Question

Find the equation of the line through the point $(-1,3)$ and parallel to the line joining the points $(0,-2)$ and $(4,5)$.

Answer 1

$(-1,3)$ & 11 to $(0,-2)$ & $(4,5)$

Equation of line = $y=mx+c$

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{5+2}{4-0}=\frac{7}{4}$

So, $e{q}^n$ becomes,

$y=\frac{7}{4}x+c$ __- (1)

$\because$ this lines passes through $(-1,3),$ then,

$3=\frac{7}{4}(-1)+c$

$3=\frac{-7}{4}+c$

$3+\frac{7}{4}=c=\frac{19}{4}$

Now $e{q}^n$ (1) becomes

$4y=7x+19$ Answer