Two vertices of an isosceles triangle are (−2, 5) and (4, −1). Find the third vertex if the length of equal sides is
3 root 2 units.
Answers
Step-by-step explanation:
Let ΔABC be the isosceles triangle, the third vertex be C(a,b).
and A(2,0) and B(2,5)
Let AC and BC be equal sides of the triangle
By distance formula we have,
D=
(x
2
−x
1
)+(y
2
−y
1
)
∴AB=
(2−2)
2
+(5−0)
2
=
25
=5
Now,
AC=3 [length of equal sides=3 ]
AC
2
=3
2
=9..........(a)
but by distance formula,
AC
2
=
(a−2)
2
+b
2
...........(b)
Combining (a) and (b)
⇒(a−2)
2
+b
2
=9
⇒a
2
+4−4a+b
2
=9
⇒a
2
+b
2
−4a=5 ---------- (1)
Consider BC
BC
2
=
(a−2)
2
+(b−5)
2
but BC=3
∴(a−2)
2
+(b−5)
2
=9
⇒a
2
+4−4a+b
2
+25−10b=9
⇒a
2
+b
2
−4a−10b+20=0 ------- (2)
⇒−10b+20+5=0 ---------(from 1 )
⇒10b=25
⇒b=
10
25
=
2
5
Now, a
2
+
4
25
−4a=5
⇒a
2
−4a=
4
−5
⇒a
2
−4a+
4
5
=0
⇒a=
2
4±
16−4∗1∗
4
5
=
2
4±
11
=
2
2+
11
,
2
2−
11
∴ Coordinate of third vertex (a,b) = (
2
2+
11
,
2
5
) and (
2
2−
11
,
2
5
)
Answer:
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..