Question

4. Find the point on the y-axis which is equidistant from the points (5, -2) and (-3, 2).​

Answer 1

☯ Let the given points be P(5, -2) and Q(-3,2) is equidistant from a point R.

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Given that,

• The points are equidistant from y - axis.

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$\therefore$ It's y - coordinate will be 0.

Therefore, Coordinates of point R is (y,0).

⠀⠀⠀⠀$\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}$

• Point R is equidistant from the points P & Q.

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$\dag\;{\underline{\frak{Using\:Distance\:Formula,}}}$

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$\star\;{\boxed{\sf{\pink{d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}}}}$

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Therefore,

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$:\implies\sf\sqrt{\Big(0 -5 \Big)^2 + \Big(y + 2 \Big)^2} = \sqrt{\Big(0 + 3 \Big)^2 + \Big(y - 2 \Big)^2} \\\\\\:\implies\sf 25 + y^2 + 4y + 4 = 9 + y^2 - 4y + 4 \\\\\\:\implies\sf 4y + 29 = -4y + 13\\\\\\:\implies\sf 4y + 4y = 13 - 29\\\\\\:\implies\sf y = \cancel\dfrac{-16}{\:8}\\\\\\:\implies{\underline{\boxed{\frak{\purple{y = -2}}}}}$

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$\therefore\:{\underline{\sf{Hence,\:Required\:point\:is\: {\textsf{\textbf{(-2,0)}}}.}}}$

Answer 2

$\; \; \; \; \; \; \; \; \;{\large{\bold{\sf{\underbrace{\underline{Let's \; understand \; the \; question \; 1^{st}}}}}}}$

This question says that we have to find the point on the y-axis which is equidistant from the points (5, -2) and (-3, 2)

${\large{\bold{\sf{\underline{Given \; that}}}}}$

The y-axis which is equidistant from the points (5, -2) and (-3, 2).

${\large{\bold{\sf{\underline{To \; find}}}}}$

Point on which the y-axis is equidistant from the points (5, -2) and (-3, 2).

${\large{\bold{\sf{\underline{Solution}}}}}$

(-2,0) is the point on which the y-axis is equidistant from the points (5, -2) and (-3, 2).

${\large{\bold{\sf{\underline{Assumption}}}}}$

Let A(5,-2) and B(-3,2) is equidistant from the point C(y,0)

Note : It's because C(y,0) that time it is constant.

${\large{\bold{\sf{\underline{Using \; concepts}}}}}$

Distance formula.

${\large{\bold{\sf{\underline{Using \; formula}}}}}$

${\bold{\sf{\green{d = \sqrt{(x_{2} - x_{1})^{2} +( y _{2} - y_{1})^{2}}}}}}$

${\large{\bold{\sf{\underline{Full \; Solution}}}}}$

${\bullet}$ ${\bold{\sf{d = \sqrt{(x_{2} - x_{1})^{2} +( y _{2} - y_{1})^{2}}}}}$

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

${\bullet}$ 25 + y² + 4y + 4 = 9 + y² - 4y + 4

${\bullet}$ 4y + 29 = -4y + 13

${\bullet}$ 4y + 4y = 13 - 29

${\bullet}$ 8y = -16

${\bullet}$ y = ${\sf{\dfrac{-16}{8}}}$

${\bullet}$ y = (-2,0)

${\purple{\frak{Henceforth, \; (-2,0) \; is \; the \; point \; which \; is\; equidistant \; from \; given \; points}}}$