At what angle with the line x+y=4, a line through (1,2) be drawn so that the?

Question

distance between the point of intersection of the lines and the point (1,2) is 6/(root 3)

 

Answer 1

The one line equation is x + y = 4 and the other line is passing through the point (1, 2) with slope m.

So, the other line equation is y - 2 = m(x - 1)

⇒ y = mx - m + 2.

Find the point of intersection of the lines x + y = 4 and y = mx - m + 2.

Substitute y = mx - m + 2 in x + y = 4 for x.

x + mx - m + 2 = 4

x(m + 1) = m - 2 + 4

⇒ x = (m + 2)/(m + 1).

Substitute x = (m + 2)/(m + 1) in y = mx - m + 2 for y.

y = m[(m + 2)/(m + 1)] - m + 2

⇒ y = (3m + 2)/(m + 1).

Therefore, the point of intersection of the lines x + y = 4 and y = mx - m + 2 is [(m + 2)/(m + 1), (3m + 2)/(m + 1)].

From the given data : Distance between (1, 2) and [(m + 2)/(m + 1), (3m + 2)/(m + 1)] is 6/√3.

√[ ([(m + 2)/(m + 1)] - 1)² + ([(3m + 2)/(m + 1)] - 2)² ] = 6/√3

Squaring on both sides.

1 + m² = 12(m + 1)²

1 + m² = 12(m² + 2m + 1)

1 + m² = 12m² + 24m + 12

11m² + 24m + 11 = 0

⇒ m = - 1.5269 and m = - 0.6549.

So, m₁ = - 1.5269 and m₂ = - 0.6549.

tan A = (m₂ - m₁)/(1 + m₁m₂).

tan A = (- 0.6549 + 1.5269)/[1 +( (0.6549)(1.5269))]

tan A = 0.872/[1 + 0.9999]

tan A = 0.872/1.9999

tan A = 0.4360

A= tan^-1(0.4360)

A = 23.56⁰ or 0.411151 radians.

∴ The angle with the line x + y = 4 and a line through the point (1, 2) is 23.56⁰ or 0.411151 radians.