Find the equation of the perpendicular bisector of AB whereAand B the points (3,6) -3,4) .aldo find its point of intersection with X-axis
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hey here's your answer
let point A(3,6) and C(-3,4)
let midpoint B(x,y)
x=(3+(-3))÷2=0
y=(4+6)÷2=5
B(0,5)
equation to AC
by two point form
x-3y+15=0
we find equation of perpendicular line by bx-ay+k=0
when given line is ax+by+c=0
we will get 3x+y-k=0
we know that line is passing through B(0,5)
putting x and y as 0 and 5 we get k=5
so equation 3x+y-5=0
this is answer to first question
now when it cuts x axis y-coordinate =0
putting y=0 in equation we get x=5/3
so point of intersection with x axis is (0,5/3)
let point A(3,6) and C(-3,4)
let midpoint B(x,y)
x=(3+(-3))÷2=0
y=(4+6)÷2=5
B(0,5)
equation to AC
by two point form
x-3y+15=0
we find equation of perpendicular line by bx-ay+k=0
when given line is ax+by+c=0
we will get 3x+y-k=0
we know that line is passing through B(0,5)
putting x and y as 0 and 5 we get k=5
so equation 3x+y-5=0
this is answer to first question
now when it cuts x axis y-coordinate =0
putting y=0 in equation we get x=5/3
so point of intersection with x axis is (0,5/3)
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