Question

The circumcentre of a triangle whose vertices are $A(1,3),B(5,5),C(7,5)$

• A. $(2,\frac{-1}{2})$
• B. $(\frac{3}{2},\frac{\sqrt{3}-3}{6})$
• C. $(6,-2)$
• D. $(\frac{5}{4},\frac{-1}{2})$

Answer 1

The correct option is C $(6,-2)$

Step 1 :

Given coordinates $A(1,3),B(5,5),C(7,5)$

Finding the midpoint of AB

$=(\frac{5+1}{2},\frac{3+5}{2})$

$=(3,4)$

Now by finding the slope of $AB=(\frac{5-3}{5-1}=\frac{1}{2})$

Now , slope of perpendicular bisector is negative reciprocal of AB that is slope of the bisector $=-2.$

Step 2 :

Finding the equation of AB with slope -2 and the point (3, 4) is

$y-4=(-2)(x-3)$

$y-4=-2x+6$

$2x+y=4+6$

$2x+y=10$ is equation of the bisector of AB.

Similarly, if we find the equation of bisector of AC we have

Equation of bisector of $AC=>3x+y=16$

Step 3 :

When we solve the given set of equations , we have

$2x+y=10$

$3x+y=16$

$x=6$ and $y=-2$

Step 4 :

Hence the coordinates of the circumcenter of the given triangle is $(6,-2)$

Answer :

$(6,-2)$