Question

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

Answer 1

Answer:

The value of y are 13 and 3 .

Step-by-step explanation:

Formula

$Distance\ formula = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

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

$25\times 2 =25+(y-8)^{2}$

(As (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

Therefore the value of y are 13 and 3 .

Answer 2

Answer:The value of y are 13 and 3 .Step-by-step explanation:

Formula=√(x1-x2)+(y1-y2)

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

Taking square on both sides

As (√2 )² = 2 Put in the above

(As (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

Therefore the value of y are 13 and 3 .