If each of the points (x₁, 4), (-2, y₁) lies on the line joining the points (2, -1) and (5, -3), then the point P(x₁,y₁) lies on the line

The equation of the line joining the points (2, -1) and (5, -3) is given by

$\begin{array}{rl} & \mathrm{y}+1=\frac{-1+3}{2-5}(\mathrm{x}-2)\\ \mathrm{or} & 2\mathrm{x}+3\mathrm{y}-1=0---\left(\mathrm{i}\right)\end{array}$

Since $({\mathrm{x}}_1,4) \text{and} (-2,{\mathrm{y}}_1) \text{lie on }2\mathrm{x}+3\mathrm{y}-1=0$ we have

$2{\mathrm{x}}_1+12-1=0 \text{or }{\mathrm{x}}_1=-\frac{11}{2}$

and $-4+3{\mathrm{y}}_1-1=0 \text{or }{\mathrm{y}}_1=\frac{5}{3}$

Thus, $({\mathrm{x}}_1,{\mathrm{y}}_1)$ satisfies 2x + 6y + 1= 0