Question

what is an equation of the line that passes through the point (6, -8) and is perpendicular to the line 2x - y = 7?

Explanation:

Step1: Find the slope of the given line

Rewrite $2x - y=7$ to slope - intercept form $y = 2x - 7$. The slope of this line is $m_1 = 2$.

Step2: Find the slope of the perpendicular line

If two lines are perpendicular, the product of their slopes is $- 1$. Let the slope of the required line be $m_2$. Then $m_1m_2=-1$, so $2m_2=-1$, and $m_2=-\frac{1}{2}$.

Step3: Use the point - slope form to find the equation of the line

The point - slope form is $y - y_1=m(x - x_1)$. Here, $(x_1,y_1)=(6,-8)$ and $m = -\frac{1}{2}$. Substitute these values: $y+8=-\frac{1}{2}(x - 6)$.

Step4: Simplify the equation

Expand the right - hand side: $y+8=-\frac{1}{2}x + 3$. Then move the terms to get the general form: $y=-\frac{1}{2}x-5$ or $x + 2y=-10$.

Answer:

$y=-\frac{1}{2}x - 5$ (or $x + 2y=-10$)