Question

Find the values of y for which the distance between the points P(2. – 3) and Q(10, y) is
10 units.
24​

Answer 1

Question -

Find the values of y for which the distance between the points P(2,-3) and Q(10,y) is 10 units.

Solution -

Let the points be P(2,-3) and Q(10,y)

Given that PQ= 10 unit

By distance formula,

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

Here,

$x_{1} = 2$

$x_{2} = 10$

$y_{1} = - 3$

$y_{2} = y$

Now putting the values we will get -

$10 = \sqrt{( {10 - 2)}^{2} + {(y - ( - 3)}^{2} }$

$10 = \sqrt{ {8}^{2} + {(y + 3)}^{2} }$

Now Squaring both sides we will get -

${(10)}^{2} = ( { \sqrt{ {8}^{2} + (y + 3) {}^{2} }) }^{2}$

$100 = {8}^{2} + (y + 3) {}^{2}$

$100 = 64 + {y}^{2} + 9 + 6y$

$0 = {y}^{2} + 6y + 64 + 9 - 100$

$0 = {y}^{2} + 6y - 27$

${y}^{2} + 6y - 27 = 0$

${y}^{2} + (9 - 3)y - 27 = 0$

${y}^{2} + 9y - 3y - 27 = 0$

$y(y + 9) - 3(y + 9) = 0$

$(y + 9)(y - 3) = 0$

Either,

$y + 9 = 0 \\ y = - 9$

Or,

$y - 3 = 0 \\ y = 3$

Therefore the value of y are - 9 and 3