Question

Find the circumcentre of the triangle formed by the points $(2,3),(1,-5)$ and $(-1,4)$.

• A. $(\frac{7}{10},\frac{9}{10})$
• B. $(\frac{-7}{10},\frac{9}{10})$
• C. $(\frac{7}{10},\frac{-9}{10})$
• D. $(\frac{-7}{10},\frac{-9}{10})$

Answer 1

Answer: (D)

$(\frac{-7}{10},\frac{-9}{10})$

Given three points A(2,3) , B(1,-5) and C(-1,4)

Mid point of $AB=(\frac{2+1}{2},\frac{3-5}{2})=(\frac{3}{2},-1)$

Slope of $AB=\frac{y_2-y_1}{x_2-x_1}=\frac{-5-3}{1-2}=\frac{-8}{-1}=8$

$\therefore$ Slope of perpendicular bisector $=\frac{-1}{8}$

$\therefore$ Equation of AB with respect to slope $=\frac{-1}{8}$ and coordinate $(\frac{3}{2},-1)$

$y-y_1=m(x-x_1)\Rightarrow y+1=\frac{-1}{8}\left(x\frac{-3}{2}\right)y+1=\frac{-1}{16}(2x-3)⟹16y+16=-2x+32x+16y+13=0⟶\left(1\right)$

Slope of $=(\frac{2-1}{2},\frac{3+4}{2})=(\frac{1}{2},\frac{7}{2})$

Slope of $=\frac{y_2-y_1}{x_2-x_1}=\frac{4-3}{-1-2}=\frac{-1}{3}$

$\therefore$ Slope of perpendicular bisector $=3$

$\therefore$ Equation of AC with respect to slope and 3 coordinate $(\frac{1}{2},\frac{7}{2})$

$\Rightarrow y-y_1=m(x-x_1)y-\frac{7}{2}=3(x-\frac{1}{2})2y-7=6x-3\Rightarrow 6x-2y+4=0\therefore 3x-y+2=0⟶\left(2\right)2x+16y=-13⟶\times 33x-y=-2⟶\times 2\frac{6x+48y=-39-6x\pm 2y=\pm -4}{50y=-35}\therefore y=\frac{-35}{50}=\frac{-7}{10}\therefore x=\frac{-9}{10}$

$\therefore$ The circumcenter of triangle $=(\frac{-9}{10},\frac{-7}{10})$