If 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

(B) Let the third vertex of $\Delta ABC$ be $C(a, b)$.Let $A(5, -1)$ and $B(-2, 3)$ be the other two vertices.The orthocentre $H$ is at $(0, 0)$.Since

Vedclass · Accepted Answer

$(-4, -7)$

Vedclass · Answer

(B) Let the third vertex of $\Delta ABC$ be $C(a, b)$.Let $A(5, -1)$ and $B(-2, 3)$ be the other two vertices.The orthocentre $H$ is at $(0, 0)$.Since $AH \perp BC$,the product of their slopes is $-1$:$\left( \frac{-1 - 0}{5 - 0} ight) 	imes \left( \frac{b - 3}{a - (-2)} ight) = -1$$\Rightarrow \left( \frac{-1}{5} ight) 	imes \left( \frac{b - 3}{a + 2} ight) = -1$$\Rightarrow b - 3 = 5(a + 2)$ $\Rightarrow b - 3 = 5a + 10$ $\Rightarrow 5a - b + 13 = 0 \dots (1)$Similarly,since $BH \perp AC$,the product of their slopes is $-1$:$\left( \frac{3 - 0}{-2 - 0} ight) 	imes \left( \frac{b - (-1)}{a - 5} ight) = -1$$\Rightarrow \left( \frac{3}{-2} ight) 	imes \left( \frac{b + 1}{a - 5} ight) = -1$$\Rightarrow 3(b + 1) = 2(a - 5)$ $\Rightarrow 3b + 3 = 2a - 10$ $\Rightarrow 2a - 3b - 13 = 0 \dots (2)$Multiplying equation $(1)$ by $3$:$15a - 3b + 39 = 0 \dots (3)$Subtracting equation $(2)$ from $(3)$:$(15a - 2a) + (-3b - (-3b)) + (39 - (-13)) = 0$$13a + 52 = 0 \Rightarrow a = -4$Substituting $a = -4$ into equation $(1)$:$5(-4) - b + 13 = 0$ $\Rightarrow -20 - b + 13 = 0$ $\Rightarrow b = -7$Thus,the third vertex is $(-4, -7)$.

If two vertices of a triangle are $(5, -1)$ and $(-2, 3)$ and its orthocentre is at $(0, 0)$,then the third vertex is:

Similar Questions

$P(a, b)$ is the mid-point of a line segment between the axes. Show that the equation of the line is $\frac{x}{a} + \frac{y}{b} = 2$.

If $A(4, -3)$,$B(3, -2)$,and $C(2, 8)$ are the vertices of a triangle,then its centroid will be

Let $A \equiv (4,4), B \equiv (8,4), C \equiv (4,8)$. If $P, Q, R$ are the midpoints of sides $AB, BC, CA$ respectively and $(\alpha, \beta)$ are the coordinates of the orthocentre of $\Delta PQR$,then the value of $\alpha + \beta$ is

If the midpoints of the sides of a triangle are $(0, 1)$,$(1, 1)$,and $(1, 0)$,find the $x$-coordinate of its incentre.

The orthocentre and centroid of a triangle are $A(-3, 5)$ and $B(3, 3)$ respectively. If $C$ is the circumcentre of this triangle,then the radius of the circle having line segment $AC$ as a diameter is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module