Question

Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.

Answer 1

Let the equation of line AB be, 3x – 4y – 16 = 0      ....…(i)

or y = `3/4"x" - 4`

Slope of line AB = `3/4`

The perpendicular drawn from point C(−1, 3) to AB is CD.

∴ AB ⊥ CD

∴ Slope of CD = `(-1)/(("Slope of line AB")"`

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

= `(-4)/3`

Hence, the equation of line CD,

y – y₁ = m(x – x₁)

y – 3 = `(-4)/3 ("x" + 1)`

or 3y – 9 = –4x – 4

or 4x + 3y – 5 = 0     ....…(ii)

Multiplying equation (i) by 3 and equation (ii) by 4,

9x – 12y = 48

16x + 12y = 20

on adding these

25x = 68 or x = `68/25`

Putting the value of x in (i),

`3 xx 68/25 - 4"y" = 16`

∴ `4"y" = 204/25 - 16`

= `(204 - 400)/25`

∴ y = `-196/25 xx 1/4 = -49/25`

Hence, the coordinates of perpendicular foot D are `(68/25, -49/25)`.