Question

A line, with a slope greater than one, passes through point A(4, 3) and intersects the line x – y – 2 = 0 at the point B. If the length of the line segment AB is \(\frac{\sqrt{29}}{3}\) then B also lies on the line :

• A. 2x + y = 9
• B. 3x – 2y = 7
• C. x + 2y = 6
• D. 2x – 3y = 3

Answer 1

The Correct Option is C

Let the inclination of the required line be θ,
So the coordinates of point B can be assumed as
(4−\(\frac{\sqrt{29}}{3}\)cos⁡θ,3−\(\frac{\sqrt{29}}{3}\)sinθ)
Which satisfices x – y – 2 = 0
4−\(\frac{\sqrt{29}}{3}\)cos⁡θ−3+\(\frac{\sqrt{29}}{3}\)sin⁡θ−2=0
sin⁡θ−cos⁡θ=\(\frac{3}{\sqrt{29}}\)
By squaring
sin⁡2θ=\(\frac{20}{29}\)=\(\frac{2tan\theta}{1+tan2\theta}\)
tan⁡θ=\(\frac{5}{2}\) only (because the slope is greater than 1)
sin⁡θ=\(\frac{5}{\sqrt{29}}\), cos⁡θ=\(\frac{2}{\sqrt{29}}\)
Point B:(\(\frac{10}{3}\),\(\frac{4}{3}\))
Which also satisfies x + 2y = 6