If the orthocentre of the triangle formed by | vedclass

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

Question

(A) Let the lines be $L_1: y = x + 1$,$L_2: y = 4x - 8$,and $L_3: y = mx + c$. The orthocentre $H$ is $(3, -1)$.First,find the vertex $P$ by solving $

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let the lines be $L_1: y = x + 1$,$L_2: y = 4x - 8$,and $L_3: y = mx + c$. The orthocentre $H$ is $(3, -1)$.First,find the vertex $P$ by solving $L_1$ and $L_2$: $x + 1 = 4x - 8$ $\Rightarrow 3x = 9$ $\Rightarrow x = 3$. Then $y = 3 + 1 = 4$. So,$P = (3, 4)$.The altitude from $P$ to $QR$ passes through $H(3, -1)$. Since $P$ and $H$ both have $x = 3$,the altitude is the vertical line $x = 3$. Thus,$QR$ must be a horizontal line,so $m = 0$.The line $QR$ is $y = c$. Since $H(3, -1)$ is the orthocentre,the line $QH$ is perpendicular to $PR$. The slope of $PR$ (line $L_2$) is $4$,so the slope of $QH$ is $-1/4$.The line $QH$ passes through $H(3, -1)$ and $Q$. $Q$ is the intersection of $L_1$ and $L_3$. $y = x + 1$ and $y = c \Rightarrow x = c - 1$. So $Q = (c - 1, c)$.Slope of $QH = \frac{c - (-1)}{(c - 1) - 3} = \frac{c + 1}{c - 4} = -\frac{1}{4}$.$4c + 4 = -c + 4$ $\Rightarrow 5c = 0$ $\Rightarrow c = 0$.Thus,$m - c = 0 - 0 = 0$.

If the orthocentre of the triangle formed by the lines $y = x + 1$,$y = 4x - 8$,and $y = mx + c$ is at $(3, -1)$,then $m - c$ is :

Similar Questions

The area of the triangle enclosed by the straight lines $x = 0$,$y = 0$ and $x + 2y + 3 = 0$ in sq. units is:

The circumcentre of the triangle formed by the lines $x = -2$,$y = -2$ and $x + y + 2 = 0$ is

The equations of two altitudes of an equilateral triangle are $\sqrt{3}x - y + 8 - 4\sqrt{3} = 0$ and $\sqrt{3}x + y - 12 - 4\sqrt{3} = 0$. The equation of the third altitude is

Let $\alpha, \beta, \gamma, \delta \in \mathbb{Z}$ and let $A(\alpha, \beta), B(1, 0), C(\gamma, \delta)$ and $D(1, 2)$ be the vertices of a parallelogram $ABCD$. If $AB = \sqrt{10}$ and the points $A$ and $C$ lie on the line $3y = 2x + 1$,then $2(\alpha + \beta + \gamma + \delta)$ is equal to

The straight lines $x+y=0$, $5x+y=4$, and $x+5y=4$ form

Vedclass Test Series

Exam Paper Generator

Online Exam Module