Let the equations of two adjacent sides of a | vedclass

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

Question

$A$ and $C$ point will be $(-4,5)$ and $(3,2)$ mid point of $AC$ will be $\left(-\frac{1}{2}, \frac{7}{2}\right)$ equation of diagonal $BD$ is

$y-\

Vedclass · Accepted Answer

$529$

Vedclass · Answer

$A$ and $C$ point will be $(-4,5)$ and $(3,2)$ mid point of $AC$ will be $\left(-\frac{1}{2}, \frac{7}{2}\right)$ equation of diagonal $BD$ is

$y-\frac{7}{2}=\frac{\frac{7}{2}}{-\frac{1}{2}} \quad\left(x+\frac{1}{2}\right)$

$7 x+y=0$

Distance of $A$ from diagonal $BD$

$= d =\frac{23}{\sqrt{50}}$

$50 d ^2=(23)^2$

$50 d ^2=529$

Let the equations of two adjacent sides of a parallelogram $A B C D$ be $2 x-3 y=-23$ and $5 x+4 y$ $=23$. If the equation of its one diagonal $AC$ is $3 x +$ $7 y=23$ and the distance of A from the other diagonal is $d$, then $50 d ^2$ is equal to $........$.

Similar Questions

If a variable line drawn through the point of intersection of straight lines $\frac{x}{\alpha } + \frac{y}{\beta } = 1$and $\frac{x}{\beta } + \frac{y}{\alpha } = 1$ meets the coordinate axes in $A$ and $B$, then the locus of the mid point of $AB$ is

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

The equation of the base of an equilateral triangle is $x + y = 2$ and the vertex is $(2, -1)$. The length of the side of the triangle is

Without using the Pythagoras theorem, show that the points $(4,4),(3,5)$ and $(-1,-1)$ are vertices of a right angled triangle.

$ABC$ is an isosceles triangle . If the co-ordinates of the base are $(1, 3)$ and $(- 2, 7) $, then co-ordinates of vertex $A$ can be :