If the sum of squares of all real values of , for which the lines 2x-y+3=0,6x+3y+1=0 and x…

Question

If the sum of squares of all real values of $\alpha $, for which the lines $2x-y+3=0,6x+3y+1=0$ and $\alpha x+2y-2=0$ do not form a triangle is $p$, then the greatest integer less than or equal to $p$ is \_\_\_\_\_\_\_\_\_.

Answer: 32

Accepted Answer

The three lines fail to form a triangle if any two are parallel, or all three are concurrent.

Parallel to $2x-y+3=0$ (slope 2): $-\dfrac{\alpha}2=2 \implies \alpha=-4$.

Parallel to $6x+3y+1=0$ (slope $-2$): $-\dfrac{\alpha}2=-2\implies\alpha=4$.

Concurrency: Solve $2x-y+3=0$ and $6x+3y+1=0$: from the first, $y=2x+3$; substitute: $6x+3(2x+3)+1=0 \implies 12x+10=0 \implies x=-\dfrac56,\ y=\dfrac23$. Substituting into $\alpha x+2y-2=0$:
$$\alpha\left(-\frac56\right)+\frac43-2=0 \implies -\frac{5\alpha}6=\frac23 \implies \alpha=-\frac45$$

$$p = (-4)^2+4^2+\left(-\frac45\right)^2 = 16+16+\frac{16}{25} = 32.64$$

$$\lfloor p\rfloor = 32$$

Solutions · 1