Question

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.​

Answer 1

Answer:

PQ =[(5-0)+(-3-1)]^1/2

=[5-4]^1/2

1^1/2

=1

QR=PQ=1

1=[(X-0)+(6-1)]^1/2

1=[X+5]^1/2

1^2 =X+5

1=x+5

X=-4

R(-4,6)

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Answer 2

Given:-

Q (0, 1) is equidistant from P (5, – 3)

R (x, 6), which means PQ = QR

To Find :-

The values of x

The distances QR and PR.​

Solution :-

Given that,

Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), which means PQ = QR

Step I: Find the distance between PQ and QR using distance formula,

$\sf PQ=\sqrt{(5-0)^{2}+(-3-1)^{2}}$

$\sf =\sqrt{(-5)^{2}+(-4)^{2}}$

$\sf =\sqrt{25+16}$

$\sf =\sqrt{41}$

$\sf QR=\sqrt{(0-x)^{2}+(1-6)^{2}}$

$\sf =\sqrt{(-5)^{2}+(-4)^{2}}$

$\sf =\sqrt{x^{2}+25}$

Step II: Use PQ = QR

$\longrightarrow \sf \sqrt{41} =\sqrt{x^{2}+25}$

Squaring both the sides, to omit square root

$\longrightarrow \sf 41 = x^{2}+25$

$\implies \sf x^{2} = 16$

$\implies \sf x = \pm 4$

$\longrightarrow \sf x = 4 \ or \ x = -4$

Coordinates of Point R will be R (4, 6) OR R (-4, 6),

If R (4, 6), then QR

$\sf QR=\sqrt{(0-4)^{2}+(1-6)^{2}}$

$\sf =\sqrt{(4)^{2}+(-5)^{2}}$

$\sf =\sqrt{16+25}$

$\sf =\sqrt{41}$

$\sf PR=\sqrt{(5+4)^{2}+(-3+-6)^{2}}$

$\sf =\sqrt{(9)^{2}+(9)^{2}}$

$\sf =\sqrt{81+81}$

$\sf =9\sqrt{2}$