If the point A (2, – 4) is equidistant from P (3, 8) and Q (–10, y), find the values of y. Also find distance PQ
Solution:
Given, the point A(2, -4) is equidistant from P(3, 8) and Q(-10, y)
We have to find the value of y and the distance of PQ.
The point A is equidistant from P and Q
So, PA = QA
The distance between two points P (x₁ , y₁) and Q (x₂ , y₂) is
√[(x₂ - x₁)² + (y₂ - y₁)²]
Distance of P(3, 8) and A(2, -4) = √[(2 - 3)² + (-4 - 8)²]
= √[(-1)² + (-12)²]
= √(1 + 144)
= √145
Distance of Q(-10, y) and A(2, -4) = √[(2 - (-10))² + (y - (-4))²]
= √[(12)² + (y + 4)²]
= √144 + (y + 4)²]
Now, √145 = √144 + (y + 4)²]
On squaring both sides,
145 = 144 + (y + 4)²
By using algebraic identity,
(a - b)² = a² + b² - 2ab
145 = 144 + y² + 8y + 16
145 - 144 - 16 = y² + 8y
-15 = y² + 8y
So, y² + 8y + 15 = 0
On factoring,
y² + 5y + 3y + 15 = 0
y(y + 5) + 3(y + 5) = 0
(y + 3) (y + 5) = 0
Now, y + 3 = 0
y = - 3
Also, y + 5 = 0
y = -5
Therefore, the values of y are -3 and -5.
Distance between P(3, 8) and Q(-10, -3) = √[(-10 - 3)² + (-3 - 8)²]
= √[(-13)² + (-11)²]
= √(169 + 121)
= √290 units
Distance between P(3, 8) and Q(-10, -5) = √[(-10 - 3)² + (-5 - 8)²]
= √[(-13)² + (-13)²]
= √(169 + 169)
= √338 units
Therefore, the distance of PQ is √290 and √338 units.
✦ Try This: Q (0, 1) is equidistant from P (5, - 3) and R (x, 6), find the values of x. Also, find the distances QR and PR.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7
NCERT Exemplar Class 10 Maths Exercise 7.3 Problem 8
If the point A (2, – 4) is equidistant from P (3, 8) and Q (–10, y), find the values of y. Also find distance PQ
Summary:
If the point A (2, – 4) is equidistant from P (3, 8) and Q (–10, y), then the values of y are -3 and -5. Also distance PQ is √290 and √338 units
☛ Related Questions:
visual curriculum