Question

Find the coordinates of point on the y axis which is equidistant from which is equidistant from M(-5,-2) and N (3,2)​

Answer 1

Given :-

Coordinate of point which is equidistant from y-axis is M(-5,-2) and N(3,2)

To Find

We have to find the required point which is equidistant from them

Solution:-

Let the point be P(0,y) we know that on y-axis the value of (x) is 0

Now , by using distance formula

$\sf\ Distance= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ \\ \\ :\implies\sf \ MP= NP\ \ \ \ \big[\sf\ They\ are\ equidistant\big]\\ \\ \\ :\implies\sf\ (MP)^2=(NP)^2\\ \\ \\ :\implies\sf{\small{ \bigg\lgroup\sqrt{\big\{0-(-5)\big\}^2+\big\{y-(-2)\big\}^2}\bigg\rgroup^2= \bigg\lgroup\sqrt{\big\{0-3)\big\}^2+\big\{y-2)\big\}^2}\bigg\rgroup^2}}\\ \\ \\:\implies\sf\big(0+5 \big)^2+\big(y+2\big)^2= \big(0-3\big)^2+\big(y-2\big)^2\\ \\ \\ :\implies\sf\ (5)^2+(y^2+4y+4)= (-3)^2+(y^2-4y+4)\\ \\ \\ :\implies\sf\ 25+ y^2+4y+4= 9+y^2-4y+4\\ \\ \\ :\implies\sf\ \ 29+\cancel{y^2}+4y= 13+\cancel{y^2}-4y\\ \\ \\ :\implies\sf\ 4y+4y=13-29\\ \\ \\ :\implies\sf\ 8y= (-16)\\ \\ \\ :\implies\sf\ y= \cancel{\dfrac{(-16)}{8}}\\ \\ \\ :\implies\underline{\boxed{\purple{\sf\ y= -2}}}$

$\underline{\therefore{\sf\ The\ required\ point\ is\ \ P(0,-2)}}$

Answer 2

Given :-

Two points M(-5,-2) and N (3,2)​

To Find :-

Coordinates

Solution :-

Let us assume the point be P(0,y) we know that on y-axis the value of (x) is 0

Here

$\sf MP ^{2} = NP^{2}$

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

According to question

(0 + 5)² + (y + 2)² = (0 - 3)² + (y - 2)²

5²+ (y² + 4y + 4) = -3² + (y² - 4y + 4)

25 + y² + 4y + 4 = 9 + y² - 4y + 4

25 + 4 = 4y + 4y + 13

13 - 29 = 8y

-16 = 8y

-2 = y