Find the values of y for which the distance between the points P(2, –3)

Solution:

Distance between P(2, – 3) and Q(10, y) = 10 units

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

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

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

$\Rightarrow y^{2}+6 y+9=36$

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

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

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

$\Rightarrow(\mathrm{y}+9)(\mathrm{y}-3)=0$

$\Rightarrow y+9=0$ or $y-3=0$

$\Rightarrow y=-9$ or 3

Hence, there can be two values of y which are – 9 and 3.