Find the value of $ a $, if the distance between the points $ \mathrm{A}(-3,-14) $ and $ \mathrm{B}(a,-5) $ is 9 units.

AcademicMathematicsNCERTClass 10

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 $-3$.

raja
Updated on 10-Oct-2022 13:28:28

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