Animated Solution for Mathematics - Straight Lines: The equations to a pair of opposite sides of parallelogram are $x^2-5x+6=0$ and $y^2-6y+5=0$, the equations to its diagonals are

The problem begins with two quadratic equations that define the boundaries of our geometric figure. First, consider the equation:

By factoring this quadratic, we obtain $(x-2)(x-3)=0$. This reveals that the parallelogram is bounded by the two vertical lines:

Next, we examine the second equation:

Factoring this yields $(y-1)(y-5)=0$. Consequently, the figure is also bounded by two horizontal lines:

The intersection of these vertical and horizontal lines defines the four vertices of the parallelogram. By pairing the $x$ and $y$ values, we identify the coordinates as follows:

The intersection of $x=2$ and $y=1$ gives vertex $A(2,1)$.

The intersection of $x=3$ and $y=1$ gives vertex $B(3,1)$.

The intersection of $x=3$ and $y=5$ gives vertex $C(3,5)$.

The intersection of $x=2$ and $y=5$ gives vertex $D(2,5)$.

The diagonals are formed by connecting opposite vertices. We calculate the equations for these lines using the slope formula $m=\frac{y_2-y_1}{x_2-x_1}$ and the point-slope form $y-y_1=m(x-x_1)$.

For diagonal $AC$ connecting $(2,1)$ and $(3,5)$:

Applying the point-slope form:

For diagonal $BD$ connecting $(3,1)$ and $(2,5)$:

Applying the point-slope form:

By decomposing the joint equations, we have successfully determined the geometry of the parallelogram. The diagonals of the figure are defined by the linear equations $y=4x-7$ and $4x+y=13$.

Remember, the key to mastering JEE problems is not just memorizing formulas, but visualizing the geometry behind the algebra. Keep practicing, and you will find that even the most complex problems have an elegant, simple solution.