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sqdancefan

There are a couple of ways to do this. One is to find the slope of the given segment, and from that the slope of the perpendicular. You would also find the midpoint of the given segment, then write the equation of the line with the found slope through the found midpoint.

Another way to do this is to write the equation for the set of points that is equidistant from the two endpoints. We'll do that one here.

The square of the distance from (x, y) to the top point is

... d² = (x +5)² +(y -5)²

The square of the distance from (x, y) to the bottom point is

... d² = (x +2)² +(y +4)²

Subtracting the latter from the former gives 0.

... (x +5)² +(y -5)² - (x +2)² -(y +4)² = 0

Simplifying gives ...

... 10x +25 -10y +25 -4x -4 -8y -16 = 0 . . . . realize x² and y² terms cancel

...6x -18y = -30 . . . . . . collect terms, then divide by 6

... x -3y = -5 . . . . . . . standard form equation

Answer with explanation:

Equation of line passing through , (-5,5) and (-2,-4) is, given by

[tex]\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ \frac{y-5}{x+5}=\frac{-4-5}{-2+5}\\\\ \frac{y-5}{x+5}=\frac{-9}{3}\\\\ y-5=-3 \times (x+5)\\\\ y-5=-3 x-15\\\\3 x +y+10=0[/tex]

Equation of line perpendicular to line , a x + b y +c=0 , is given by

→ b x -a y + k=0

→So, equation of line perpendicular to line, 3 x + y +10=0, is

x-3 y +k=0  

→→Mid point of segment that is , (-5,5) and (-2,-4) ,is

[tex]=[\frac{(x_{1}+x_{2})}{2},\frac{(y_{1}+y_{2})}{2}]\\\\=[\frac{(-5-2)}{2},\frac{(5-4)}{2}]\\\\=(\frac{-7}{2},\frac{1}{2})[/tex]

⇒Perpendicular line will pass through the point

    [tex](\frac{-7}{2},\frac{1}{2}).[/tex]

[tex]\frac{-7}{2}-3\times\frac{1}{2}+k=0\\\\k=\frac{10}{2}\\\\k=5[/tex]

→Equation of perpendicular bisector of the segment will be

 x - 3 y +5=0  

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