Find the circumcentre of the triangle with ve | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the circumcentre be $O(x, y)$. The circumcentre is equidistant from the vertices $A, B$,and $C$. Thus,$OA^2 = OB^2 = OC^2$.Given $A(1, 2)$,$B(

Vedclass · Accepted Answer

$(-0.5, 7.0)$

Vedclass · Answer

(C) Let the circumcentre be $O(x, y)$. The circumcentre is equidistant from the vertices $A, B$,and $C$. Thus,$OA^2 = OB^2 = OC^2$.Given $A(1, 2)$,$B(-2, 2)$,and $C(1, 5)$.$OA^2 = (x - 1)^2 + (y - 2)^2$$OB^2 = (x + 2)^2 + (y - 2)^2$$OC^2 = (x - 1)^2 + (y - 5)^2$Equating $OA^2 = OB^2$:$(x - 1)^2 + (y - 2)^2 = (x + 2)^2 + (y - 2)^2$$(x - 1)^2 = (x + 2)^2$$x^2 - 2x + 1 = x^2 + 4x + 4$$-6x = 3 \implies x = -0.5$Equating $OA^2 = OC^2$:$(x - 1)^2 + (y - 2)^2 = (x - 1)^2 + (y - 5)^2$$(y - 2)^2 = (y - 5)^2$$y^2 - 4y + 4 = y^2 - 10y + 25$$6y = 21 \implies y = 3.5$Thus,the circumcentre is $(-0.5, 3.5)$ or $\left(-\frac{1}{2}, \frac{7}{2}\right)$.

Find the circumcentre of the triangle with vertices $A(1, 2)$,$B(-2, 2)$,and $C(1, 5)$.

Similar Questions

The coordinates of the midpoint of a line segment joining $A (3, -2)$ and $B (-1, 4)$ are:

The area of a triangle having vertices $A(0,0), B(4,0)$ and $C(0,6)$ is:

Obtain the circumcentre of $\Delta ABC$ with vertices $A (-3, -1)$,$B (-1, 3)$ and $C (6, 2)$.

If $P(x, y)$ is equidistant from the points $A(a+b, b-a)$ and $B(a-b, a+b),$ then prove that $bx = ay.$

$P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$ are the given points. If $\overline{PQ}$ is parallel to the $Y$-axis,then $PQ = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module