Find all possible values of y for which the distance between the points

Question:

Find all possible values of $y$ for which the distance between the points $A(2,-3)$ and $B(10, y)$ is 10 units.

 

Solution:

The given points are $A(2,-3)$ and $B(10, y)$.

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

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

$=\sqrt{64+9+y^{2}+6 y}$

$\because A B=10$

$\therefore \sqrt{64+9+y^{2}+6 y}=10$

$\Rightarrow 73+y^{2}+6 y=100 \quad$ (Squaring both sides)

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

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

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

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

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

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

Hence, the possible values of $y$ are $-9$ and 3 .

 

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