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

(i) the equation of line through A and perpendicular to BC.

(ii) the co-ordinates of the point P, where the perpendicular through A, as obtained in (i), meets BC.

$\text{Slope of BC } = \dfrac{y$2 - y1}{x2 - x1} \\[1em] = \dfrac{7 - (-2)}{11 - (-1)} \\[1em] = \dfrac{7 + 2}{11 + 1} \\[1em] = \dfrac{9}{12} \\[1em] = \dfrac{3}{4}

Then the equation of the line BC is

⇒ y - y₁ = m(x - x₁)

⇒ y - (-2) = $\dfrac{3}{4}$[x - (-1)]

⇒ 4(y + 2) = 3(x + 1)

⇒ 4y + 8 = 3x + 3

⇒ 3x - 4y = 8 - 3

⇒ 3x - 4y = 5 ……… (1)

(i) Let slope of line perpendicular to BC = m₁

As product of slope of perpendicular lines is -1

∴ m x m₁ = -1

⇒ $\dfrac{3}{4}$ x m₁ = -1

⇒ m₁ = $-\dfrac{4}{3}$

So, the required equation of the line through A (0, 5) and perpendicular to BC is given by

⇒ y - y₁ = m(x - x₁)

⇒ y - 5 = $-\dfrac{4}{3}$(x - 0)

⇒ 3y - 15 = -4x

⇒ 4x + 3y = 15 ……… (2)

Hence, equation of the required line is 4x + 3y = 15.

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

3x - 4y = 5 ……… (1)

4x + 3y = 15 ……… (2)

Multiplying equation (1) by 3 and (2) by 4 and adding we get,

⇒ 9x - 12y + 16x + 12y = 15 + 60

⇒ 25x = 75

⇒ x = $\dfrac{75}{25}$

⇒ x = 3

Substituting the value of x in equation (1) we get,

3(3) - 4y = 5

⇒ 4y = 3(3) – 5

⇒ 4y = 9 - 5

⇒ 4y = 4

⇒ y = 1

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