Question

Determine the equation of the line in graph.

Answer 1

Points of the given line are

$(12,-3),(-4,2)$

slope(m) will be=$\frac{y_2-y_1}{x_2-x_1}=\frac{2-(-3)}{-4-12}=\frac{5}{-16}=-\frac{5}{16}$

Standard equation of the line is ${y-y}_1=m(x-x_1)$

Substituting the value of $m$ and point $(12,-3)$ in the equation we get:

$y-(-3)=-\frac{5}{16}(x-12)$

$y+3=-\frac{5}{16}(x-12)$

$16y+48=-5x+60$

$16y=-5x+60-48$

$16y=-5x+12$