Question

find the circumtance of a triangle ABC with virtices A(5,1) , B(-3,-7) and C(7,-11).


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Answer 1

Step-by-step explanation:

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Answer 2

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The circumcentre of∆ABC is (2,-4)

The circumcentre of∆ABC is (2,-4)The vertices of ∆ABC are A(5,1) B(-3,-7) and C(7,-1)

Circumcentre: It is point of intersection of perpendicular bisector of each sides of triangle.

# $Midpoint\: of \:side\: AB\: at\: P, =$ $(\frac{5-3}{2},\frac{1-7}{2})$

$Midpoint\: of\: side \:AB \:at \:P,\: =\:(1,-3)$

$Slope \:of \:line \:AB, \:m =$$\dfrac{-7-1}{-3-5}=1$

•$Equation\: of \:perpendicular\: line \:bisector \:of\: AB\: passing\: through \:point \:P$

$\dfrac{-1}{1}(x-1)y+3= 1−1(x−1)$

y=−x−2 ---------- (1)

# $Midpoint\: of \:side \:BC \:at \:Q, =$$(\frac{7-3}{2},\frac{-1-7}{2})$

$Midpoint \:of \:side \:BC\: at \:Q, =(2,-4)$

$Slope\: of \:line\: BC, \:m=$$\dfrac{-7+1}{7+3}=-\dfrac{3}{5}$

• $Equation\: of \:perpendicular\: line \:bisector \:of\: BC\: passing\: through \:point \:Q$

y+4=$-\dfrac{5}{3}(x-2)$

y=$-\dfrac{5}{3}x-\dfrac{2}{3}$------------(2)

$Point \:of\: intersection\: of\: equation (1) \:and \:equation (2)\: is \:circumcentre\: of\: triangle.$

$Using \:substitution\: method:-$

$-\dfrac{5}{3}x-\dfrac{2}{3}=-x-2$

$x=2$

$Put \:x=2 \:into \:y=-x-2$

$y=-4$