Find the equation of a line passing through the points (7, -3) and (2, -2). If this line meets x-axis at point P and y-axis at point Q; find the co-ordinates of points P and Q.

By two-point formula,

Equation of line : $y - y$1 = \dfrac{y2 - y1}{x2 - x1}(x - x1)

∴ Equation of line passing through (7, -3) and (2, -2) is

$\Rightarrow y - (-3) = \dfrac{-2 - (-3)}{2 - 7} \times (x - 7) \\[1em] \Rightarrow y + 3 = \dfrac{-2 + 3}{-5} \times (x - 7) \\[1em] \Rightarrow y + 3 = -\dfrac{1}{5}(x - 7) \\[1em] \Rightarrow 5(y + 3) = -(x - 7) \\[1em] \Rightarrow 5y + 15 = -x + 7 \\[1em] \Rightarrow x + 5y + 15 - 7 = 0 \\[1em] \Rightarrow x + 5y + 8 = 0.$

Given,

Line meets x-axis at point P. We know that, y-coordinate = 0 at x-axis.

Substituting, y = 0 in x + 5y + 8 = 0, we get :

⇒ x + 5(0) + 8 = 0

⇒ x + 8 = 0

⇒ x = -8.

∴ P = (-8, 0).

Line meets y-axis at point Q. We know that, x-coordinate = 0 at y-axis.

Substituting, x = 0 in x + 5y + 8 = 0, we get :

⇒ 0 + 5y + 8 = 0

⇒ 5y + 8 = 0

⇒ y = -$\dfrac{8}{5}$.

∴ Q = $\Big(0, -\dfrac{8}{5}\Big)$.

Hence, equation of line is x + 5y + 8 = 0, P = (-8, 0) and Q = $\Big(0, -\dfrac{8}{5}\Big)$.