Coordinate Geometry Question 83
Question: If the vertices of a triangle be $ (am_1^{2},2am_1),(am_2^{2},2am_2) $ and $ (am_3^{2},2am_3), $ then the area of the triangle is
Options:
A) $ a(m_2-m_3)(m_3-m_1)(m_1-m_2) $
B) $ (m_2-m_3)(m_3-m_1)(m_1-m_2) $
C) $ a^{2}(m_2-m_3)(m_3-m_1)(m_1-m_2) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Area $ =\frac{1}{2} \begin{vmatrix} am_1^{2} & 2am_1 & 1 \\ am_2^{2} & 2am_2 & 1 \\ am_3^{2} & 2am_3 & 1 \\ \end{vmatrix} =\frac{1}{2}a^{2}\times 2 \begin{vmatrix} m_1^{2} & m_1 & 1 \\ m_2^{2} & m_2 & 1 \\ m_3^{2} & m_3 & 1 \\ \end{vmatrix} $
$ =a^{2} \begin{vmatrix} m_1^{2}-m_2^{2} & m_1-m_2 & 0 \\ m_2^{2}-m_3^{2} & m_2-m_3 & 0 \\ m_3^{2} & m_3 & 1 \\ \end{vmatrix} $ , by $ \begin{vmatrix} R_1\to R_1-R_2 \\ R_2\to R_2-R_3 \\ \end{vmatrix} $
$ =a^{2}(m_2^{2}-m_3^{2})(m_1-m_2)-(m_2-m_3)(m_1^{2}-m_2^{2}) $
$ =a^{2}(m_1-m_2)(m_2-m_3)(m_3-m_1) $ .
Trick : Let $ a=2,m_1=0,m_2=1,m_3=2, $ then the coordinates are (0, 0), (2, 4), (8, 8).
$ \therefore \Delta =\frac{1}{2} \begin{vmatrix} 0 & 0 & 1 \\ 2 & 8 & 1 \\ 4 & 8 & 1 \\ \end{vmatrix} =\frac{1}{2}(16-32)=8sq.units $ .
BETA
