Question

If the circumcentre of the triangle with
vertices (a, -1), (6, -9) and (10, b) is (6, -5),
find the values of a and b.​

Answer 1

GIVEN:-

• Circumcentre of the triangle is (6, -5).
• Vertices of triangle are (a, -1), (6, -9) and (10, b).

To Find:-

Values of a and b.

SOLUTION:-

We know the formula of circumcentre of a triangle is,

$\large\boxed{\sf{For\:x=\dfrac{x_{1}+x_{2}+x_{3}}{3}}}$

$\large\boxed{\sf{For\:y=\dfrac{y_{1}+y_{2}+y_{3}}{3}}}$

According to the question,

$\large{\sf{Let\:x_{1},\:x_{2}\:and\:x_{3}\:be\:a,6,10\:respectively.}}$

$\large{\sf{Let\:y_{1},\:y_{2}\:and\:y_{3}\:be\:-1,-9,b\:respectively.}}$

Now,

$\large\Rightarrow{\sf{For\:x=\dfrac{x_{1}+x_{2}+x_{3}}{3}}}$

$\large\Rightarrow{\sf{6=\dfrac{a+6+10}{3}}}$

$\large\Rightarrow{\sf{6\times3=16+a}}$

$\large\Rightarrow{\sf{18=16+a}}$

$\large\Rightarrow{\sf{18-16=a}}$

$\large\therefore\boxed{\sf{\pink{a=2}}}$

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$\large\Rightarrow{\sf{For\:y=\dfrac{y_{1}+y_{2}+y_{3}}{3}}}$

$\large\Rightarrow{\sf{-5=\dfrac{(-1)+(-9)+b}{3}}}$

$\large\Rightarrow{\sf{-5=\dfrac{-1-9+b}{3}}}$

$\large\Rightarrow{\sf{-5\times3=-10+b}}$

$\large\Rightarrow{\sf{-15=-10+b}}$

$\large\Rightarrow{\sf{-15+10=b}}$

$\large\therefore\boxed{\sf{\red{b=-5}}}$

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$\large{\blue{\underline{\boxed{\therefore{\sf{\blue{Value\:of\:a=2\:and\:Value\:of\:b=-5.}}}}}}}$