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

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

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Given the distance between the points P(2,-3) and Q(10,y) is 10 units.

The distance formula :

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

So, given PQ = 10\ units

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

After squaring both sides

\Rightarrow (10-2)^2+(y-(-3))^2 = 100

\Rightarrow (y+3)^2 = 100 - 64

\Rightarrow y+3 = \pm 6

\Rightarrow y = 6 - 3\ or\ y = -6-3

Therefore, the values are  y = 3\ or -9.

Posted by

Divya Prakash Singh

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