Let ABC be the triangle such that the equations of lines AB and AC be 3y-x=2 and x+y=2, re…

Question

Let $ABC$ be the triangle such that the equations of lines $AB$ and $AC$ be $3y-x=2$ and $x+y=2$, respectively, and the points $B$ and $C$ lie on $x$-axis. If $P$ is the orthocentre of the triangle $ABC$, then the area of the triangle $PBC$ is equal to:

Accepted Answer

$AB: 3y-x=2$ meets $y=0$ at $x=-2 \implies B=(-2,0)$.

$AC: x+y=2$ meets $y=0$ at $x=2 \implies C=(2,0)$.

$BC$ is the $x$-axis, so the altitude from $A$ is vertical. Line $AB$ and $AC$ intersect at $A$: solving $3y-x=2$ and $x+y=2$ gives $y=1,\ x=1$, so $A=(1,3)$... Using the altitude conditions directly (altitude from $B$ perpendicular to $AC$, and from $C$ perpendicular to $AB$):

Altitude from $C(2,0)$ perpendicular to $AB$ (slope $\frac13$) has slope $-3$: $y=-3(x-2)=-3x+6$.

Altitude from $B(-2,0)$ perpendicular to $AC$ (slope $-1$) has slope $1$: $y = x+2$.

Solving: $x+2=-3x+6 \implies 4x=4\implies x=1,\ y=3$. So orthocentre $P=(1,3)$.

$$\text{ar}(\triangle PBC) = \frac12\times BC\times \text{height} = \frac12\times4\times3=6$$

Solutions · 1