Question

1. the endpoints of the diameter of a circle are m(-3, 5) and n(9, 7). determine a) the coordinates of the centre of the circle b) the radius of the circle 9. determine whether the triangle with vertices a(-3, 4), b(-1, -2), and c(3, 2) is isosceles.

Explanation:

Step1: Find center of circle

The center of a circle is the mid - point of its diameter. The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For points $M(-3,5)$ and $N(9,7)$, we have $x=\frac{-3 + 9}{2}=\frac{6}{2}=3$ and $y=\frac{5 + 7}{2}=\frac{12}{2}=6$.

Step2: Calculate radius of circle

The radius is the distance from the center to a point on the circle. Using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, with the center $(3,6)$ and point $N(9,7)$. We have $r=\sqrt{(9 - 3)^2+(7 - 6)^2}=\sqrt{6^2+1^2}=\sqrt{36 + 1}=\sqrt{37}$.

Step3: Check if triangle is isosceles

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For $AB$ with $A(-3,4)$ and $B(-1,-2)$: $AB=\sqrt{(-1+3)^2+(-2 - 4)^2}=\sqrt{2^2+(-6)^2}=\sqrt{4 + 36}=\sqrt{40}=2\sqrt{10}$.
For $BC$ with $B(-1,-2)$ and $C(3,2)$: $BC=\sqrt{(3 + 1)^2+(2 + 2)^2}=\sqrt{4^2+4^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$.
For $AC$ with $A(-3,4)$ and $C(3,2)$: $AC=\sqrt{(3 + 3)^2+(2 - 4)^2}=\sqrt{6^2+(-2)^2}=\sqrt{36 + 4}=\sqrt{40}=2\sqrt{10}$.
Since $AB = AC=2\sqrt{10}$, the triangle is isosceles.

Answer:

a) $(3,6)$
b) $\sqrt{37}$

1. Yes, the triangle is isosceles.