Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

## Practice Set 5.1 Geometry 10th Std Maths Part 2 Answers Chapter 5 Co-ordinate Geometry

Practice Set 5.1 Geometry Class 10 Question 1. Find the distance between each of the following pairs of points.

i. A (2, 3), B (4,1)

ii. P (-5, 7), Q (-1, 3)

iii. R (0, -3), S (0,\(\frac { 5 }{ 2 } \))

iv. L (5, -8), M (-7, -3)

v. T (-3, 6), R (9, -10)

vi. W(\(\frac { -7 }{ 2 } \),4), X(11, 4)

Solution:

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

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

By distance formula,

∴ d(A, B) = 2\(\sqrt { 2 }\) units

∴ The distance between the points A and B is 2\(\sqrt { 2 }\) units.

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

∴ x_{1} = -5, y_{1} = 7, x_{2} = -1, y_{2} = 3

By distance formula,

∴ d(P, Q) = 4\(\sqrt { 2 }\) units

∴ The distance between the points P and Q is 4\(\sqrt { 2 }\) units.

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

∴ x_{1} = 0, y_{1} = -3, x_{2} = 0, y_{2} = \(\frac { 5 }{ 2 } \)

By distance formula,

∴ d(R, S) = \(\frac { 11 }{ 2 } \) units

∴ The distance between the points R and S is \(\frac { 11 }{ 2 } \) units.

iv. Let L (x_{1}, y_{1}) and M (x_{2}, y_{2}) be the given points.

∴ x_{1} = 5, y_{1} = -8, x_{2} = -7, y_{2} = -3

By distance formula,

∴ d(L, M) = 13 units

∴ The distance between the points L and M is 13 units.

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

∴ x_{1} = -3, y_{1} = 6,x_{2} = 9,y_{2} = -10

By distance formula,

∴ d(T, R) = 20 units

∴ The distance between the points T and R 20 units.

vi. Let W (x_{1}, y_{1}) and X (x_{2}, y_{2}) be the given points.

∴ d(W, X) = \(\frac { 29 }{ 2 } \) units

∴ The distance between the points W and X is \(\frac { 29 }{ 2 } \) units.

Practice Set 5.1 Geometry 10th Question 2. Determine whether the points are collinear.

i. A (1, -3), B (2, -5), C (-4, 7)

ii. L (-2, 3), M (1, -3), N (5, 4)

iii. R (0, 3), D (2, 1), S (3, -1)

iv. P (-2, 3), Q (1, 2), R (4, 1)

Solution:

i. By distance formula,

∴ d(A, B) = \(\sqrt { 5 }\) …(i)

On adding (i) and (iii),

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

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

∴ Points A, B and C are collinear.

ii. By distance formula,

On adding (i) and (iii),

d(L, M) + d(L, N) = 3\(\sqrt { 5 }\) + 5\(\sqrt { 2 }\) ≠ \(\sqrt { 65 }\)

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

∴ Points L, M and N are not collinear.

iii. By distance formula,

On adding (i) and (ii),

∴ d(R, D) + d(D, S) = \(\sqrt { 8 }\) + \(\sqrt { 5 }\) ≠ 5

∴ d(R, D) + d(D, S) ≠ d(R, S) … [From (iii)]

∴ Points R, D and S are not collinear.

iv. By distance formula,

On adding (i) and (ii),

d(P, Q) + d(Q, R) = \(\sqrt { 10 }\) + \(\sqrt { 10 }\) = 2\(\sqrt { 10 }\)

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

∴ Points P, Q and R are collinear.

Coordinate Geometry Class 10 Practice Set 5.1 Question 3. Find the point on the X-axis which is equidistant from A (-3,4) and B (1, -4).

Solution:

Let point C be on the X-axis which is equidistant from points A and B.

Point C lies on X-axis.

∴ its y co-ordinate is 0.

Let C = (x, 0)

C is equidistant from points A and B.

∴ AC = BC

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

∴ x^{2} + 6x + 9 + 16 = x^{2} – 2x + 1 + 16

∴ 8x = – 8

∴ x = – \(\frac { 8 }{ 8 } \) = -1

∴ The point on X-axis which is equidistant from points A and B is (-1,0).

10th Geometry Practice Set 5.1 Question 4. Verify that points P (-2, 2), Q (2, 2) and R (2, 7) are vertices of a right angled triangle.

Solution:

Distance between two points

Consider, PQ^{2} + QR^{2} = 42 + 52 = 16 + 25 = 41 … [From (i) and (ii)]

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

∴ ∆PQR is a right angled triangle. … [Converse of Pythagoras theorem]

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

Proof:

Distance between two points

PQ = RS … [From (i) and (iii)]

QR = PS … [From (ii) and (iv)]

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

∴ □ PQRS is a parallelogram.

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

Question 6.

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

Proof:

Distance between two points

∴ AB = BC = CD = AD …[From (i), (ii), (iii) and (iv)]

In a quadrilateral, if all the sides are equal, then it is a rhombus.

∴ □ ABCD is a rhombus.

∴ Points A, B, C and D are the vertices of rhombus ABCD.

Practice Set 5.1 Question 7. Find x if distance between points L (x, 7) and M (1,15) is 10.

Solution:

X_{1} = x, y_{1} = 7, x_{2} = 1, y_{2} = 15

By distance formula,

∴ 1 – x = ± 6

∴ 1 – x = 6 or l – x = -6

∴ x = – 5 or x = 7

∴ The value of x is – 5 or 7.

Geometry 5.1 Question 8. Show that the points A (1, 2), B (1, 6), C (1 + 2\(\sqrt { 3 }\), 4) are vertices of an equilateral triangle.

Proof:

Distance between two points

∴ AB = BC = AC … [From (i), (ii) and (iii)]

∴ ∆ABC is an equilateral triangle.

∴ Points A, B and C are the vertices of an equilateral triangle.

**Maharashtra Board Class 10 Maths Chapter 5 Coordinate Geometry Intext Questions and Activities**

Question 1.

In the figure, seg AB || Y-axis and seg CB || X-axis. Co-ordinates of points A and C are given. To find AC, fill in the boxes given below. (Textbook pa. no. 102)

Solution:

In ∆ABC, ∠B = 900

∴ (AB)^{2} + (BC)^{2} = [(Ac)^{2} …(i) … [Pythagoras theorem]

seg CB || X-axis

∴ y co-ordinate of B = 2

seg BA || Y-axis

∴ x co-ordinate of B = 2

∴ co-ordinate of B is (2, 2) = (x_{1},y_{1})

co-ordinate of A is (2, 3) = (x_{2}, Y_{2})

Since, AB || to Y-axis,

d(A, B) = Y_{2} – Y_{1}

d(A,B) = 3 – 2 = 1

co-ordinate of C is (-2,2) = (x_{1},y_{1})

co-ordinate of B is (2, 2) = (x_{2}, y_{2})

Since, BC || to X-axis,

d(B, C) = x_{2} – x_{1}

d(B,C) = 2 – -2 = 4

∴ AC^{2} = 12 + 42 …[From (i)]

= 1 + 16 = 17

∴ AC = \(\sqrt { 17 }\) units …[Taking square root of both sides]