Question

The vertices of a triangle are A(10, 4), B(4, −9) and C(−2, −1). Find the equation of the altitude through A.

Answer 1

The vertices of the side BC are B(4, −9) and C(−2, −1).

Using the slope formula:

`m = (y_2 - y_1)/(x_2 - x_1)`

`m_(BC) = (-1 - (-9))/(-2 - 4)`

`m_(BC) = (-1 + 9)/-6`

`m_(BC) = 8/-6`

∴ `m_(BC) = - 4/3`

Since the altitude is perpendicular to BC, the product of their slopes must be −1 (m_A × m_BC = −1):

`m_A xx (- 4/3) = -1`

`m_A = 3/4`

Using the point-slope formula with point A(10, 4) and slope `m_A = 3/4`:

y − y₁​ = m(x − x₁​)

`y - 4 = 3/4 (x - 10)`

4(y − 4) = 3(x − 10)    ...[Multiplied both sides by 4]

4y − 16 = 3x − 30

Let’s rearrange into the general form (Ax + By + C = 0):

3x − 4y − 30 + 16 = 0

3x − 4y − 14 = 0

Hence, the equation of the altitude through A is 3x − 4y − 14 = 0.