**Chapter 5 Co-ordinate Geometry Set 5.3**

## Chapter 5 Co-ordinate Geometry Set 5.3

**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° = √3

∴ The slope of the line is √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.

**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 −38

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 52

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 54

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 12.

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.

**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[-2,52], 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.

**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:**

**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 = K+1/-3

∴ k + 1 = 6

∴ k = 6 – 1

∴ k = 5

**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 = 7/1-K

∴ 7(1 – k) = 7

∴ 1 – k = 77

∴ 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 = k-1/2

∴ 4 = k – 1

∴ k = 4 + 1

∴ k = 5