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

Question

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

My Attempt:

Given equations of sides are $4x+5y=0$ and $7x+2y=0$. Solving them, we get: $$x=0, y=0$$.

If $y=mx+c$ is the equation of other diagonal, then $c=0$ and hence $y=mx$.

How do we find $m$?

Answer 1

Given the sides are $4x + 5y = 0 ….(1)$ and $7x + 2y = 0….(2)$

Equation of the diagonal,$11x + 7y = a….(3)$

Solving $(1)$ and $(3)$ coordinates of $A$ are $(x_1 , y_1)$

Solving $(2)$ and $(3)$ coordinates of$ B$ are $(y_1 , y_2)$

Then find the Midpoint of $AB$ is $M (x , y )$

Then you will get equation of the diagonal $OC$

Note that the coordinates of $O$ are $(0,0)$

Answer 2

1. Find intersection of the consective sides
2. Find intersection of sides with diagonal 1
3. Find mid point of diagonal 1
4. As we know diagonals bisect each other in parralellogram

5. Mid point of diagonal 1 should also pass through diagonal 2 which we are trying to find And you have the point of intersection of consective sides(0,0) which also lies on diagonal.

   6.Use 2 point form to find slope($m$) $(y_2 - y_1)/(x_2 - x_1)$

6. Put in $y=mx$

Answer 3

The direction vectors of the sides are $(-5,4)$ and $(-2,7)$; the diagonal’s direction vector is $(-7,11)$. Given any point $\gamma(-7,11)$ for a real $\gamma$ on the diagonal, it must be a linear combination of the side vectors, i. e., we have to solve $$\alpha(-5,4)+\beta(-2,7)=\gamma(-7,11)$$ for $\alpha$ and $\beta$. An easy calculation gives $\alpha=\beta=\gamma$, hence $(-5\gamma,4\gamma)$ and $(-2\gamma,7\gamma)$ are two points of the other diagonal. The usual calculation gives $$y=x+9\gamma$$ as an equation of the straight line through these points.