## Maharashtra State Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3

Practice Set 5.3 Geometry Class 10 Question 1. Angles made by the line with the positive direction of X-axis are given. Find the slope of these lines.

i. 45°

ii. 60°

iii. 90°

Solution:

i. Angle made with the positive direction of

X-axis (θ) = 45°

Slope of the line (m) = tan θ

∴ m = tan 45° = 1

∴ The slope of the line is 1.

ii. Angle made with the positive direction of X-axis (θ) = 60°

Slope of the line (m) = tan θ

∴ m = tan 60° = \(\sqrt { 3 }\)

∴ The slope of the line is \(\sqrt { 3 }\).

iii. Angle made with the positive direction of

X-axis (θ) = 90°

Slope of the line (m) = tan θ

∴ m = tan 90°

But, the value of tan 90° is not defined.

∴ The slope of the line cannot be determined.

Practice Set 5.3 Geometry Question 2. Find the slopes of the lines passing through the given points.

i. A (2, 3), B (4, 7)

ii. P(-3, 1), Q (5, -2)

iii. C (5, -2), D (7, 3)

iv. L (-2, -3), M (-6, -8)

v. E (-4, -2), F (6, 3)

vi. T (0, -3), s (0,4)

Solution:

i. A (x_{1}, y_{1}) = A (2, 3) and B (x_{2}, y_{2}) = B (4, 7)

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

∴ The slope of line AB is 2.

ii. P (x_{1}, y_{1}) = P (-3, 1) and Q (x_{2}, y_{2}) = Q (5, -2)

Here, x_{1} = -3, x_{2} = 5, y_{1} = 1, y_{2} = -2

∴ The slope of line PQ is \(\frac { -3 }{ 8 } \)

iii. C (x_{1}, y_{1}) = C (5, -2) and D (x_{2}, y_{2}) = D (7, 3)

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

∴ The slope of line CD is \(\frac { 5 }{ 2 } \)

iv. L (x_{1}, y_{1}) = L (-2, -3) and M (x_{2},y_{2}) = M (-6, -8)

Here, x_{1} = -2, x_{2} = – 6, y_{1} = – 3, y_{2} = – 8

∴ The slope of line LM is \(\frac { 5 }{ 4 } \)

v. E (x_{1}, y_{1}) = E (-4, -2) and F (x_{2}, y_{2}) = F (6, 3)

Here,x_{1} = -4, x_{2} = 6, y_{1} = -2, y_{2} = 3

∴ The slope of line EF is \(\frac { 1 }{ 2 } \).

vi. T (x_{1}, y_{1}) = T (0, -3) and S (x_{2}, y_{2}) = S (0, 4)

Here, x_{1} = 0, x_{2} = 0, y_{1} = -3, y_{2} = 4

∴ The slope of line TS cannot be determined.

5.3.5 Practice Question 3. Determine whether the following points are collinear.

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

ii. D (- 2, -3), E (1, 0), F (2, 1)

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

iv. P (2, -5), Q (1, -3), R (-2, 3)

v. R (1, -4), S (-2, 2), T (-3,4)

vi. A(-4,4),K[-2,\(\frac { 5 }{ 2 } \)], N (4,-2)

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 DE = slope of line EF

∴ line DE || line EF

Also, point E is common to both the lines.

∴ Both lines are the same.

∴ Points D, E and F are collinear.

∴ slope of line LM ≠ slope of line MN

∴ Points L, M and N are not 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 RS = slope of line ST

∴ line RS || line ST

Also, point S is common to both the lines.

∴ Both lines are the same.

∴ Points R, S and T are collinear.

∴ slope of line AK = slope of line KN

∴ line AK || line KN

Also, point K is common to both the lines.

∴ Both lines are the same.

∴ Points A, K and N are collinear.

Practice Set 5.3 Geometry 9th Standard Question 4. If A (1, -1), B (0,4), C (-5,3) are vertices of a triangle, then find the slope of each side.

Solution:

∴ The slopes of the sides AB, BC and AC are -5, \(\frac { 1 }{ 5 } \) and \(\frac { -2 }{ 3 } \) respectively.

Geometry 5.3 Question 5. Show that A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are the vertices of a parallelogram.

Proof:

∴ Slope of side AB = Slope of side CD … [From (i) and (iii)]

∴ side AB || side CD

Slope of side BC = Slope of side AD … [From (ii) and (iv)]

∴ side BC || side AD

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

꠸ABCD is a parallelogram.

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

Question 6.

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

Solution:

R(x_{1}, y_{1}) = R (1, -1), S (x_{2}, y_{2}) = S (-2, k)

Here, x_{1} = 1, x_{2} = -2, y_{1} = -1, y_{2} = k

But, slope of line RS is -2. … [Given]

∴ -2 = \(\frac { k+1 }{ -3 } \)

∴ k + 1 = 6

∴ k = 6 – 1

∴ k = 5

5.3 Class 10 Question 7. Find k, if B (k, -5), C (1, 2) and slope of the line is 7.

Solution:

B(x_{1}, y_{1}) = B (k, -5), C (x_{2}, y_{2}) = C (1, 2)

Here, x_{1} = k, x_{2} = 1, y_{1} = -5, y_{2} = 2

But, slope of line BC is 7. …[Given]

∴ 7 = \(\frac { 7 }{ 1-k } \)

∴ 7(1 – k) = 7

∴ 1 – k = \(\frac { 7 }{ 7 } \)

∴ 1 – k = 1

∴ k = 0

Question 8.

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

Solution:

But, line PQ || line RS … [Given]

∴ Slope of line PQ = Slope of line RS

∴ 2 = \(\frac { k-1 }{ 2 } \)

∴ 4 = k – 1

∴ k = 4 + 1

∴ k = 5