Orthocentre of a triangle with vertices (0,0),(3,4) and (4,0) is

(3, \frac{5}{4})

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(3,12)

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(3, \frac{3}{4})

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(3,9)

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Solution

The correct option is C $(3, \frac{3}{4})$
We know that orthocentre is the point of intersection of altitudes of a $Δ .$

Equation of altitude AD
$\Rightarrow$ AD line is parallel to Y - axis through (3,4)
$\Rightarrow x = 3 (1)$
Similarly equation of OE is
y = - $\frac{3 - 4}{4 - 0} x$
$\Rightarrow y = x / 4 (2)$
Solving (1) and (2), we get orthocentre as (3,3/4).

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