Question

The coordinates of A for which area of triangle, whose vertices are $A(a,2a),B( - 2,6)$ and $C(3,1)$ is $10$ square units are:

A. $(0,3)$
B. $(5,8)$
C. $\left( {3,\dfrac{8}{3}} \right)$
D. None of these

Answer 1

Hint : As we know that the above question is related to the coordinate geometry. We know the formula of area of triangle is $\dfrac{1}{2}\left[ {{a_1}({b_1} - {b_2}) + {b_1}({c_1} - {c_2}) + {c_1}({a_1} - {a_2})} \right]$ , where $\left[ {({b_1} - {b_2}),({c_1} - {c_2})and({a_1} - {a_2})} \right]$ are the coordinates of vertices of triangle. We will apply this formula of area of the triangle and find the value of $a$ .

Complete step by step solution:
Here in this question, the area is $10$ square units. We have
${a_1} = a,{a_2} = 2a,{b_1} = - 2,{b_2} = 6$ and ${c_1} = 3$ , ${c_2} = 1$ .
Are of triangle is
$\dfrac{1}{2}\left[ {{a_1}({b_1} - {b_2}) + {b_1}({c_1} - {c_2}) + {c_1}({a_1} - {a_2})} \right]$ , by putting the values we have:
$10 = \dfrac{1}{2}\left[ {a\left( {-2 - 6} \right) + ( - 2)\left( {3 - 1} \right) + 3\left( {a - 2a} \right)} \right]$ .
We will now solve it,
$-8a - 4 - 3a = 20 \\\ \Rightarrow -11a = 24$
On further solving we have $a = -\dfrac{{24}}{{11}}$ and then $2a = 2 \times -\dfrac{24}{11} = -\dfrac{{48}}{11}$ .
Hence the correct option is (D).
So, the correct answer is “Option D”.

Note : Before solving this kind of question we should have the proper knowledge of triangles, their area and the vertices. Sometimes we get a quadratic equation while finding the value of required missing number, then we can find the value of roots by the method of quadratic formula i.e. $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .