The equation of a straight line AC is 2y=-3x+6. Given the point A(-2,6) and AB:BC = 1:5. Find the equation of straight line that passes through point B and perpendicular to line AC.
+130071 Unless we know some specific value for C, there could be many answers to this
Here' s one thing that would work
Construct two circles with radii of 1 and 6 that are centered at ( -2,6)
This will guarantee that if B lies the first circle and the line, and C lies on the second circle and the line, then AB : BC =1:5
Rearrangng the equation of the line we haxe that y = (-3/2)x + 3 (1)
The equation of the first circle is ( x +2)^2 + ( y - 6) = 1 (2)
Putting (1) into (2) and solving for either x we have that
(x + 2)^2 + (-3x/2 + 3 - 6)^2 = 1
x = 2/sqrt (13) - 2 = [ 2 - 2sqrt (13)] /sqrt (13)
And y = (-3/2) [ 2/sqrt (13) - 2 ] + 3 = 6 - 3/sqrt (13) = [6 sqrt (13) - 3 ] /sqrt (13)
A perpendicular line at B will have a slope of 2/3
And the equation of this line is
y = (2/3) ( x -[ 2/sqrt/13 -2]) + (6 - 3/sqrt(13) )
See here : https://www.desmos.com/calculator/zu8eebvbso
