of AB = slope of CD

∴ line AB || line CD

slope of BC = slope of AD

∴ line BC || line AD

Both the pairs of opposite sides of ∆ABCD are parallel.

∴ ꠸ ABCD is a parallelogram.

∴ The quadrilateral formed by joining the points A, B, C and D is 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 co-ordinates of A, P, R, B.

Solution:

Points P, Q, R and S divide seg AB in five congruent parts.

Let A (x_{1}, y_{1}), B (x_{2}, y_{2}), P (x_{3}, y_{3}) and

R (x_{4}, y_{4}) be the given points.

Point R is the midpoint of seg QS.

By midpoint formula,

x co-ordinate of R = \(\frac { 12+4 }{ 2 } \) = \(\frac { 16 }{ 2 } \) = 8

y co-ordinate of R = \(\frac { 14+18 }{ 2 } \) = \(\frac { 32 }{ 2 } \) = 16

∴ co-ordinates of R are (8, 16).

Point Q is the midpoint of seg PR.

By midpoint formula,

∴ 28 = y_{3} + 16

∴ y_{3} = 12

∴ P(x_{3},y_{3}) = (16, 12)

∴ co-ordinates of P are (16, 12).

Point P is the midpoint of seg AQ.

By midpoint formula,

∴ co-ordinates of A are (20, 10).

Point S is the midpoint of seg RB.

By midpoint formula,

∴ 36 = y_{2} + 16

∴ y_{2} = 20

∴ B(x_{2}, y_{2}) = (0, 20)

∴ co-ordinates of B are (0, 20).

∴ The co-ordinates of points A, P, R and B are (20, 10), (16, 12), (8, 16) and (0, 20) respectively.

Question 20.

Find the co-ordinates of the centre of the circle passing through the points P (6, -6), Q (3, -7) and R (3,3).

Solution:

Suppose O (h, k) is the centre of the circle passing through the points P, Q and R.

∴ (h – 6)^{2} + (k + 6)^{2} = (h – 3)^{2} + (k + 7)^{2}

∴ h^{2} – 12h + 36 + k^{2} + 12k + 36

= h^{2} – 6h + 9 + k^{2} + 14k + 49

∴ 6h + 2k = 14

∴ 3h + k = 7 …(i)[Dividing both sides by 2]

OP = OR …[Radii of the same circle]

∴ (h – 6)^{2} + (k + 6)^{2} = (h – 3)^{2} + (k – 3)^{2}

∴ h^{2} – 12h + 36 + k^{2} + 12k + 36

= h^{2} – 6h + 9 + k^{2} – 6k + 9

∴ 6h – 18k = 54

∴ 3h – 9k = 27 …(ii)[Dividing both sides by 2]

Subtracting equation (ii) from (i), we get

Substituting the value of k in equation (i), we get

3h – 2 = 7

∴ 3h = 9

∴ h = \(\frac { 9 }{ 3 } \) = 3

∴ The co-ordinates of the centre of the circle are (3, -2).

Question 21.

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

Solution:

Let the points A (5, 6), B (1, -2) and C (3, -2) be the three vertices of a parallelogram.

The fourth vertex can be point D or point Di or point D_{2} as shown in the figure.

Let D(x_{1},y_{1}), D, (x_{2}, y_{2}) and D_{2} (x_{3},y_{3}).

Consider the parallelogram ACBD.

The diagonals of a parallelogram bisect each other.

∴ midpoint of DC = midpoint of AB

Co-ordinates of point D(x_{1}, y_{1}) are (3, 6).

Consider the parallelogram ABD_{1}C.

The diagonals of a parallelogram bisect each other.

∴ midpoint of AD_{1} = midpoint of BC

∴ Co-ordinates of D_{1}(x_{2},y_{2}) are (-1,-10).

Consider the parallelogram ABCD_{2}.

The diagonals of a parallelogram bisect each other.

∴ midpoint of BD_{2} = midpoint of AC

∴ co-ordinates of point D_{2} (x_{3}, y_{3}) are (7, 6).

∴ The possible pairs of co-ordinates of the fourth vertex D of the parallelogram are (3, 6), (-1,-10) 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).

Solution:

Suppose ABCD is the given quadrilateral.

∴ The slopes of the diagonals of the quadrilateral are 10 and 0.