Find the value of $ a $, if the distance between the points $ \mathrm{A}(-3,-14) $ and $ \mathrm{B}(a,-5) $ is 9 units.
Given:
The distance between the points \( \mathrm{A}(-3,-14) \) and \( \mathrm{B}(a,-5) \) is 9 units.
To do:
We have to find the value of \( a \).
Solution:
The distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
The distance between $A(-3,-14)$ and $B(a,-5) is,
$A B=\sqrt{(a+3)^{2}+(-5+14)^{2}}$
$9=\sqrt{(a+3)^{2}+(9)^{2}}$
Squaring on both sides, we get
$81=(a+3)^{2}+81$
$(a+3)^{2} =81-81$
$(a+3)^2=0$
$a+3=0$
$a=-3$
The value of \( a \) is $-3$.
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