Two vertices of a triangle are (0, 2) and (4, | vedclass

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(B) Let the third vertex be $A(h, k)$. Let the other two vertices be $B(0, 2)$ and $C(4, 3)$. The orthocenter $H$ is at $(0, 0)$.Since $AH \perp BC$,t

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(B) Let the third vertex be $A(h, k)$. Let the other two vertices be $B(0, 2)$ and $C(4, 3)$. The orthocenter $H$ is at $(0, 0)$.Since $AH \perp BC$,the slope of $AH 	imes$ slope of $BC = -1$.Slope of $BC = \frac{3 - 2}{4 - 0} = \frac{1}{4}$.Slope of $AH = \frac{k - 0}{h - 0} = \frac{k}{h}$.Therefore,$\frac{k}{h} 	imes \frac{1}{4} = -1 \implies k = -4h$.Since $BH \perp AC$,the slope of $BH 	imes$ slope of $AC = -1$.Slope of $AC = \frac{k - 3}{h - 4}$.Slope of $BH = \frac{2 - 0}{0 - 0}$ is undefined (vertical line $x = 0$).Thus,$AC$ must be a horizontal line,so $k = 3$.Substituting $k = 3$ into $k = -4h$,we get $3 = -4h$,so $h = -\frac{3}{4}$.The third vertex is $(-\frac{3}{4}, 3)$,which lies in the second quadrant.

Two vertices of a triangle are $(0, 2)$ and $(4, 3)$. If its orthocenter is at the origin $(0, 0)$,then its third vertex lies in which quadrant?

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