Question

a point of y axis at distance 5 unit form A(5, -3)
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Answer 1

Answer:

Point is on y axis so x coordinate is 0

So point=(0,y)

Using distance formula=√{(x2-x1)+(y2-y1)}

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

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

$5 = \sqrt{25 + {y}^{2} - 6y+ 9 }$

$5 = \sqrt{ {y}^{2} - 6y + 34 }$

Squaring both the sides

$25 = {y}^{2} - 6x + 34$

${y}^{2} - 6y + 9 = 0$

Product=9

Sum= -6

-3,-3

${y}^{2} - 3y - 3y + 9$

$( {y}^{2} - 3y) - (3y - 9)$

$y(y - 3) - 3(y - 3)$

$(y - 3)(y - 3)$

y-3=0

y=3

Point on y axis=3

Coordinate=(0,3)