Question

22) Find the value of y for which the distance between the points P (2. -3) and (10 y is 10 units.
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Answer 1

Given:

The given points are P(2,-3) and Q(10,y). The distance between these points is 10 units.

To Find:

We have to find the value of y for which the distance between the point P and Q is 10 units.

Solution:

We know, the distance between the two points is given by the following distance formula:

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

Now, we have,

$PQ=10 units$

$x_{1} =2$

$y_{1} =-3$

$x_{2} =10$

$y_{2} =y$

Then, using the above values in the distance formula given above, we get,

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

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

Now, on squaring both sides, we get,

$100=(10-2) ^{2}+(y+3)^{2}$

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

$100=64+(y+3)^{2}$

$(y+3)^{2}=100-64$

$(y+3)^{2}=36$

$(y+3)^{2}=(6)^{2}$

Taking square root on both sides, we get,

$(y+3)=6$

$y=6-3$

$\therefore y=3$

Hence, for the value of  $y=3$, the distance between the points P and Q will be 10 units.