Question

What is Perpendicular to 5x-3y=30 and passing through (-2,7)

Answer 1

$k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\k\ \perp\ l\iff m_1\cdot m_2=-1\\\\k\ ||\ l\iff m_1=m_2$

We have:

$k:5x-3y=30\ \ \ \ |-5x\\\\-3y=-5x+30\ \ \ \ |:(-3)\\\\y=\dfrac{5}{3}x-10\to m_1=\dfrac{5}{3}\\\\l:y=m_2x+b\\\\k\ \perp l\Rightarrow \dfrac{5}{3}m_2=-1\ \ \ \ |\cdot\dfrac{3}{5}\\\\m_2=-\dfrac{3}{5}$

$l:y=-\dfrac{3}{5}x+b$

The line l is passing through (-2, 7). Substitute the coordinates of the point to the equation of a line l:

$7=-\dfrac{3}{5}\cdot(-2)+b\\\\\dfrac{6}{5}+b=7\ \ \ \ |-\dfrac{6}{5}\\\\b=\dfrac{35}{5}-\dfrac{6}{5}\\\\b=\dfrac{29}{5}$

$l:y=-\dfrac{3}{5}x+\dfrac{29}{5}\ \ \ \ |\cdot5\\\\5y=-3x+29\ \ \ \ |+3x\\\\3x+5y=29$

Answer: 3x + 5y = 29