Question

find the values of y for which the distance between the points p(2, -3) and q (10,y) is 10.
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Answer 1

Question -

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

Solution -

Given points P(2,-3) and Q (10,y)

Let,

$x_1 =2$

$x_2=10$

$y_1= - 3$

$y_2= y$

Now,

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

According to the question -

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

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

Squaring both sides,

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

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

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

$100 = 73 + {y}^{2} + 6y$

${y}^{2} + 6y + 73 - 100 = 0$

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

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

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

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

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

Either,

y+9=0

y = - 9

Or,

y-3=0

y=3

Therefore the value of y is - 9 and 3