Question

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

Explanation:

Step1: Find slope of given line

Rewrite $2x + y = 2$ to slope-intercept form $y = mx + b$:
$y = -2x + 2$
Slope of given line $m_1 = -2$

Step2: Find perpendicular slope

Perpendicular slope $m_2 = -\frac{1}{m_1}$
$m_2 = -\frac{1}{-2} = \frac{1}{2}$

Step3: Use point-slope form

Point-slope formula: $y - y_1 = m(x - x_1)$ with $(x_1,y_1)=(6,-2)$ and $m=\frac{1}{2}$
$y - (-2) = \frac{1}{2}(x - 6)$

Step4: Simplify to slope-intercept form

Simplify the equation:
$y + 2 = \frac{1}{2}x - 3$
$y = \frac{1}{2}x - 5$
Or rearrange to standard form:
$x - 2y = 10$

Answer:

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