Question

Find the values of y for which the distance between the points P(2, -5) and Q(10, y) is 10 units."

Answer 1

I think answer will be 13

Answer 2

The value of y = 1 or - 11

Given :

The distance between the points P(2, -5) and Q(10, y) is 10 units

To find :

The value of y

Formula :

For the given two points ( x₁ , y₁) & (x₂ , y₂) the distance between the points

$= \sf{ \sqrt{ {(x_2 -x_1 )}^{2} + {(y_2 -y_1 )}^{2} } }$

Solution :

Step 1 of 2 :

Form the equation to find the value of y

Here it is given that distance between the points P(2, - 5) and Q(10, y) is 10 units

By the given condition

$\displaystyle \sf \sqrt{ {(10 - 2)}^{2} + { \{ y - ( - 5) \}}^{2} } = 10$

Step 2 of 2 :

Find the value of y

$\displaystyle \sf \sqrt{ {(10 - 2)}^{2} + { \{ y - ( - 5) \}}^{2} } = 10$

$\displaystyle \sf{ \implies } \sqrt{ {8}^{2} + {(y + 5)}^{2} } = 10$

$\displaystyle \sf{ \implies } {8}^{2} + {(y + 5)}^{2} = {10}^{2}$

$\displaystyle \sf{ \implies } 64 + {(y + 5)}^{2} = 100$

$\displaystyle \sf{ \implies } {(y + 5)}^{2} = 100 - 64$

$\displaystyle \sf{ \implies } {(y + 5)}^{2} = 36$

$\displaystyle \sf{ \implies } {(y + 5)}^{2} = {6}^{2}$

$\displaystyle \sf{ \implies }y + 5 = \pm 6$

Now ,

y + 5 = + 6 gives y = 1

y + 5 = - 6 gives y = - 11

Hence the required value of y = 1 or - 11

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