Two vertices of a triangle are (5, -1) and (- | vedclass

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

Question

(C) Let the vertices of the triangle be $A(h, k)$,$B(5, -1)$,and $C(-2, 3)$. The origin $O(0, 0)$ is the orthocentre. Since $AO \perp BC$,the slope of

Vedclass · Accepted Answer

$(-4, -7)$

Vedclass · Answer

(C) Let the vertices of the triangle be $A(h, k)$,$B(5, -1)$,and $C(-2, 3)$. The origin $O(0, 0)$ is the orthocentre. Since $AO \perp BC$,the slope of $AO 	imes$ slope of $BC = -1$. Slope of $BC = \frac{3 - (-1)}{-2 - 5} = \frac{4}{-7}$. Slope of $AO = \frac{k - 0}{h - 0} = \frac{k}{h}$. Thus,$\frac{k}{h} 	imes \left(-\frac{4}{7}ight) = -1$ $\Rightarrow \frac{k}{h} = \frac{7}{4}$ $\Rightarrow 7h - 4k = 0$ (Eq. $1$). Since $BO \perp AC$,the slope of $BO 	imes$ slope of $AC = -1$. Slope of $BO = \frac{-1 - 0}{5 - 0} = -\frac{1}{5}$. Slope of $AC = \frac{k - 3}{h - (-2)} = \frac{k - 3}{h + 2}$. Thus,$\left(-\frac{1}{5}ight) 	imes \left(\frac{k - 3}{h + 2}ight) = -1$ $\Rightarrow k - 3 = 5(h + 2)$ $\Rightarrow 5h - k + 13 = 0$ (Eq. $2$). From Eq. $1$,$k = \frac{7h}{4}$. Substituting this into Eq. $2$: $5h - \frac{7h}{4} + 13 = 0$ $\Rightarrow \frac{20h - 7h}{4} = -13$ $\Rightarrow 13h = -52$ $\Rightarrow h = -4$. Then $k = \frac{7(-4)}{4} = -7$. Therefore,the third vertex is $(-4, -7)$.

Two vertices of a triangle are $(5, -1)$ and $(-2, 3)$. If the origin $(0, 0)$ is the orthocentre of this triangle,then the coordinates of the third vertex of that triangle are

Similar Questions

The orthocentre of the triangle whose vertices are $(0, 0)$,$(3, 0)$,and $(0, 4)$ is

If $P(2, 2)$,$Q(-2, 4)$,and $R(3, 4)$ are the vertices of $\triangle PQR$,then the equation of the median through vertex $R$ is $........$

If $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$ are the vertices of a triangle with side lengths $a, b, c$ opposite to vertices $A, B, C$ respectively,then the excentre with respect to vertex $B$ is:

The circumcentre of the triangle formed by the points $A(1, \sqrt{3})$,$B(-1, -\sqrt{3})$,and $C(3, -\sqrt{3})$ is

If $P(6,1)$ is the orthocentre of the triangle whose vertices are $A(5,-2)$,$B(8,3)$,and $C(h, k)$,then the point $C$ lies on the circle:

Vedclass Test Series

Exam Paper Generator

Online Exam Module