Question

Let ABC be a triangle with A(–3, 1) and ∠ACB = θ, 0 < θ < `π/2`. If the equation of the median through B is 2x + y – 3 = 0 and the equation of angle bisector of C is 7x – 4y – 1 = 0, then tan θ is equal to ______.

Options

• `4/3`

• `1/2`

• 2

• `3/4`

Answer 1

Let ABC be a triangle with A(–3, 1) and ∠ACB = θ, 0 < θ < `π/2`. If the equation of the median through B is 2x + y – 3 = 0 and the equation of angle bisector of C is 7x – 4y – 1 = 0, then tan θ is equal to `underlinebb(4/3)`.

Explanation:

Let C`(α, (7α - 1)/4)` because C lies on 7x – 4y – 1 = 0

Midpoint of AC = `"M"((α - 3)/2, (7α + 3)/8)`

Now, M lies on 2x + y – 3 = 0,

`2((α - 3)/2) + ((7α + 3)/8) - 3` = 0

⇒ α = 3 and `(7α - 1)/4 = (7(3) - 1)/4` = 5

So, C = (3, 5)

Now, A = (–3, 1) and C = (3, 5)

Slope of AC, `("y"_2 - "y"_1)/(x_2 - x_1) = (5 - 1)/(3 + 3) = 4/6 = 2/3`

and slope of 7x – 4y – 1 = 0 is `7/4`  ...`[∵ "y" = 7/4x - 1]`

 ∴ `tan  θ/2 = |("m"_2 - "m"_1)/(1 + "m"_2"m"_1)|`

= `(7/4 - 2/3)/(1 + 7/4(2/3))`

= `1/2`

So, tanθ = `(2tan  θ/2)/(1 - tan^2  θ/2)`

= `(2(1/2))/(1 - 1/4)`

= `4/3`

⇒ tanθ = `4/3`