Question

The vertices of a triangle ABC are A(0, 5), B(−1, −2) and C(11, 7). Write down the equation of BC. Find:

1. the equation of line through A and perpendicular to BC.
2. the co-ordinates of the point P, where the perpendicular through A, as obtained in (i), meets BC.

Answer 1

Slope of BC = `(7 + 2)/(11 + 1) = 9/12 = 3/4`

Equation of the line BC is given by

y – y₁ = m₁(x – x₁)

`y + 2 = 3/4 (x + 1)`

4y + 8 = 3x + 3

3x − 4y = 5     ...(1)

i. Slope of line perpendicular to BC = `(-1)/(3/4) = (-4)/3`

Required equation of the line through A(0, 5) and perpendicular to BC is

y – y₁ =  m₁(x − x₁)

`y - 5 = (-4)/3 (x + 0)`

3y − 15 = −4x

4x + 3y = 15   ...(2)

ii. The required point will be the point of intersection of lines (1) and (2).

(1) `=>` 9x − 12y = 15

(2) `=>` 16x + 12y = 60

Adding the above two equations, we get,

25x = 75

x = 3

So, 4y = 3x − 5

= 9 − 5

= 4

y = 1

Thus, the co-ordinates of the required point is (3, 1).