Explanation

(c)
Now, we take $BD\bot AC$ and $AE\bot BC$
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Slope of $BD = - \frac{3}{4}$
$Equation\ of\ BD,\ y - 0 = \frac{- 3}{4}(x - 4)$
$\Rightarrow 4y = - 3x + 12$
$\Rightarrow 3x + 4y - 12 = 0$ …(i)
and slope of $AE = \frac{1}{4}$
$\therefore Equation\ of\ AE,\ y - 0 = \frac{1}{4}(x - 0)$
$\Rightarrow x - 4y = 0$ …(ii)
On solving Eqs. (i) and (ii), we get
$x = 3,\ \ y = \frac{3}{4}$
$\therefore Orthocentre\ of\ the\ traingle\ is\ \left( 3,\frac{3}{4} \right)$