The equations of the sides AB, BC and CA of a triangle ABC are 2x+y=0, x+py=39 and x-y=3 r…

Question

The equations of the sides $AB, BC$ and $CA$ of a triangle $ABC$ are $2x+y=0,\ x+py=39$ and $x-y=3$ respectively and $P(2,3)$ is its circumcentre. Then which of the following is NOT true?

Accepted Answer

$A = AB\cap CA$: solving gives $A=(1,-2)$.

Perpendicular bisector of $AB$ (through circumcentre): $x-2y=-4$. Perpendicular bisector of $AC$: $x+y=5$.

$B$ = image of $A$ in the perpendicular bisector line $x-2y+4=0$: $B=\left(-\dfrac{13}5,\dfrac{26}5\right)$.

Similarly $C=(7,4)$.

Line $BC$: $x+8y=39$, so $p=8$.

$$AC = \sqrt{(7-1)^2+(4+2)^2}=\sqrt{36+36}=6\sqrt2 \implies (AC)^2=72$$

Check: $9p=72$  (matches A). $(AC)^2+p^2 = 72+64=136$  (matches B).

Area of $\triangle ABC$: computed as $32.4$, which lies in $(32,36)$ [matches C] but not in $(34,38)$ [does NOT match D].

So D is the incorrect statement (as required by the question).

Solutions · 1