The coordinates of the orthocentre of the tri | vedclass

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(D) Let the vertices of the triangle be $A(0, 0)$,$B(3, 4)$,and $C(4, 0)$.The orthocentre is the intersection of the altitudes of the triangle.$1$. Th

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let the vertices of the triangle be $A(0, 0)$,$B(3, 4)$,and $C(4, 0)$.The orthocentre is the intersection of the altitudes of the triangle.$1$. The altitude from $B(3, 4)$ to the side $AC$ (which lies on the $x$-axis,$y=0$) is a vertical line passing through $x=3$. Thus,the equation of this altitude is $x=3$.$2$. The altitude from $C(4, 0)$ to the side $AB$ passes through $(4, 0)$. The slope of $AB$ is $m_{AB} = \frac{4-0}{3-0} = \frac{4}{3}$. The slope of the altitude perpendicular to $AB$ is $m_{\perp} = -\frac{3}{4}$.The equation of the altitude from $C$ is $y - 0 = -\frac{3}{4}(x - 4)$.Since the orthocentre lies on $x=3$,we substitute $x=3$ into the equation of the second altitude:$y = -\frac{3}{4}(3 - 4) = -\frac{3}{4}(-1) = \frac{3}{4}$.Therefore,the coordinates of the orthocentre are $\left(3, \frac{3}{4}\right)$.

The coordinates of the orthocentre of the triangle with vertices $A(0, 0)$,$B(3, 4)$,and $C(4, 0)$ are:

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