If O, G, S are respectively the orthocentre,c | vedclass

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

Question

(B) The vertices of the triangle are $A(2,3)$,$B(2,4)$,and $C(4,3)$.Since $AB$ is vertical (along $x=2$) and $AC$ is horizontal (along $y=3$),the tria

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) The vertices of the triangle are $A(2,3)$,$B(2,4)$,and $C(4,3)$.Since $AB$ is vertical (along $x=2$) and $AC$ is horizontal (along $y=3$),the triangle is a right-angled triangle with the right angle at $A$.For a right-angled triangle,the orthocentre $O$ is the vertex at the right angle. Thus,$O = A = (2,3)$.Therefore,$AO^2 = (2-2)^2 + (3-3)^2 = 0$.The centroid $G$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}ight) = \left(\frac{2+2+4}{3}, \frac{3+4+3}{3}ight) = \left(\frac{8}{3}, \frac{10}{3}ight)$.We calculate $9BG^2 = 9 	imes \left[ (2 - \frac{8}{3})^2 + (4 - \frac{10}{3})^2 ight] = 9 	imes [(-\frac{2}{3})^2 + (\frac{2}{3})^2] = 9 	imes [\frac{4}{9} + \frac{4}{9}] = 9 	imes \frac{8}{9} = 8$.The centroid $G$ divides the line segment joining the orthocentre $O$ and the circumcentre $S$ in the ratio $2:1$. Using the section formula,$G = \frac{1 \cdot O + 2 \cdot S}{3}$ $\Rightarrow 3G = O + 2S$ $\Rightarrow S = \frac{3G - O}{2}$.$S = \frac{3(\frac{8}{3}, \frac{10}{3}) - (2,3)}{2} = \frac{(8-2, 10-3)}{2} = \frac{(6, 7)}{2} = (3, \frac{7}{2})$.Now,$4CS^2 = 4 	imes [(4-3)^2 + (3-\frac{7}{2})^2] = 4 	imes [1^2 + (-\frac{1}{2})^2] = 4 	imes [1 + \frac{1}{4}] = 4 	imes \frac{5}{4} = 5$.Finally,$AO^2 + 9BG^2 + 4CS^2 = 0 + 8 + 5 = 13$.

If $O, G, S$ are respectively the orthocentre,centroid and circumcentre of a triangle whose vertices are $A(2,3), B(2,4)$ and $C(4,3)$,then $AO^2 + 9BG^2 + 4CS^2 =$

Similar Questions

$A$ triangle has a vertex at $(1, 2)$ and the midpoints of the two sides through it are $(-1, 1)$ and $(2, 3)$. Then the centroid of this triangle is

The distance of the origin from the centroid of the triangle whose two sides have the equations $x - 2y + 1 = 0$ and $2x - y - 1 = 0$ and whose orthocenter is $\left(\frac{7}{3}, \frac{7}{3}\right)$ is

What is the distance of the orthocenter from the vertex of a triangle with vertices $(2, 3), (4, 5),$ and $(-1, 10)$?

The distance (in units) between the circumcentre and the centroid of the triangle formed by the vertices $(1, 2)$,$(3, -1)$ and $(4, 0)$ is

Let $A(6,8)$,$B(10 \cos \alpha, -10 \sin \alpha)$ and $C(-10 \sin \alpha, 10 \cos \alpha)$ be the vertices of a triangle. If $L(a, 9)$ and $G(h, k)$ are its orthocenter and centroid respectively,then $(5a - 3h + 6k + 100 \sin 2\alpha)$ is equal to . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module