Question

The co-ordinates of the vertices of Triangle ABC are $A(4,1),B(–3,2)$ and $C(0,k)$. Given that the area of $△ABC$ is 12 sq. units. Find the value(s) of $k$.

• A. 5 only
• B. only
• C. 5 and
• D. 4

Answer 1

Answer: (A), (B)

The correct options are
A

5 only

B

only

Area of a triangle formed by $(x_1,y_1),(x_2,y_2)(x_3,y_3)$ = $\frac{1}{2}$|$x_1$($y_2$ - $y_3$) + $x_2$($y_3$ - $y_1$)+ $x_3$($y_1$ - $y_2$)|
Thus, the area of $\Delta ABC$ formed by the given points A(4, 1), B(-3, 2) and C(0, k) is

$=\frac{1}{2}|4(2-k)+(-3)(k-1)+0(1-2)|$

$=\frac{1}{2}|8-4k-3k+3|=\frac{1}{2}|11-7k|$

But given that the area of $\Delta ABC=12\text{sq. unit}$

$\Rightarrow |11-7k|=24$

$\pm (11-7k)=24$ $\Rightarrow 11-7k=24$ or $-(11-7k)=24$

$If11-7k=24,then-7k=24-11=13$ $\Rightarrow k=\frac{-13}{7}$

$-(11-7k)=24⟹-11+7k=24$

$\Rightarrow 7k=24+11=35$ $\Rightarrow k=\frac{35}{7}=5$

Hence the values of k are : 5, $\frac{-13}{7}$