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

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

Question

(d) Let the third vertex be $(h, k)$. Obviously the equation of $AB$ is $\sqrt {\frac{5}{2}} $$ \Rightarrow 4x + 7y - 13 = 0$.
Thus equation of $CE$

Vedclass · Accepted Answer

$(-4, -7)$

Vedclass · Answer

(d) Let the third vertex be $(h, k)$. Obviously the equation of $AB$ is $\sqrt {\frac{5}{2}} $$ \Rightarrow 4x + 7y - 13 = 0$.
Thus equation of $CE$ is $7x - 4y + \lambda = 0$, but it passes through origin, so $\lambda = 0$. Hence $7x - 4y = 0$ ....(i)
Also equation of $CE$ is $y - 0 = \frac{k}{h}(x - 0)$
==> $ - kx + hy = 0$......(ii)
Here (i) and (ii) are same lines, therefore $(h, k)$ is $( - 4, - 7)$.

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

Similar Questions

The equation of straight line passing through $( - a,\;0)$ and making the triangle with axes of area ‘$T$’ is

Let the points of intersections of the lines $x-y+1=0$, $x-2 y+3=0$ and $2 x-5 y+11=0$ are the mid points of the sides of a triangle $A B C$. Then the area of the triangle $\mathrm{ABC}$ is .... .

The opposite vertices of a square are $(1, 2)$ and $(3, 8)$, then the equation of a diagonal of the square passing through the point $(1, 2)$, is

The triangle formed by the lines $x + y - 4 = 0,\,$ $3x + y = 4,$ $x + 3y = 4$ is

Let $A B C$ be a triangle formed by the lines $7 x-6 y+3=0, x+2 y-31=0$ and $9 x-2 y-19=0$, Let the point $( h , k )$ be the image of the centroid of $\Delta A B C$ in the line $3 x+6 y-53=0$. Then $h^2+k^2+h k$ is equal to