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

If the points $A(1, 2)$,$O(0, 0)$,and $C(a, b)$ are collinear,then:

Prove that the centroid of the triangle with vertices $(1, a), (2, b), (c^2, -3)$ does not lie on the $Y$-axis.

The points $A(2, 9)$,$B(a, 5)$,and $C(5, 5)$ are the vertices of a triangle $ABC$ right-angled at $B$. Find the value of $a$ and the area of $\triangle ABC$.

$A$ median of a triangle divides it into two triangles of equal areas. Verify this result for $\Delta ABC$ whose vertices are $A (4, -6), B (3, -2)$ and $C (5, 2)$. Find the area of each triangle (in square units).

Find the area of a quadrilateral having the vertices $(-1, -1), (-4, -2), (-5, -4),$ and $(-2, -3)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module