The orthocentre of the triangle formed by the | vedclass

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Question

(A) Let the vertices of the triangle be $O(0, 0)$,$A(8, 0)$,and $B(4, 6)$.The slope of side $AB$ is $m_{AB} = \frac{6 - 0}{4 - 8} = \frac{6}{-4} = -\f

Vedclass · Accepted Answer

$\left( 4, \frac{8}{3} \right)$

Vedclass · Answer

(A) Let the vertices of the triangle be $O(0, 0)$,$A(8, 0)$,and $B(4, 6)$.The slope of side $AB$ is $m_{AB} = \frac{6 - 0}{4 - 8} = \frac{6}{-4} = -\frac{3}{2}$.The altitude from $O(0, 0)$ to $AB$ is perpendicular to $AB$,so its slope is $m_1 = -\frac{1}{m_{AB}} = \frac{2}{3}$.The equation of this altitude is $y - 0 = \frac{2}{3}(x - 0)$,which simplifies to $y = \frac{2}{3}x$.The slope of side $OB$ is $m_{OB} = \frac{6 - 0}{4 - 0} = \frac{6}{4} = \frac{3}{2}$.The altitude from $A(8, 0)$ to $OB$ is perpendicular to $OB$,so its slope is $m_2 = -\frac{1}{m_{OB}} = -\frac{2}{3}$.The equation of this altitude is $y - 0 = -\frac{2}{3}(x - 8)$,which simplifies to $3y = -2x + 16$.To find the orthocentre,we solve the system of equations $y = \frac{2}{3}x$ and $3y = -2x + 16$.Substituting $y = \frac{2}{3}x$ into the second equation: $3(\frac{2}{3}x) = -2x + 16 \implies 2x = -2x + 16 \implies 4x = 16 \implies x = 4$.Then $y = \frac{2}{3}(4) = \frac{8}{3}$.Thus,the orthocentre is $\left( 4, \frac{8}{3} \right)$.

The orthocentre of the triangle formed by the vertices $(0, 0)$,$(8, 0)$,and $(4, 6)$ is

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