Question

The equation of a pair of opposite sides of parallelogram are $x^2-5x+6=0$ and $y^2-5y+6=0$. The equation to its diagonals are____?

• A. $x-y=0$ and $y+x=1$
• B. $x+y=0$ and $y+x=1$
• C. $x-y=0$ and $y-x=1$
• D. $x+y=1$ and $y-3=x-2$

Answer 1

The correct option is D $x+y=1$ and $y-3=x-2$

REF.Image.

Now,

$x^2-5x+6=0$

or,$(x-3)(x-2)=0$

or, $x=3,x=2$

Also, $y^2-5y+6=0$

or, $y=3,y=2.$

According to the problem the sides

of the parallelogram one

$x=2,y=2,x=3,y=3$

So, the vertices of the parallelogram

are respectively A(2,2), B(3, 2)

C(3,3),D(2,3).

opposite vertices are A,C & B,D.

so, the equation of diagonal A, C

is, $\frac{y-3}{2-3}=\frac{x-3}{2-3}$

or, x = y

Equation of BD is

$\frac{y-3}{2-3}=\frac{x-2}{3-2}$

or, $y-3=2-x$

or, $x+y=1$