Let A(a, 0), B(b, 2b + 1) and C(0, b), b $\ne$ 0, |b| $\ne$ 1, be points such that the area of triangle ABC is 1 sq. unit, then the sum of all possible values of a is :
Solution
$$\left| {{1 \over 2}\left| {\matrix{
a & 0 & 1 \cr
b & {2b + 1} & 1 \cr
0 & b & 1 \cr
} } \right|} \right| = 1$$<br><br>$$ \Rightarrow \left| {\matrix{
a & 0 & 1 \cr
b & {2b + 1} & 1 \cr
0 & b & 1 \cr
} } \right| = \pm \,2$$<br><br>$\Rightarrow a(2b + 1 - b) - 0 + 1({b^2} - 0) = \pm \,2$<br><br>$\Rightarrow a = {{ \pm \,2 - {b^2}} \over {b + 1}}$<br><br>$\therefore$ $a = {{2 - {b^2}} \over {b + 1}}$ and $a = {{ - 2 - {b^2}} \over {b + 1}}$<br><br>Sum of possible values of 'a' is <br><br>$= {{ - 2{b^2}} \over {a + 1}}$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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