# What is an equation in slope-intercept form of the line that is perpendicular to the graph of  y= -4x+5 and passes through (1, 1)?

Jun 24, 2018

The equation of line in slope-intercept form is $y = \frac{1}{4} x + \frac{3}{4}$

#### Explanation:

Slope of the line,  y= -4 x +5 or ; [y=mx+c]

is ${m}_{1} = - 4$ [Compared with slope-intercept form of equation]

The product of slopes of the pependicular lines is ${m}_{1} \cdot {m}_{2} = - 1$

$\therefore {m}_{2} = \frac{- 1}{-} 4 = \frac{1}{4}$. The equation of line passing through

$\left({x}_{1} = 1 , {y}_{1} = 1\right)$ having slope of ${m}_{2}$ is $y - {y}_{1} = {m}_{2} \left(x - {x}_{1}\right)$.

$\therefore y - 1 = \frac{1}{4} \left(x - 1\right) \mathmr{and} y = \frac{1}{4} x - \frac{1}{4} + 1 \mathmr{and} y = \frac{1}{4} x + \frac{3}{4}$.

Equation of line in slope-intercept form is $y = \frac{1}{4} x + \frac{3}{4}$ [Ans]

Jun 24, 2018

$y = \frac{1}{4} x + 34$

#### Explanation:

The given equation $y = \textcolor{g r e e n}{- 4} x + \textcolor{red}{5}$
is in slope-vertex form with
slope $\textcolor{g r e e n}{m} = \textcolor{g r e e n}{- 4}$, and
y-intercept $\textcolor{red}{b} = \textcolor{red}{5}$

If a line has a slop of $\textcolor{g r e e n}{m}$
then every line perpendicular to it has a slope of $\textcolor{b l u e}{- \frac{1}{m}}$.

Therefore all lines perpendicular to $y = \textcolor{g r e e n}{- 4} + 5$
will have a slope of $\textcolor{b l u e}{\frac{1}{4}}$
and will have an equation in slope-intercept form:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{b l u e}{\frac{1}{4}} x + \textcolor{m a \ge n t a}{c}$ for some constant (the y-intercept) $\textcolor{m a \ge n t a}{c}$

If $\left(\textcolor{b r o w n}{x} , \textcolor{\lim e}{y}\right) = \left(\textcolor{b r o w n}{1} , \textcolor{\lim e}{1}\right)$ is a solution for the required line with this form,
then
$\textcolor{w h i t e}{\text{XXX}} \textcolor{\lim e}{1} = \textcolor{b l u e}{\frac{1}{4}} \cdot \textcolor{b r o w n}{1} + \textcolor{m a \ge n t a}{c}$

$\textcolor{w h i t e}{\text{XXX}} \Rightarrow \textcolor{m a \ge n t a}{c} = \frac{3}{4}$

Therefore the resolved equation for the given line is
$\textcolor{w h i t e}{\text{XXX}} y = \frac{1}{4} x + \frac{3}{4}$