Question

Two vertices of an isosceles triangle are (2,0) and (2,5).Find the third vertex if the length of the equal sides is 3

Answer 1

PA= PB= 3

By Distance Formula
√(x1-x2)² + (y1- y²)²

PA= PB

√ (x-2)² +(y-0)² = √(x-2)²+(y-5)²

Squaring both sides

(x-2)² +y² = (x-2)²+(y-5)²

(x-2)² -(x-2)² +y² =( y-5)²

y² = y²+5²-2×y×5

[ (a-b)² = a²+b²-2ab]

y² = y²+25-10y

10y = 25

y=25/10= 5/2

y= 5/2


√ (x-2)² +y² = 3

Squaring both sides

(x-2)² +y² = 9

[ (a-b)² = a²+b²-2ab]

x²+2²-2×x×2+y²= 9

x²+4-4x +y² =9

x²+4-9-4x+y²=0

x²-5-4x+y²=0

Put the value of y

x²- 4x-5+(5/2)²=0

x²-4x-5+25/4=0

x²-4x (-20+25)/4=0

x²-4x +5/4=0

On Comparing

a= 1, b= -4, c= 5/4

By quadratic Formula

X= -b ±√b²-4ac/ 2a

X= -(-4)±√-4²-4×1×5/4 /2×1

X= 4 ±√16-5/2

X= 4/2 ±√11/2

X= 2 ±√11/2

Hence the third vertex P(x,y)
= (2 ±√11/2, 5/2)

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Hope this will help you..

Answer 2

Answer:

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