Two vertices of an isosceles triangle are 2 0
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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
Answered by admin @
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)
============================°====================================
Hope this will help you..
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