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Answer 1

Given :-

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

To Find :-

Value of x

PR and QR

Solution :-

As

Q = Equidistant

QP = QR

For QP

$\sf QP=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

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

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

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

$\sf QP=\sqrt{41}$

For QR

$\sf QR=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

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

$\sf QR=\sqrt{x^2+5^2}$

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

As QP = QR

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

On squaring both the sides

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

$\sf 41=x^2+25$

$\sf 41-25=x^2$

$\sf 16=x^2$

$\sf\sqrt{16}=x$

$\sf\pm 4=x$

For QR

QP = QR

QP = QR = √41

For PR

$\sf PR=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

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

$\sf PR=\sqrt{x^2-10x+25+(6+3)^2}$

$\sf PR=\sqrt{x^2-10x+25+(9)^2}$

$\sf PR=\sqrt{x^2-10x+106}$

If x = 4

$\sf PR=\sqrt{(4)^2-10(4)+106}$

$\sf PR=\sqrt{16-40+106}$

$\sf PR=\sqrt{122-40}$

$\sf PR=\sqrt{82}$

If x = -4

$\sf PR=\sqrt{(-4)^2-10(-4)+106}$

$\sf PR=\sqrt{16+40+106}$

$\sf PR=\sqrt{162}$

$\sf PR=9\sqrt{2}$

Answer 2

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

⇒ P Q = QR

→ √(0-5)² + (1 − (−3))²

=√(x −0)² + (16)² [By using disatnce

$formula = \sqrt{(x2-x₁)² + (y2 − ₁)² ]}$

⇒ 25 + 16 = x² + 25

⇒ 41 = x² + 25

⇒ x² = 16

⇒x=4

Distance between Q(0, 1) and R(4, 6)

QR = √(40)² + (6 − 1)² = √√4² +5²: √16 +25 = √41

Distance between P (5, -3) and R(4, 6)

PR= √(4 − 5)² + (6 + 3)² = √1 +81

= √82

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