Question

A(-3,2),b(3,2) and c(-3,-2) are the vertices of the right angled tringle

Answer 1

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Answer:


Step-by-step explanation:

Formula used:


Midpoint formula:


The midpoint of the line joining\begin{lgathered}(x_1,y_1) and (x_2,y_2) is\\\\(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\end{lgathered}
(x
1
​
,y
1
​
)and(x
2
​
,y
2
​
)is
(
2
x
1
​
+x
2
​

​
,
2
y
1
​
+y
2
​

​
)
​



Distance formula:


The distance between two points\begin{lgathered}(x_1,y_1) and (x_2,y_2) is\\\\d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\end{lgathered}
(x
1
​
,y
1
​
)and(x
2
​
,y
2
​
)is
d=
(x
1
​
−x
2
​
)
2
+(y
1
​
−y
2
​
)
2

​

​




Let S be the midpoint of hypotenuse BC.

Then, S is

(\frac{3-3}{2},\frac{2-2}{2})(
2
3−3
​
,
2
2−2
​
)

(0,0)


Now,

\begin{lgathered}SA=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\SA=\sqrt{(0+3)^2+(0-2)^2}\\SA=\sqrt{9+4}\\SA=\sqrt{13}\\\\SB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\SB=\sqrt{(0-3)^2+(0-2)^2}\\SB=\sqrt{9+4}\\SB=\sqrt{13}\\\\SC=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\SC=\sqrt{(0+3)^2+(0+2)^2}\\SC=\sqrt{9+4}\\SC=\sqrt{13}\end{lgathered}
SA=
(x
1
​
−x
2
​
)
2
+(y
1
​
−y
2
​
)
2

​

SA=
(0+3)
2
+(0−2)
2

​

SA=
9+4
​

SA=
13
​

SB=
(x
1
​
−x
2
​
)
2
+(y
1
​
−y
2
​
)
2

​

SB=
(0−3)
2
+(0−2)
2

​

SB=
9+4
​

SB=
13
​

SC=
(x
1
​
−x
2
​
)
2
+(y
1
​
−y
2
​
)
2

​

SC=
(0+3)
2
+(0+2)
2

​

SC=
9+4
​

SC=
13
​

​


This implies

SA=SB=SC


Hence S is equidistant from A, B and C