Two sides of a parallelogram are along the li | vedclass

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

Question

Both the lines pass through origin.
point $D$ is equal of intersection of $4 x+5 y=0\,  \,11 x+7 y=9$
So, coordinates of point $D=\left(\frac{5

Vedclass · Accepted Answer

$(2,2)$

Vedclass · Answer

Both the lines pass through origin.
point $D$ is equal of intersection of $4 x+5 y=0\,  \,11 x+7 y=9$
So, coordinates of point $D=\left(\frac{5}{3},-\frac{4}{3}\right)$
Also, point $B$ is point of intersection of $7 x+2 y=0\, \, 11 x+7 y=9$
So, coordinates of point $B=\left(-\frac{2}{3}, \frac{7}{3}\right)$
diagonals of parallelogram intersect at middle let middle point of $B,D$
$\Rightarrow\left(\frac{\frac{5}{3}-\frac{2}{3}}{2}, \frac{-4}{3}+\frac{7}{3}\right)=\left(\frac{1}{2}, \frac{1}{2}\right)$
equation of diagonal $AC$
$\Rightarrow(y-0)=\frac{\frac{1}{2}-0}{\frac{1}{2}-0}(x-0)$
$y=x$
diagonal $AC$ passes through $(2,2)$

Two sides of a parallelogram are along the lines $4 x+5 y=0$ and $7 x+2 y=0$. If the equation of one of the diagonals of the parallelogram is $11 \mathrm{x}+7 \mathrm{y}=9$, then other diagonal passes through the point:

Similar Questions

The equation of perpendicular bisectors of the sides $AB$ and $AC$ of a triangle $ABC$ are $x - y + 5 = 0$ and $x + 2y = 0$ respectively. If the point $A$ is $(1,\; - \;2)$, then the equation of line $BC$ is

For a point $P$ in the plane, let $d_1(P)$ and $d_2(P)$ be the distance of the point $P$ from the lines $x-y=0$ and $x+y=0$ respectively. The area of the region $R$ consisting of all points $P$ lying in the first quadrant of the plane and satisfying $2 \leq d_1(P)+d_2(P) \leq 4$, is

If $A$ and $B$ are two points on the line $3x + 4y + 15 = 0$ such that $OA = OB = 9$ units, then the area of the triangle $OAB$ 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

Two consecutive sides of a parallelogram are $4x + 5y = 0$ and $7x + 2y = 0.$ If the equation to one diagonal is $11x + 7y = 9,$ then the equation of the other diagonal is