Question 798046
Part (c) is not clear.  Part (a) does not ask for an exact numeric answer.  Part (b) seems clear enough.  


Part (b):
Find the intersection, a point, of L1 and L2.  Find the slope of PQ and determine the slope of a perpendicular line to PQ.  Use the point-slope form for a line with this slope and point of intersection of L1 and L2 to find the equation for the line.  



L1:  x+7y=23
L2:  5x-4y=-15


L1 is also {{{5x+35y=5*23}}}, or {{{5x+35y=115}}}.
L1-L2 is {{{(5x+35y)-(5x-4y)=115-(-15)}}}, {{{39y=130}}}, {{{y=13*10/(3*13)}}}, {{{y=10/3}}}

From L1, {{{x=23-7y}}}, so {{{x=23-7(10/3)}}}, {{{x=23*3/3-70/3}}}, {{{69/3-70/3}}}, {{{x=-1/3}}}

Intersection of L1 and L2 is at {{{x=-1/3}}}, {{{y=10/3}}}.


Slope of PQ:  {{{(4-0)/(6-(-2))=1/2}}}.
A line perpendicular to PQ must have a slope of {{{-2}}}.


The line perpendicular to PQ and containing the point (-1/3, 10/3) is, using the point-slope form, {{{y-10/3=-2(x-(-1/3))}}}
{{{y-10/3=-2x-2/3}}}
{{{3y-10=-6x-2}}}
{{{6x+3y=10-2}}}
{{{6x+3y=8}}} Standard Form; or if you prefer, {{{6x+3y-8=0}}}