Question

What is the y-intercept of the equation of the line that is perpendicular to the line y = 3/5 x + 10 and passes through the point (15, –5)?
The equation of the line in slope-intercept form is y = -5/3 x +

Answer 1

Answer:

$\large\boxed{y-intercept=20}$

Step-by-step explanation:

$\text{Let}\ k:y=_1x+b_1\ \text{and}\ l:y=m_2x+b_2.\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\============================\\\\\text{We have}\ y=\dfrac{3}{5}x+10\to m_1=\dfrac{3}{5}.\\\\\text{Therefore}\ m_2=-\dfrac{1}{\frac{3}{5}}=-\dfrac{5}{3}.\\\\\text{The equation of the searched line:}\ y=-\dfrac{5}{3}x+b.\\\\\text{The line passes through }(15,\ -5).$

$\text{Put thecoordinates of the point to the equation.}\ x=15,\ y=-5:\\\\-5=-\dfrac{5}{3}(15)+b\\\\-5=(-5)(5)+b\\\\-5=-25+b\qquad\text{add 25 to both sides}\\\\b=20\\\\\boxed{y=-\dfrac{5}{3}x+20}$