Question

in each part that follows, you are given an equation of a line and a point. find the equation of the line through the given point that is perpendicular to the given line (the slope negative reciprocal of the slope of the given line if the given line is neither vertical nor horizontal )
(a) ( y = 8x ), ( p(0,0) )
(b) ( y = 5x + 8 ), ( q(1,3) )
(a) ( y = square )
(simplify your answer. type your answer in slope - intercept form. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Find slope of given line

The given line is $y=8x$, so its slope $m_1=8$.

Step2: Find perpendicular slope

Perpendicular slope $m_2 = -\frac{1}{m_1} = -\frac{1}{8}$.

Step3: Use point-slope form

Point $P(0,0)$ is the y-intercept ($b=0$). Slope-intercept form is $y=mx+b$.
Substitute $m=-\frac{1}{8}$ and $b=0$: $y = -\frac{1}{8}x + 0$

Step4: Simplify the equation

$y = -\frac{1}{8}x$

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Step1: Find slope of given line

The given line is $y=5x+8$, so its slope $m_1=5$.

Step2: Find perpendicular slope

Perpendicular slope $m_2 = -\frac{1}{m_1} = -\frac{1}{5}$.

Step3: Use point-slope formula

Point-slope form: $y - y_1 = m(x - x_1)$. Substitute $Q(1,3)$, $m=-\frac{1}{5}$:
$y - 3 = -\frac{1}{5}(x - 1)$

Step4: Convert to slope-intercept form

Expand and solve for $y$:
$y - 3 = -\frac{1}{5}x + \frac{1}{5}$
$y = -\frac{1}{5}x + \frac{1}{5} + 3$
$y = -\frac{1}{5}x + \frac{1}{5} + \frac{15}{5}$
$y = -\frac{1}{5}x + \frac{16}{5}$

Answer:

(a) $\boldsymbol{y = -\frac{1}{8}x}$
(b) $\boldsymbol{y = -\frac{1}{5}x + \frac{16}{5}}$