If 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

Solution:

We know that the distance between the two points is given by the Distance Formula = √[( x₂ ^_- x₁ )² + (y₂ - y₁)²]

Q (0, 1) is equidistant from P (5, - 3) and R (x, 6).

So, PQ = QR

Hence by applying the distance formula we get,

√(5 - 0)² + (-3 - 1)² = √(0 - x)² + (1 - 6)²

√(5)² + (- 4)² = √(- x)² + (- 5)²

By squaring both the sides,

25 + 16 = x² + 25

16 = x²

x = ± 4

Therefore, point R is (4, 6) or (- 4, 6).

Case (1): When point R is (4, 6),

Distance between P (5, - 3) and R (4, 6) can be calculated using the Distance Formula as,

PR = √(5 - 4)² + (- 3 - 6)²

= √1² + (- 9)²

= √1 + 81

= √82

Distance between Q (0, 1) and R (4, 6) can be calculated using the distance formula as,

QR = √(0 - 4)² + (1 - 6)²

= √(- 4)² + (- 5)²

= √16 + 25

= √41

Case (2): When point R is (- 4, 6)

Distance between P (5, - 3) and R (- 4, 6) can be calculated using the distance formula as,

PR = √(5 - (- 4))² + (- 3 - 6)²

= √(9)² + (- 9)²

= √81 + 81

= 9√2

Distance between Q (0, 1) and R (- 4, 6) can be calculated using the distance formula as,

QR = √(0 - (- 4))² + (1 - 6)²

= √(4)² + (- 5)²

= √16 + 25

= √41

Thus, we see that using R (- 4, 6) we get PR = QR. Thus, the point is R (- 4, 6). Hence, x = - 4.

☛ Check: NCERT Solutions for Class 10 Maths Chapter 7

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Video Solution:

If 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

NCERT Class 10 Maths Solutions Chapter 7 Exercise 7.1 Question 9

Summary:

The value of x for which Q (0, 1) is equidistant from P (5, - 3) and R (x, 6) is - 4. Also, the distances QR and PR are √41 units.

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☛ Related Questions:

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