Question

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Answer 1

Question

The line joining A(a,3) to B(2,-3) is perpendicular to the line joining C(10,1) to B.Find the value of a

Given

Points A(a,3) , B(2,-3) , C(10,1)

To Find

Value of a

Points A(a,3) B(2,-3)

Slope of AB =$\sf m_{1}\dfrac{ - 3 - 3}{2 - a } = \dfrac{ - 6}{2 - a}$

Points C(10,1) ,B(2,-3)

Slope of BC=$\sf m_{2} \dfrac{ - 3 - 1}{2 - 10} = \dfrac{ - 4}{ - 8} = \dfrac{1}{2}$

Since the both lines are perpendicular to each other so ,cross product of their slope is equal to -1.

$\sf\boxed{ m_{1}\times m_{2}=-1}$

$\sf= > \dfrac{ - 6}{2 - a} \times \dfrac{1}{2} = - 1$

$\sf= > \dfrac{6}{2 - a} \times \dfrac{1}{2} = 1$

$\sf = > 6 = 2(2 - a)$

$\sf= > 3 = 2 - a$

$\sf = > - a = 1$

$\sf\bold{a = - 1}$