Find the value of y so that the points (-5,10) and (3,y) lie on a line with slope -3/8
Find the value of y so that the points (-5,10) and (3,y) lie on a line with slope -3/8
$$slope = \frac{\Delta y}{\Delta x} = \frac
{y-10}
{3-(-5)}
=
\frac
{y-10}
{8}
\\\\
\text{also slope } = -\frac{3}{8}$$
$$so\ we\ have \quad
\frac{y-10}{8} =
-\frac{3}{8} \qquad or\ \quad y-10=-3$$
the value of y = 7
$$slope=\frac{y_a-y_b}{x_a-x_b}=\frac{y-10}{3--5}=\frac{y-10}{8}=-\frac{3}{8}$$
Now solve for y.
Good luck! :)
Find the value of y so that the points (-5,10) and (3,y) lie on a line with slope -3/8
$$slope = \frac{\Delta y}{\Delta x} = \frac
{y-10}
{3-(-5)}
=
\frac
{y-10}
{8}
\\\\
\text{also slope } = -\frac{3}{8}$$
$$so\ we\ have \quad
\frac{y-10}{8} =
-\frac{3}{8} \qquad or\ \quad y-10=-3$$
the value of y = 7