Two lines passing through the points (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line

Solution:

It is given that the slope of the first line, m\(_1\) = 2.

Let the slope of the other line be m₂.

The angle between the two lines is 60°.

tan θ = |(m\(_1\) - m\(_2\))/(1 + m\(_1\)m\(_2\))|

tan 60° = |(2 - m\(_2\))/(1 + 2m\(_2\))|

√3 = ± |(2 - m\(_2\))/(1 + 2m\(_2\))|

√3 (1 + 2m\(_2\)) = (2 - m\(_2\))

√3 + 2√3m\(_2\) + m\(_2\) = 2

√3 + m\(_2\)(2√3 + 1) = 2

m\(_2\) = (2 - √3)/(2√3 + 1)

or

√3 = - (2 - m\(_2\))/(1 + 2m\(_2\))

√3 (1 + 2m\(_2\)) = - (2 - m\(_2\))

√3 + m\(_2\)(2√3 - 1) = - 2

m\(_2\) = - (2 + √3)/(2√3 - 1)

Case 1:

m\(_2\) = (2 - √3)/(2√3 + 1)

The equation of the line passing through the point (2, 3) and having a slope of (2 - √3)/(2√3 + 1) is

(y - 3) = [(2 - √3)/(2√3 + 1)] (x - 2)

(2√3 + 1)y - 3(2√3 + 1) = (2 - √3)x - 2(2 - √3)

(√3 - 2)x + (2√3 + 1)y = - 1 + 8√3

In this case, the equation of the other line is (√3 - 2)x + (2√3 + 1)y = - 1 + 8√3

Case 2:

m\(_2\) = - (2 + √3)/(2√3 - 1)

The equation of the line passing through the point (2, 3) and having a slope of - (2 + √3)/(2√3 - 1) is

(y - 3) = [- (2 + √3)/(2√3 - 1)] (x - 2)

(2√3 - 1)y - 3(2√3 - 1) = - (2 + √3)x + 2(2 - √3)

(√3 + 2)x + (2√3 - 1)y = 4 - 2√3 + 6√3 - 3

(√3 + 2)x + (2√3 - 1)y = 1 + 8√3

In this case, the equation of the other line is (√3 + 2)x + (2√3 - 1)y = 1 + 8√3

Thus, the required equation of the other line is (√3 - 2)x + (2√3 + 1)y = - 1 + 8√3 or (√3 + 2)x + (2√3 - 1)y = 1 + 8√3

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NCERT Solutions Class 11 Maths Chapter 10 Exercise 10.3 Question 12

Two lines passing through the points (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line.

Summary:

Two lines passing through the points (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, then the equation of the other line can be (√3 - 2)x + (2√3 + 1)y = - 1 + 8√3 or (√3 + 2)x + (2√3 - 1)y = 1 + 8√3