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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