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

$\huge{\orange{\underline{\red{\mathscr{ANSWER:-}}}}}$

$\large{\orange{\underline{\purple{\mathscr{formula:-}}}}}$

$distance \: \: formula = \sqrt{(x2 - x1) {}^{2} + (y2 - y1) {}^{2} }$

$\large{\orange{\underline{\pink{\mathscr{step \: by \:step \:explanation :-}}}}}$

As given

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

As the point on Y-axis is in the form (0, y) .

Thus put all the values in the formula

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

Taking square on both sides

$(5 \sqrt{2} ) {}^{2} = (\sqrt{( - 5) {}^{2} + (y - 8) {}^{2} } ) {}^{2}$

As (√2 )² = 2

Put in the above

25×2=25+(y-8)²

$\huge{\orange{\underline{\red{\mathscr{we \: know \:that}}}}}$

(a - b)² = a² + b² - 2ab

50 = 25 + y² + 64 - 2 × y × 8

50 = 25 + y² + 64 - 16y

= y² - 16y + 89 - 50

= y² - 16y + 39

= y² - 13y - 3y + 39

= y (y - 13) -3 (y - 13)

= (y- 13)(y-3)

y = 13 , y = 3

.°. the value of y are 13 and 3 .

$\huge{\orange{\underline{\red{\mathscr{THANKS.}}}}}$

Answer 2

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

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

◖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;

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