Coordinate Geometry Question 134
Question: The area of the triangle formed by the lines $ y=m_1x+c_1, $
$ y=m_2x+c_2 $ and $ x=0 $ is
Options:
A) $ \frac{1}{2}\frac{{{(c_1+c_2)}^{2}}}{(m_1-m_2)} $
B) $ \frac{1}{2}\frac{{{(c_1-c_2)}^{2}}}{(m_1+m_2)} $
C) $ \frac{1}{2}\frac{{{(c_1-c_2)}^{2}}}{(m_1-m_2)} $
D) $ \frac{{{(c_1-c_2)}^{2}}}{(m_1-m_2)} $
Show Answer
Answer:
Correct Answer: C
Solution:
On solving the equation of lines, we get the vertices of triangle $ (0,c_1),(0,c_2) $ , and $ ( \frac{c_2-c_1}{m_1-m_2},\frac{m_1c_2-m_2c_1}{m_1-m_2} ) $
Hence, the area $ =\frac{1}{2} \begin{vmatrix} 0 & c_1 & 1 \\ 0 & c_2 & 1 \\ \frac{c_2-c_1}{m_1-m_2} & \frac{m_1c_2-m_2c_1}{m_1-m_2} & 1 \\ \end{vmatrix} $
$ =\frac{1}{2}[ 0+c_1( \frac{c_2-c_1}{m_1-m_2} )-{c _{2}}( \frac{c_2-c_1}{m_1-m_2} ) ] $
$ =\frac{1}{2}\frac{(c_2-c_1)(c_1-c_2)}{m_1-m_2}=\frac{1}{2}\frac{{{(c_1-c_2)}^{2}}}{(m_1-m_2)} $ . (sign is not considered) Note : Students should remember this question as a formula.
BETA
