Answer:
y = -5x + 7
Step-by-step explanation:
Given that the location of the points are:
A(-4, 1), B(6, 3), C(-1, -5) and D(5, -1)
The location of the bisector of a line is given as:
[tex]x=\frac{x_1+x_2}{2} \\\\y=\frac{y_1+y_2}{2}[/tex]
If E(x1, y1) is the bisector of segment AB then the location of E is at:
[tex]x_1 =\frac{-4+6}{2}=1 \\\\y_1=\frac{1+3}{2}=2[/tex]
Hence E is at (1, 2)
If F(x2, y2) is the bisector of segment CD then the location of F is at:
[tex]x_2 =\frac{-1+5}{2}=2 \\\\y_2=\frac{-5-1}{2}=-3[/tex]
Hence F is at (2, -3)
Hence the line that is a segment bisector of both AB and CD pass through the point (1, 2) and (2, -3).
The equation of the line is given by:
[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\y-2=\frac{-3-2}{2-1}(x-1)\\ \\y-2=-5(x-1)\\\\y-2=-5x+5\\\\y=-5x+7[/tex]
Answer: The equation is: y = -5x + 7.
Step-by-step explanation:
Given that the location of the points are:
A(-4, 1), B(6, 3), C(-1, -5) and D(5, -1).
Midpoint of A(-4, 1), B(6, 3) is = [tex]\left(\frac{\left(-4+6\right)}{2},\frac{\left(1+3\right)}{2}\right)=(1,2)[/tex].
Midpoint of C(-1, -5) and D(5, -1) is [tex]\left(\frac{\left(-1+5\right)}{2},\frac{\left(-5-1\right)}{2}\right)=(2,-3)[/tex].
Using point slope formula:
[tex]\frac{y-2}{x-1}=\frac{-3-2}{2-1}\\\frac{y-2}{x-1}=-5\\y-2=-5x+5\\y=-5x+7[/tex]
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