Question

6.) prove that c is on the perpendicular bisector of segment ab.

Explanation:

Step1: Recall the property

A point is on the perpendicular - bisector of a segment if and only if it is equidistant from the endpoints of the segment. So, we need to show that \(CA = CB\).
Let the coordinates of \(A=(x_1,y_1)\), \(B=(x_2,y_2)\) and \(C=(x_0,y_0)\).
The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
The distance \(CA=\sqrt{(x_0 - x_1)^2+(y_0 - y_1)^2}\) and \(CB=\sqrt{(x_0 - x_2)^2+(y_0 - y_2)^2}\).
We need to use the given information (not provided in this problem - statement, but in a general case) to show that \(\sqrt{(x_0 - x_1)^2+(y_0 - y_1)^2}=\sqrt{(x_0 - x_2)^2+(y_0 - y_2)^2}\), or by squaring both sides (since both sides are non - negative), \((x_0 - x_1)^2+(y_0 - y_1)^2=(x_0 - x_2)^2+(y_0 - y_2)^2\).
Expand the equations:
\(x_0^2-2x_0x_1 + x_1^2+y_0^2-2y_0y_1 + y_1^2=x_0^2-2x_0x_2 + x_2^2+y_0^2-2y_0y_2 + y_2^2\).
Cancel out \(x_0^2\) and \(y_0^2\) on both sides:
\(-2x_0x_1 + x_1^2-2y_0y_1 + y_1^2=-2x_0x_2 + x_2^2-2y_0y_2 + y_2^2\).
If we are given geometric properties such as congruent triangles, we can also prove \(CA = CB\) using triangle - congruence criteria. For example, if we can show that \(\triangle ACD\cong\triangle BCD\) (where \(D\) is the mid - point of \(AB\)) using SSS (Side - Side - Side), SAS (Side - Angle - Side), ASA (Angle - Side - Angle) or AAS (Angle - Angle - Side) criteria, then \(CA = CB\).
Once we have shown that \(CA = CB\), by the converse of the perpendicular - bisector theorem, we can conclude that \(C\) is on the perpendicular bisector of segment \(AB\).

Answer:

We need to show that the distance from \(C\) to \(A\) is equal to the distance from \(C\) to \(B\) (i.e., \(CA = CB\)) using distance formula or triangle - congruence criteria. If \(CA = CB\), then by the converse of the perpendicular - bisector theorem, \(C\) is on the perpendicular bisector of segment \(AB\).