Answer :
The coordinates of the required point are (-0.148,1.904).
We know that,
slope= (y2-y1) / (x2-x1)
The slope of BC= (-2-0)/(6-1) = -2/5
Then the equation of the line BC is
y -0=(x-1) when x1 = 1 and y1 = 0
5y = -2x+2...........(1)
The slope of the line perpendicular to BC will be = 1 / (-2/5) = 5/-2
So, the required equation of the line through A(-2,5) and perpendicular to BC is given by
y-y1=m(x-x1)
y-5=(x+2)
-2y+10 = 5x+10
-2y = 5x....................(2)
Hence, the required point will be the point of intersection of lines (1) and (2).
Solving (1) & (2).
(1)⇒5y = -2x+2
(2)⇒-2y = 5x
Now, adding the above two equations, we get
10 y = -4x + 4
-10 y = 25x +0
21x= -4
x = -4/21
And,
y=-2y = 5x => -2y = 5*-4/21
y = 40/21
Thus, the coordinates of the required point are (-0.148,1.904).
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