Question

write an equation in slope intercept form of a line that passes through the point (14, 8) and is (a) parallel (v) perpendicular to the line that passes through the points (4,9) and (-3, 6)

Answer 1

Answer:

$y-8=-\frac{7}{3} (x-14)$

Step-by-step explanation:

We can write the equation of a line in 3 different forms including slope intercept, point-slope, and standard depending on the information we have. We have a point and a slope from the equation. We will chose point-slope since we have a point and can find the slope.

Point slope:$y-y_1=m(x-x_1)$

We must find the slope using the slope formula.

Slope:$m=\frac{y_2-y_1}{x_2-x_1}$

We substitute $x_1=4\\y_1=9$ and $x_2=-3\\y_2=6$

$m=\frac{6-9}{-3-4}$

$m=\frac{6-9}{-3-4}=\frac{-3}{-7} =\frac{3}{7}$

$m\neq \frac{3}{7}$ in our new equation because it is perpendicular to it. This means we will need to change it into its negative reciprocal which is $m=-\frac{7}{3}$.

We will substitute $m=-\frac{7}{3}$ and $x_1=14\\y_1=8$.

$y-8=-\frac{7}{3} (x-14)$

This is the equation of the line perpendicular to the equation line through the points given that crosses through (14,8).