The equations of sides AB,BC,and CA of a tria | vedclass

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(A) In $	riangle ABC$,the equations of sides $AB$,$BC$,and $CA$ are $2x+y=0$,$x+py=q$,and $x-y=3$ respectively. $P(2,3)$ is the orthocenter.$1$. Find

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$50$

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(A) In $	riangle ABC$,the equations of sides $AB$,$BC$,and $CA$ are $2x+y=0$,$x+py=q$,and $x-y=3$ respectively. $P(2,3)$ is the orthocenter.$1$. Finding vertex $A$ by solving $AB$ and $CA$:$2x+y=0 \Rightarrow y=-2x$Substitute into $x-y=3$: $x-(-2x)=3$ $\Rightarrow 3x=3$ $\Rightarrow x=1, y=-2$.So,$A = (1, -2)$.$2$. Finding $p$:The altitude from $A$ to $BC$ passes through $P(2,3)$.Slope of $AP = \frac{3-(-2)}{2-1} = 5$.Since $AP \perp BC$,the slope of $BC$ is $-\frac{1}{5}$.The equation of $BC$ is $x+py=q$,so its slope is $-\frac{1}{p}$.$-\frac{1}{p} = -\frac{1}{5} \Rightarrow p=5$.$3$. Finding $q$:Vertex $B$ is the intersection of $AB$ $(2x+y=0)$ and $BC$ $(x+5y=q)$.$y=-2x$ $\Rightarrow x+5(-2x)=q$ $\Rightarrow -9x=q$ $\Rightarrow x=-\frac{q}{9}, y=\frac{2q}{9}$.So,$B = \left(-\frac{q}{9}, \frac{2q}{9}ight)$.The altitude from $B$ to $AC$ passes through $P(2,3)$.Slope of $AC$ (from $x-y=3$) is $1$.Since $BP \perp AC$,the slope of $BP$ is $-1$.Slope of $BP = \frac{\frac{2q}{9}-3}{-\frac{q}{9}-2} = \frac{2q-27}{-q-18} = -1$.$2q-27 = q+18 \Rightarrow q=45$.$4$. Final value:$p+q = 5+45 = 50$.

The equations of sides $AB$,$BC$,and $CA$ of a $\triangle ABC$ are $2x+y=0$,$x+py=q$,and $x-y=3$ respectively. If $P(2,3)$ is its orthocenter,then the value of $p+q$ equals

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