Question
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-
A.
$$x+4y=13,\,y=4x-7$$
B.
$$4x+y=13, \,4y=x-7$$
C.
$$4x+y=13, \,y=4x-7$$
D.
$$y-4x=13, \,y+4x=7$$
Answer :
$$4x+y=13, \,y=4x-7$$
Solution :
The sides of parallelogram are $$x=2, \,x=3, \,y=1, \,y=5.$$
$$\therefore $$ Diagonal $$AC$$ is $$\frac{{y - 1}}{{5 - 1}} = \frac{{x - 2}}{{3 - 2}}\,\,{\text{or}}\,\,y = 4x - 7$$
Equation diagonal $$BD$$ is $$\frac{{x - 2}}{{3 - 2}} = \frac{{y - 5}}{{1 - 5}}\,\,{\text{or}}\,\,4x + y = 13$$
The sides of parallelogram are $$x=2, \,x=3, \,y=1, \,y=5.$$
$$\therefore $$ Diagonal $$AC$$ is $$\frac{{y - 1}}{{5 - 1}} = \frac{{x - 2}}{{3 - 2}}\,\,{\text{or}}\,\,y = 4x - 7$$
Equation diagonal $$BD$$ is $$\frac{{x - 2}}{{3 - 2}} = \frac{{y - 5}}{{1 - 5}}\,\,{\text{or}}\,\,4x + y = 13$$