Question

Find the equation of the line that passes through (-1,5) and is perpendicular to y – 5x = 1.

Answer 1

The answer is

$y = - \frac{1}{5} x + \frac{24}{5}$

Step-by-step explanation:

Equation of a line is y = mx + c

where

m is the slope

c is the y intercept

y - 5x = 1

y = 5x + 1

Comparing with the above formula

The slope / m of the line is 5

Since the is perpendicular to y = 5x + 1 it's slope it's the negative inverse of y = 5x + 1

That's

Slope of the perpendicular line = - 1/5

Equation of the line using point (-1,5) is

$y - 5 = - \frac{1}{5} (x + 1)$

$y - 5 = - \frac{1}{5} x - \frac{1}{5}$

$y = - \frac{1}{5} x - \frac{1}{5} + 5$

We have the final answer as

$y = - \frac{1}{5} x + \frac{24}{5}$

Hope this helps you

Answer 2

Answer:

$\huge\boxed{y=-\dfrac{1}{5}x+\dfrac{24}{5}\to x+5y=24}$

Step-by-step explanation:

The slope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept

Let

$k:y=m_1x+b_1;\ l:y=m_2x+b_2$

therefore

$k||l\iff m_1=m_2\\k\perp l\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}$

We have the equation of a line in the standard form. Convert it to the slope-intercept form:

$y-5x=1$      add 5x to both sides

$y-5x+5x=1+5x\\\\y=5x+1\to m_1=5;\ b_1=1$

Calculate the slope:

$m_2=-\dfrac{1}{5}$

Substitute the value of a slope and the coordinates of the given point (-1, 5) to the equation of a line:

$y=m_2x+b$

$5=\left(-\dfrac{1}{5}\right)(-1)+b$

$5=\dfrac{1}{5}+b$           subtract 1/5 from both sides

$5-\dfrac{1}{5}=\dfrac{1}{5}-\dfrac{1}{5}+b$

$\dfrac{25}{5}-\dfrac{1}{5}=b\\\\\dfrac{24}{5}=b\to b=\dfrac{24}{5}$

Final answer:

$y=-\dfrac{1}{5}x+\dfrac{24}{5}$

convert to the standard form (Ax + By = C):

$y=-\dfrac{1}{5}x+\dfrac{24}{5}$       multiply both sides by 5

$5y=(5)\left(-\dfrac{1}{5}x\right)+(5)\left(\dfrac{24}{5}\right)$

$5y=-x+24$        add x to both sides

$x+5y=24$