Question

find the equation of a line which is perpendicular bisectors of the line joining points (8, -2) &(6, 4) ​

Answer 1

Answer:

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Step-by-step explanation:

The line joining (-3,8) and (5,-6) has slope of (-6-8)/(5-(-3))=-14/8=-7/4.

The line perpendicular to that has slope (4/7)

The midpoint of the first line is the average of -3 and 5 for x (which is 1) and 8 and -6 for y (or 1). The midpoint is (1,1)

We now have the slope and one point for the perpendicular bisector, and with the point-slope formula, y-y1=m(x-x1), where m is the slope and (x1,y1) a point,

y-1=(4/7)(x-1)

That is y=(4/7)x-(4/7)+1, or y=(4/7)x+(3/7).

The original line has an equation, again by the point slope formula,

of y-(-6)=(-7/4)(x-5)

That is y+6=-(7/4)x+(35/4). 6=24/4

so y=-(7/4)x+(11/4)