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