Question

Find the coordinates of the points on Y-axis which are at a distance of 5√2 units
from the point (5,8).​

Answer 1

$\huge\star{\mathfrak{\underline{\pink{Answer}}}}$

$\large\green{\tt{\underline{\underline{Step - by - step -Explanation}}}}$

◖By using Distance formula ;

$\large\red{\boxed{\bold{ Distance \: Formula \: = \: \sqrt{ (x_2 - x_1) ^2 + (y_2 - y_1) ^2} }}}$

◖Now, consitute; the given values

$\large\implies\tt\ 5\sqrt{2} \: = \: \sqrt{ (0 \: - \: 5) ^2 \: + \: (y -8)^2}$

$\large\implies\tt\ 5\sqrt{2} \: = \: \sqrt{ (-5) ^2 \: + \: (y \: - \: 8) ^2}$

◖Squaring both sides;

$\large\implies\tt\ (5\sqrt{2}) ^2 \: = \: (\sqrt{ (-5) ^2 \: + \: (y-8)^2}^2$

$\large\implies\tt\ 50 \: = \: 25 \: + \: (y -8)^2$

◖Using identity $\sf\ (a-b)^2 \: = \: a^2 \: + \: b^2 \: - \: 2ab$ consitute using identity;

$\large\implies\tt\ 50 \: = \: 25 \: + \: y^2 \: + \: 64 \: - 16y$

◖Now solve;

$\large\implies\tt\ 50 \: = \: y^2 \: + \: 89 \:- \: 16y$

$\large\implies\tt\ y^2 \: - \: 16y \: + \: 89 \: -50 \: = \: 0$

$\large\implies\tt\ y^2 \: - \: 16y \: + \: 39$

◖By Middle term splitting ;

❖factors = 13 and 3

$\large\implies\tt\ y^2 \: - \: 13y \: - \: 3y \: + \: 39 \: = \: 0 </p><p></p><p>[tex]\large\implies\tt\ y(y \: - \: 13) \: - \: 3(y \: - \: 13)$

◖Condition 1 :-

$\large\implies\tt\ (y \: - \: 13 ) \: = \: 0$

$\large\implies\tt\ y \: = \: 13$

◖Condition 2 :-

$\large\implies\tt\ (y \: - \: 3) \: = \: 0$

$\large\implies\tt\ y \: = \: 3$

◖Your Answer;

$\large\sf\orange{(\therefore\ y\: = \: 13,3)}$

Answer 2

$\huge\star{\mathfrak{\underline{\red{Answer}}}}$

$\large\pink{\tt{\underline{\underline{Step - by - step -Explanation}}}}$

◖By using Distance formula ;

$\large\red{\boxed{\bold{ Distance \: Formula \: = \: \sqrt{ (x_2 - x_1) ^2 + (y_2 - y_1) ^2} }}}$

◖Now, consitute; the given values

$\large\implies\tt\ 5\sqrt{2} \: = \: \sqrt{ (0 \: - \: 5) ^2 \: + \: (y -8)^2}$

$\large\implies\tt\ 5\sqrt{2} \: = \: \sqrt{ (-5) ^2 \: + \: (y \: - \: 8) ^2}$

◖Squaring both sides;

$\large\implies\tt\ (5\sqrt{2}) ^2 \: = \: (\sqrt{ (-5) ^2 \: + \: (y-8)^2}^2$

$\large\implies\tt\ 50 \: = \: 25 \: + \: (y -8)^2$

◖Using identity $\sf\ (a-b)^2 \: = \: a^2 \: + \: b^2 \: - \: 2ab$ consitute using identity;

$\large\implies\tt\ 50 \: = \: 25 \: + \: y^2 \: + \: 64 \: - 16y$

◖Now solve;

$\large\implies\tt\ 50 \: = \: y^2 \: + \: 89 \:- \: 16y$

$\large\implies\tt\ y^2 \: - \: 16y \: + \: 89 \: -50 \: = \: 0$

$\large\implies\tt\ y^2 \: - \: 16y \: + \: 39$

◖By Middle term splitting ;

❖factors = 13 and 3

$\large\implies\tt\ y^2 \: - \: 13y \: - \: 3y \: + \: 39 \: = \: 0 </p><p></p><p>[tex]\large\implies\tt\ y(y \: - \: 13) \: - \: 3(y \: - \: 13)$

◖Condition 1 :-

$\large\implies\tt\ (y \: - \: 13 ) \: = \: 0$

$\large\implies\tt\ y \: = \: 13$

◖Condition 2 :-

$\large\implies\tt\ (y \: - \: 3) \: = \: 0$

$\large\implies\tt\ y \: = \: 3$

◖Your Answer;

$\large\sf\green{(\therefore\ y\: = \: 13,3)}$