Find an equation for the line with the given properties.
-Perpendicular to the line y = -3x + 4; containing the point (-4, 3)
A) $y = \frac { 1 } { 3 } x + \frac { 13 } { 3 }$
B) $y = 3 x + \frac { 13 } { 3 }$
C) $y = - \frac { 1 } { 3 } x + \frac { 13 } { 3 }$
D) $y = - 3 x + \frac { 13 } { 3 }$

Question

Quizplus · Accepted Answer

The slope of the line perpendicular to y = -3x + 4 is the negative reciprocal of -3, which is $ \frac{1}{3} $. Using the point (-4, 3) to solve for y-intercept, $y = mx + b$ becomes $3 = \frac{1}{3}(-4) + b$, solving for b gives $b = \frac{13}{3}$. Thus, the equation is $y = \frac{1}{3}x + \frac{13}{3}$.

Find an Equation for the Line with the Given Properties y=13x+133y = \frac { 1 } { 3 } x + \frac { 13 } { 3 }y=31​x+313​

Find an Equation for the Line with the Given Properties $y = \frac { 1 } { 3 } x + \frac { 13 } { 3 }$