Question 1106028
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Obviously you are talking about two different lines.,   The line through points R and S is fixed, so let's call that the "original" line L.  Then let's make M the line through P and Q parallel to line L and N the line through P and Q perpendicular to line L.<br>
The slope of the fixed line L is {{{(5-3)/(1-4) = -2/3}}}.<br>
The slope of the first line is {{{(2-4)/(1-n) = -2/(1-n) = 2/(n-1)}}}.<br>
For the lines to be parallel, the slope of the first line has to be -2/3 -- the same as the slope of line L:
{{{2/(n-1) = -2/3}}}
{{{6 = -2n+2}}}
{{{4 = -2n}}}
{{{n = -2}}}<br>
The lines are parallel when n is -2; line M passes through the points (-2,4) and (1,2).<br>
For the lines to be perpendicular, the slope of the first line has to be 3/2 -- the negative reciprocal of the slope of line L:
{{{2/(n-1) = 3/2}}}
{{{4 = 3n-3}}}
{{{7 = 3n}}}
{{{n = 7/3}}}<br>
The lines are perpendicular when n is 7/3; line N passes through the points (7/3,4) and (1,2).<br>
The graph shows the original line L (red), passing through (1,5) and (4,3); line M which is parallel to L (green), passing through (-2,4) and (1,2); and line N which is perpendicular to L, passing through (1,2) and (7/3,4).<br>
{{{graph(400,400,-10,10,-10,10,y=(-2/3)x+17/3,y=(-2/3)x+8/3,y=(3/2)x+1/2)}}}