Answer:
Option 1: CD is a perpendicular bisector of AB
Step-by-step explanation:
Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.
Formula to find slope
[tex]m= \frac{y_2-y_1}{x_2-x_1}[/tex]
Formula to Find Distance between two points
[tex]D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]
mAB ( represents , Slope of AB )
1. [tex]mAC= \frac{3-2}{2-5}=\frac{1}{-3}=-\frac{1}{3}[/tex]
2. [tex]mBC=\frac{2-1}{5-8}=\frac{1}{-3}=-\frac{1}{3}[/tex]
3. [tex]mCD=\frac{5-2}{6-5}=\frac{3}{1}=3[/tex]
4. [tex]AC=\sqrt{(3-2)^2+(2-5)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}[/tex]
5. [tex]BC=\sqrt{(2-1)^2+(5-8)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}[/tex]
mAC = mBC , and C is common point , hence these three are collinear points making a straight line whole slope is [tex]-\frac{1}{3}[/tex]
[tex]mAB=-\frac{1}{3}[/tex]
[tex]mCD=3[/tex]
[tex]mAB \times mCD = -\frac{1}{3} \times 3 = -1[/tex]
Hence CD ⊥ AB
Also
From Point 4 and point 5 above , we see that
AC = CB
Hence CD bisect AB at C, also CD ⊥ AB
There fore
CD is a perpendicular bisector of AB
Therefor option 1 is true
