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