**Question 1.Find the slope of each of the lines which passes through the following points:i. A(2, -1), B(4,3)ii. C(- 2,3), D(5, 7)iii. E(2,3), F(2, – 1)iv. G(7,1), H(- 3,1)Solution:**

i. Here, A = (2, -1) andB = (4, 3)

Alternate Method:

Points E and F have same x co-ordinates i.e. 2.

Points E and F lie on a line parallel to Y-axis.

∴ The slope of EF is not defined.

iv. Here, G = (7, 1) and H = (-3, 1)

Alternate Method:

Points G and H have same y co-ordinate i.e. 1.

∴ Points G and H lie on a line parallel to the

X-axis.

∴ The slope of GH is 0.

**Question 2.If the x and x-intercepts of line L are 2 and 3 respectively, then find the slope of line L.Solution:**

Given, x-intercept of line L is 2 and y-intercept of line L is 3

∴ The line L intersects X-axis at (2, 0) and Y-axis at (0,3).

∴ The line L passes through (2, 0) and (0, 3).

**Question 3.Find the slope of the line whose inclination is 30°.Solution:**

Given, inclination (θ) = 30°

**Question 6.Without using Pythagoras theorem, show that points A (4, 4), B (3, 5) and C (- 1, – 1) are the vertices of a right-angled triangle.Solution:**

Slope of AB x slope of AC = – 1 x 1 = – 1

∴ side AB ⊥ side AC

∴ ∆ABC is a right angled triangle right angled at A.

∴ The given points are the vertices of a right angled triangle.

**Question 7.Find the slope of the line which makes angle of 45° with the positive direction of the Y-axis measured anticlockwise.Solution:**

Since the line makes an angle of 45° with positive direction of Y-axis in anticlockwise direction,

Inclination of the line (0) = (90° + 45°)

∴ Slope of the line = tan(90° + 45°)

= – cot 45°

= -1

**Question 8.Find the value of k for which the points P(k, -1), Q(2,1) and R(4,5) are collinear.Solution:**

Given, points P(k, – 1), Q (2, 1) and R(4, 5) are collinear.

∴ Slope of PQ = Slope of QR .

∴ 4 = 4 (2 – k)

∴ 1 = 2 – k

∴ k = 2 – 1 = 1

**Question 9.Find the acute angle between the X-axis and the line joining the points A(3, -1) and B(4, – 2).Solution:**

Given, A (3, – 1) and B (4, – 2)

But, slope of a line = tan θ

∴ tan θ = – 1

= – tan 45°

= tan (180° -45°)

… [∵ tan (180° – θ) = -tan θ]

= tan 135°

∴ θ = 135°

Let α be the acute angle that line AB makes with X-axis.

Then, α + 0 = 180°

α = 180°- 135° = 45°

∴ The acute angle between the X-axis and the line joining the points A and B is 45°.

**Question 10.A line passes through points A(xi, y0 and B(h, k). If the slope of the line is m, then show**

**Question 11.Solution:**

Given, A(h, 0), B(0, k) and C(a, b)

Since the points A, B and C lie on a line, they are collinear.

∴ Slope of AB = slope of BC