Question

The equations to a pair of opposite sides of a parallelogram are $x^2-7x+6=0$ and $y^2-14y+40=0$ ; find the equations to its diagonals.

Answer 1

$x^2-7x+6=0$

factorising the equation

$x^2-6x-x+6=0x(x-6)-1(x-6)=0(x-1)(x-6)=0x=1,6$

$\therefore$ two opossite sides of parallelogram are $x=1$ and $x=6$

$y^2-14y+40=0$

Factorising the equation

$y^2-10y-4y+40=0y(y-10)-4(y-10)=0(y-4)(y-10)=0$

$\therefore$ the two opossite sides of parallelogram are $y=4$ and $y=10$

There fore the vertices of the parallelogram are$A(1,4)B(6,4)C(6,10)D(1,10)$

Equation of line joining any two points is $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Equation of $AC$

$y-4=\frac{10-4}{6-1}(x-1)y-4=\frac{6}{5}(x-1)5y-20=6x-66x-5y+14=0$

Equation of $BD$

$y-4=\frac{10-4}{1-6}(x-6)y-4=\frac{6}{-5}(x-1)-5y+20=6x-366x+5y-56=0$

So the equation of diagonals of parallelogram are $6x-5y+14=0$ and $6x+5y-56=0$