Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Locus and Straight Line Ex 5.2 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Ex 5.2

Question 1.

Find the slope of each of the following lines which pass through the points:

(a) (2, -1), (4, 3)

(b) (-2, 3), (5, 7)

(c) (2, 3), (2, -1)

(d) (7, 1), (-3, 1)

Solution:

(a) Let A = (x_{1}, y_{1}) = (2, -1) and B = (x_{2}, y_{2}) = (4, 3).

Slope of line AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

= \(\frac{3-(-1)}{4-2}\)

= \(\frac{4}{2}\)

= 2

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

Slope of line CD = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

= \(\frac{7-3}{5-(-2)}\)

= \(\frac{4}{7}\)

(c) Let E = (2, 3) = (x_{1}, y_{1}) and F = (2, -1) = (x_{2}, y_{2})

Since x_{1} = x_{2} = 2

∴ The slope of EF is not defined. ……[EF || y-axis]

(d) Let G = (7, 1) = (x_{1}, y_{1}) and H = (-3, 1) = (x_{2}, y_{2}) say.

Since y_{1} = y_{2}

∴ The slope of GH = 0 …..[GH || x-axis]

Question 2.

If the X and Y-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).

i.e. the line L passes through (2, 0) = (x_{1}, y_{1}) and (0, 3) = (x_{2}, y_{2}) say.

Slope of line L = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

= \(\frac{3-0}{0-2}\)

= \(\frac{-3}{2}\)

Question 3.

Find the slope of the line whose inclination is 30°.

Solution:

Given, inclination (θ) = 30°

Slope of the line = tan θ = tan 30° = \(\frac{1}{\sqrt{3}}\)

Question 4.

Find the slope of the line whose inclination is 45°.

Solution:

Given, inclination (θ) = 45°

Slope of the line = tan θ = tan 45° = 1

Question 5.

A line makes intercepts 3 and 3 on the co-ordinate axes. Find the slope of the line.

Solution:

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

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

i.e. the line passes through (3, 0) = (x_{1}, y_{1}) and (0, 3) = (x_{2}, y_{2}) say.

Slope of line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

= \(\frac{3-0}{0-3}\)

= -1

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:

Given, A(4, 4) = (x_{1}, y_{1}), B(3, 5) = (x_{2}, y_{2}), C(-1, -1) = (x_{3}, y_{3})

Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-4}{3-4}=-1\)

Slope of BC = \(\frac{y_{3}-y_{2}}{x_{3}-x_{2}}=\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}\)

Slope of AC = \(\frac{y_{3}-y_{1}}{x_{3}-x_{1}}=\frac{-1-4}{-1-4}=\frac{-5}{-5}=1\)

Slope of AB × slope of AC = -1 × 1 = -1

∴ side AB ⊥ side AC

∴ ΔABC 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 clockwise.

Solution:

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

∴ Inclination of the line (θ) = (90° + 45°)

∴ Slope of the line = tan(90° + 45°)

= -cot 45° …….[tan(90 + θ°) = -cot θ]

= -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

∴ \(\frac{1-(-1)}{2-k}=\frac{5-1}{4-2}\)

∴ \(\frac{2}{2-k}=\frac{4}{2}\)

∴ 1 = 2 – k

∴ k = 2 – 1 = 1

Check:

For collinear points P, Q, R,

Slope of PQ = Slope of QR = Slop of PR

For k = 1, if the given points are collinear, then our answer is correct.

P(1, -1), Q(2, 1) and R(4, 5)

Slope of PQ = \(\frac{1-(-1)}{2-1}=\frac{2}{1}=2\)

Slope of QR = \(\frac{5-1}{4-2}=\frac{4}{2}=2\)

Slope of PQ = Slope of QR

∴ The given points are collinear.

Thus, our answer is correct.