The equations of the sides AB, BC and CA of a triangle ABC are 2x+y=0, x+py=15a and x-y=3 …

Question

The equations of the sides $AB, BC$ and $CA$ of a triangle $ABC$ are $2x+y=0,\ x+py=15a$ and $x-y=3$ respectively. If its orthocentre is $(2,a),\ -\dfrac12<a<2$, then $p$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

$A = AB\cap CA$: solving $2x+y=0$ and $x-y=3$: $x=1,\ y=-2$, so $A=(1,-2)$.

Altitude from $A$ perpendicular to $BC$ (slope of $BC=-\dfrac1p$): slope of altitude $=p$. This altitude passes through orthocentre $(2,a)$: slope $=\dfrac{a-(-2)}{2-1}=a+2$.

So $p = a+2$ \ldots  (i)

$C = \left(\dfrac{18p-30}{p+1},\dfrac{15p-33}{p+1}\right)$ (intersection of $BC$ and $CA$, computed in the source).

The altitude from $B$ perpendicular to $CA$ (slope 1, so altitude slope $-1$) must pass through orthocentre and $B$. Working through the constraint on slope of $HC$ (perpendicularity condition), the source arrives at:
$$p^2-8p+15=0 \implies p=3 \text{ or } 5$$

Since $p=5$ gives $a=3$ (outside the given range $-\dfrac12<a<2$), we reject it. So $p=3$.

Solutions · 1