Question

Find the value of $a$ if area of the triangle is $17$ square units whose vertices are $(0,0),(4,a),(6,4)$.

• A. $a=-2$
  
• B. $a=-3$
  
• C. $a=3$
  
• D. none of the above

Answer 1

The correct option is B $a=-3$

vertices of the triangle are $A(0,0),B(4,a),C(6,4)$.

Then area of triangle$ABC=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

$\Rightarrow 17=\frac{1}{2}\left[0\right(a-4)+4(4-0)+6(0-a)$

$\Rightarrow 34=16-6a$

$\Rightarrow a=-\frac{18}{6}=-3$