In an equilateral triangle P Q R, let the vertex P be at (3,5) and the side Q R be along t…

Question

In an equilateral triangle $P Q R$, let the vertex $P$ be at $(3,5)$ and the side $Q R$ be along the line $x+y=4$. If the orthocentre of the triangle PQR is $(\alpha, \beta)$, then $9(\alpha+\beta)$ is equal to:

Accepted Answer

Since $QR$ lies on the line

$x+y=4,$

and $P=(3,5)$, first find the altitude from $P$ to the side $QR$.

In an equilateral triangle, all altitudes, medians and angle bisectors coincide. Also, the orthocentre lies on the altitude from each vertex and divides the altitude in the ratio $2:1$ from the vertex.

So let us proceed step by step.

The line $QR$ has equation

$x+y-4=0.$

Distance of point $P(3,5)$ from this line is

$d=\frac{|3+5-4|}{\sqrt{{1}^{2}+{1}^{2}}}=\frac{4}{\sqrt{2}}=2\sqrt{2}.$

This is the altitude of the equilateral triangle from vertex $P$.

Now, the foot of the perpendicular from $P$ to the line $x+y=4$ lies on the altitude. Since in an equilateral triangle the orthocentre lies at $\frac{2}{3}$ of the altitude from the vertex, we can find it along the perpendicular direction.

The slope of $x+y=4$ is $-1$, so the perpendicular slope is $1$.

Hence the altitude through $P(3,5)$ is

$y-5=1(x-3)$

or

$y=x+2.$

The orthocentre lies on this line.

Now find the foot of perpendicular from $P$ to $x+y=4$ by solving

$y=x+2$

and

$x+y=4.$

Substitute $y=x+2$ into $x+y=4$:

$x+(x+2)=4$

$2x=2$

$x=1,\, y=3.$

So the foot is $(1,3)$.

Thus the altitude segment is from $P(3,5)$ to $(1,3)$.

The orthocentre divides this altitude in the ratio $2:1$ from vertex $P$. Therefore,

$H=(\frac{2\cdot 1+1\cdot 3}{3},\frac{2\cdot 3+1\cdot 5}{3})$

This is incorrect ratio placement, so let us use the correct section carefully:

If $H$ divides the segment from $P$ to foot $D$ in the ratio $2:1$, then

$H=P+\frac{2}{3}(D-P).$

Here,

$D-P=(1-3,\, 3-5)=(-2,-2).$

So

$H=(3,5)+\frac{2}{3}(-2,-2)=(3-\frac{4}{3},\, 5-\frac{4}{3})=(\frac{5}{3},\, \frac{11}{3}).$

Thus,

$\alpha =\frac{5}{3},\, \beta =\frac{11}{3}.$

So,

$\alpha +\beta =\frac{5}{3}+\frac{11}{3}=\frac{16}{3}.$

Therefore,

$9(\alpha +\beta )=9\cdot \frac{16}{3}=48.$

Hence the correct answer is

$48$

So, Option D is correct.

Solutions · 1