Question

Find whether the points (-1, -2), (1, 0) and (-3, -4) lie above, below or on the line 3x + 2y + 7 = 0

Answer 1

Let the given points be P(-1, -2) Q(1, 0) and R(-3, -4)

Equation of the given line is 3x + 2y + 7 = 0   ...(1)

(i) Perpendicular distance from P(-1, -2) to the line (1) is = `+- |("a"x + "b"y + "c")/(sqrt("a"^2 + "b"^2))|`

`= |(3(-1) + 2(-2) + 7)/(sqrt(3^2 + 2^2))|`

`= |(- 3 - 4 + 7)/sqrt13|`

`= |0/sqrt13|` = 0

∴ P(-1, -2) lie on the line.

(ii) Perpendicular distance from Q(1, 0) to the line (1) is

`= (3(1) + 2(0) + 7)/(sqrt(3^2 + 2^2))`

`= (3 + 7)/sqrt13`

`= 10/sqrt13` > 0

∴ Q(1, 0) lie above the line.

(iii) Perpendicular distance from R(-3, -4) to the line (1) is

`= (3(-3) + 2(-4) + 7)/(sqrt(3^2 + 2^2))`

`= (- 9 - 8 + 7)/sqrt13`

`= (- 10)/sqrt13` < 0

∴ The point R (-3, -4) lie below the line.