Question

Find the slope-intercept form of the equation of the line that passes through the point (5,1) and is perpendicular to the line 2x + 5y = 10

Answer 1

The slope-intercept form:

$y=mx+b$

m - slope

b - y-intercept

Convert 2x + 5y = 10 to the slope0intercept form:

$2x+5y=10$        subtract 2x from both sides

$5y=-2x+10$       divide both sides by 5

$y=-\dfrac{2}{5}x+2$

Let $k:y=m_1x+b_1$ and $l:y=m_2x+b_2$

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

We have $m_1=-\dfrac{2}{5}$

Therefore

$m_2=-\dfrac{-\frac{2}{5}}=\dfac{5}{2}$

We have the equation of a line:

$y=\dfrac{5}{2}x+b$

Put the coordinates of the point (5, 1) to the equation of a line:

$1=\dfrac{5}{2}(5)+b$

$1=\dfrac{25}{2}+b$       subtract $\dfrac{25}{2}$ from both sides

$-\dfrac{23}{2}=b\to b=\dfrac{23}{2}$

Answer: $\boxed{y=\dfrac{5}{2}x+\dfrac{23}{2}}$