Question

A software designer is mapping the streets for a new racing game. All of the streets are depicted as either perpendicular or parallel lines. The equation of the lane passing through A and B is -7x + 3y = -21.5. What is the equation of the central street PQ?

Answer 1

equition of line passing trough A and B $-7x + 3y = -21.5$ can also be written as $y=\frac{7}{3}x-\frac{21.5}{3}$

comparing with standard form $y=mx+c$

slope $m=\frac{7}{3}$

PQ is perpendicular to lane AB. Product of slope of perpendicular lines will be -1.

$m_1.m_2=-1$

slope of the line$=\frac{-1}{slope \space of\space parallel\space line}=\frac{-1}{\frac{7}{3}}=\frac{-3}{7}$

equition of the line will then be

$y=mx+c\\y=-\frac{3}{7}x+c\\7y=-3x+7c\\7y+3x=7c$

dividing by 2

$3.5y+1.5x=3.5c$

find c

the line pq passes through the point (7,6) in as shown in the figure hence line could be found out using point slope form of line.

$y-y_1=m(x-4)\\slope=\frac{-3}{7}\space (7,6)\\y-6=\frac{-3}{7}(x-7)=7\\7(y-6)=-3(x-7)\\7y-42-3x+21\\=7x+3x=21+42\\7x+3x=63$ is the eqition of central line.

divide by 2

$3.5y+1.5x=31.5$