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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.


Solution

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

factorising the equation

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

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

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

Factorising the equation

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

$$\therefore $$ 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 $$y-{ y }_{ 1 }=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ x_{ 2 }-{ x }_{ 1 } } (x-{ x }_{ 1 })$$

Equation of $$AC$$

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

Equation of $$BD$$

$$y-4=\dfrac { 10-4 }{ 1-6 } (x-6)\\ y-4=\dfrac { 6 }{ -5 } (x-1)\\ -5y+20=6x-36\\ 6x+5y-56=0\\$$

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


Mathematics

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