Question

The point A(3,y) is equidistant from the points P(6,5) and Q(0,-3). find the value of y

Answer 1

Point A(3,y) is equidistant from the points P and Q

so,

AP=AQ

squaring both sides

AP^2 = AQ^2

Using distance formula,

(6-3)^2+(5-y) ^2 = (-3)^2 + (-3-y)^2

9+25-10y+y^2 = 9+9+6y+y^2

25-9=6y+10y

16=16y

y=1 [Ans]

Answer 2

Answer:

The value of y = 1

Step-by-step explanation:

Given,

A(3,y) is equidistant from the points P(6,5) and Q(0,-3)

To find,

The value of y

Solution:

Recall the formula

Distance between the points A(x₁,y₁) and B(x₂,y₂) is given by the formula

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

Since it is given that the point A(3,y)  is equidistant from the points P(6,5) and Q(0,-3), we have

AP = AQ

AP = $\sqrt{(6- 3)^2 + (5-y)^2}$

AQ = $\sqrt{(0- 3)^2 + (-3-y)^2}$

Since AP = AQ, we have

$\sqrt{(6- 3)^2 + (5-y)^2}$ =  $\sqrt{(0- 3)^2 + (-3-y)^2}$

Squaring on both sides we have

(6-3)² + (5-y)² = (0-3)²+(-3-y)²

3² + 25+y²-10y = (-3)²+9+y²+6y

9+25+y²-10y = 9+9+y²+6y

Cancelling 9+y² on both sides we get

25-10y = 9+6y

10y+6y = 25 - 9 = 16

16y = 16

y = 1

The value of y = 1

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