Question

The area of triangle with vertices (K, 0), (4, 0), (0, 2) is 4 square units, then value of K is

• A. 0 or -8
• B. 0 or 8
• C. 8
• D. 0

Answer 1

The correct option is B 0 or 8

Equation of line AC is
$y-2=\frac{0-2}{4-0}(x-0)$
$y-2=\frac{-1}{2}x$
$x+2y-4=0$
$\therefore$ perpendicular distance from B(K,0) to AC
$BD=\left|\frac{k+2\left(0\right)-4}{\sqrt{1^2+2^2}}\right|$
$BD=\left|\frac{k-4}{\sqrt{5}}\right|$
$AC=\sqrt{{(0-2)}^2+{(4-0)}^2}=\sqrt{4+16}$$AC=2\sqrt{5}$
Area of $△ABC=4$sq.units
$⟹\frac{1}{2}\times AC\times BD=4$
$⟹\frac{1}{2}\times 2\sqrt{5}\times \left|\frac{k-5}{\sqrt{5}}\right|=4$
$⟹|k-4|=4$
$⟹k-4=\pm 4$
$k-4=4,k-4=-4$
$k=8or0$
$\therefore$ option B is correct