Question

find the point on the y-axis which is equidistant from the point - 5 ,- 2 and 3 , 2

Answer 1

Answer :

Let, the point on y-axis be (0, y)

Then, the distance between the points (0, y)

and (- 5, - 2) be

$d_1 = \sqrt{(-5)^{2}+(y+2)^{2}}$

= $\sqrt{25+y^{2}+4y+4}$

= $\sqrt{y^{2}+4y+29}$

and the distance between the points (0, y)

and (3, 2) be

$d_2 = \sqrt{3^{2}+(y-2)^{2}}$

= $\sqrt{9+y^{2}-4y+4}$

= $\sqrt{y^{2}-4y+13}$

By the given condition,

d₁ = d₂

⇒ d₁² = d₂²

⇒ y² + 4y + 29 = y² - 4y + 13

⇒ 8y = - 16

⇒ y = - 2

∴ the required coordinates of the point on y-axis be (0, - 2)

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Answer 2

Answer:

⇒√(0+5)²+(y+2)²=√(0-3)²+(y-2)² (Using Distance Formula)

⇒√(25+y²+4y+4)=√(9+y²-4y+4)

⇒√(y²+4y+29)=√(y²-4y+13)

⇒y²+4y+29=y²-4y+13

⇒4y+29=13-4y

⇒8y=16

⇒y=2