Question

find the slope - intercept form for the line passing through (-1,2) and perpendicular to the line passing through (left(-3,\frac{1}{2}
ight)) and (left(-2,\frac{2}{3}
ight)). the slope - intercept form for the line passing through (-1,2) and perpendicular to the line passing through (left(-3,\frac{1}{2}
ight)) and (left(-2,\frac{2}{3}
ight)) is (y=square). (simplify your answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Calculate slope of given line

The slope formula is $m_1=\frac{y_2-y_1}{x_2-x_1}$. Substitute $(-3,\frac{1}{2})$ and $(-2,\frac{2}{3})$:
$m_1=\frac{\frac{2}{3}-\frac{1}{2}}{-2-(-3)}=\frac{\frac{4-3}{6}}{1}=\frac{1}{6}$

Step2: Find perpendicular slope

Perpendicular slope $m_2$ is negative reciprocal:
$m_2=-\frac{1}{m_1}=-6$

Step3: Use point-slope form

Point-slope formula: $y-y_0=m(x-x_0)$. Substitute $(-1,2)$ and $m_2=-6$:
$y-2=-6(x-(-1))$

Step4: Simplify to slope-intercept form

Expand and rearrange to $y=mx+b$:
$y-2=-6x-6$
$y=-6x-6+2$
$y=-6x-4$

Answer:

$y=-6x-4$