Question

Equation to a pair of opposite sides of a parallelogram are $x^2-5x+6=0$ and $y^2-6y+5=0$.

$a$ and $b$ are respectively the sum of the intercepts made by the diagonals on $x$-axis and $y$-axis respectively then $a+b=\frac{k}{2}+10$ then $k$ equals to

Answer 1

$x^2-5x+6=0$

Or $x=2,3$

$y^2-6y+5=0$

Or $y=1,5$

Hence, the vertices of the parallelogram are $A(2,1),B(3,1),C(3,5)$ and $D(2,5)$

Let $BD,AC$ be the two diagonals

$BD=\frac{y-1}{x-3}=\frac{-4}{1}$

$y-1=-4x+12$

$y+4x=13$

$b_1=13,a_1=\frac{13}{4}$ [Intercepts on $x$ axis and $y$ axis]

$AC=\frac{y-1}{x-2}=4$

$y=4x-8+1$

$y=4x-7$

$b_2=-7,a_2=\frac{7}{4}$ [Intercepts on $x$ and $y$ axis respectively]

$a+b=6+\frac{20}{4}=\frac{k}{2}+10$ [$a=a_1+a_2,b=b_1+b_2$]

$\Rightarrow 11=\frac{k}{2}+10$

$\Rightarrow k=2$