Question

Find the value of a, if the distance between the poinys A(-3 , -14) and B(a, -5) is 9 units​

Answer 1

Step-by-step explanation:

A(-3,-14) B(a,-5)

distance=√(x2-x1)²+(y2-y1)²

9=√(a+3)²+(-5+14)²

81=a²+6a+9+81

a²+6a+9=0

a²+3a+3a+9=0

a(a+3)+3(a+3)=0

(a+3)(a+3)=0

a=-3

Answer 2

SOLUTION

GIVEN

The distance between the points A(-3 , -14) and B(a, -5) is 9 units

TO DETERMINE

The value of a

FORMULA TO BE IMPLEMENTED

If two two points

$\sf{(x_1,y_1) \: \: and \: \: (x_2,y_2) \: are \: given}$

Then the distance between them

$= \sf{ \sqrt{ {(x_1 - x_2)}^{2} + {(y_1 - y_2)}^{2} } }$

EVALUATION

Here the given points are (-3 , -14) and B(a, -5)

Now the distance between them

$= \sf{ \sqrt{ {(a + 3)}^{2} + {( - 5 + 14)}^{2} } }$

$= \sf{ \sqrt{ {(a + 3)}^{2} + {( - 5 + 14)}^{2} } }$

$= \sf{ \sqrt{ {(a + 3)}^{2} + {( 9)}^{2} } }$

So by the given condition

$\sf{ \sqrt{ {(a + 3)}^{2} + {( 9)}^{2} } } = 9$

Squaring both sides

$\sf{ {(a + 3)}^{2} + {( 9)}^{2} = {9}^{2} }$

$\implies \: \sf{ {(a + 3)}^{2} = 0 }$

$\implies \: \sf{ {(a + 3)} = 0 }$

$\implies \: \sf{a = - 3 }$

FINAL ANSWER

The required value of a is - 3

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