Question

4x3=12
4)Answer the following questions
a)Find the point on y-axis which is equidistant from the point (-5,2) and(9,-2)​

Answer 1

☯ Let the given point be A (-5,2) and B (9,-2) is equidistant from point P.

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Given that,

• The given points are equidistant from y - axis.

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$\therefore$ It x - coordinate will be 0.

Therefore, Coordinates of P is (0,y).

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

• Point P is equidistant from A and B.

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$:\implies\sf AP = BP$

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

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

Therefore,

$:\implies\sf \sqrt{(0 - (-5))^2 + (y - 2)^2} = \sqrt{(0 - 9)^2 + (y - (-2))^2}\\ \\ \\ :\implies\sf \bigg( \sqrt{(0 - (-5))^2 + (y - 2)^2} \bigg)^2= \bigg(\sqrt{(0 - 9)^2 + (y - (-2))^2} \bigg)^2$

$:\implies\sf (0 - (-5))^2 + (y - 2)^2 = (0 - 9)^2 + (y - (-2))^2\\ \\ \\ :\implies\sf 25 + y^2 + 4 - 4y = 81 + y^2 + 4 + 4y\\ \\ \\ \\ :\implies\sf \cancel{y^2} - 4y + 29 = \cancel{y^2} + 4y + 85\\ \\ \\ \\ :\implies\sf - 4y + 29 = 4y + 85\\ \\ \\ \\ :\implies\sf - 4y - 4y = 85 - 29\\ \\ \\ \\ :\implies\sf - 8y = 56\\ \\ \\ :\implies\sf y = - \cancel{\dfrac{56}{8}}\\ \\ \\:\implies{\underline{\boxed{\frak{\pink{y = - 7}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Hence,\:the\:required\:point\:on\:y-axis\:is\:{\textsf{\textbf{(0,-7)}}}.}}}$

Answer 2

Given:-

• The given points are equidistant from y - axis.

X - coordinate will be 0.

So,

Coordinates of P = ( 0 , y )

Solution:-

We use distance formula:-

$\tt\implies \: \sqrt{ {(0 - ( -5)] }^{2} + {(y - 2)}^{2} } = \sqrt{{(0 - 9)}^{2} + {[ y - ( -2 ) ]}^{2}}$

$\tt\implies \: {(\sqrt{ {(0 - ( -5)] }^{2} + {(y - 2)}^{2} }) }^{2} = {(\sqrt{{(0 - 9)}^{2} + {[ y - ( -2 ) ]}^{2}} )}^{2}$

$\tt\implies \: { {[0 - ( -5)] }^{2} + {(y - 2)}^{2} } = {{(0 - 9)}^{2} + {[ y - ( -2 ) ]}^{2}}$

$\tt\implies \: 25 + {y}^{2} + 4 - 4y = 81 + {y}^{2} + 4 + 4y$

$\tt\implies \: {y}^{2} - {y}^{2} - 4y - 4y = 85 - 29$

$\tt\implies \: - 8y = 56$

$\tt\implies \: y = \cancel\frac{ \: \: 56 \: \: }{ - 8}$

$\tt\implies \: y = -7$

$\boxed{\tt \: The\: \: required \: \: point \: \: in \: \: y - axis = ( 0 , -7 )}$