The orthocentre of the triangle with vertices | vedclass

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(C) Let the vertices be $A(0, 0)$,$B(4, 0)$,and $C(3, 4)$.To find the orthocentre,we find the intersection of the altitudes.$1$. Altitude from $A$ to

Vedclass · Accepted Answer

$\left( 3, \frac{3}{4} \right)$

Vedclass · Answer

(C) Let the vertices be $A(0, 0)$,$B(4, 0)$,and $C(3, 4)$.To find the orthocentre,we find the intersection of the altitudes.$1$. Altitude from $A$ to $BC$: The slope of $BC$ is $\frac{4-0}{3-4} = -4$. The altitude is perpendicular to $BC$,so its slope is $\frac{1}{4}$. Since it passes through $A(0, 0)$,its equation is $y = \frac{1}{4}x$,or $x - 4y = 0$.$2$. Altitude from $C$ to $AB$: The side $AB$ lies on the $x$-axis $(y=0)$,so the altitude from $C(3, 4)$ is a vertical line $x = 3$.$3$. Intersection: Substituting $x = 3$ into $x - 4y = 0$,we get $3 - 4y = 0$,which gives $y = \frac{3}{4}$.Thus,the orthocentre is $\left( 3, \frac{3}{4} \right)$.

The orthocentre of the triangle with vertices $(0, 0)$,$(3, 4)$,and $(4, 0)$ is

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