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Question: The area of the triangle formed by the lines \(y = m_{1}x + c_{1},\) \(y = m_{2}x + c_{2}\) and \(x ...

The area of the triangle formed by the lines y=m1x+c1,y = m_{1}x + c_{1}, y=m2x+c2y = m_{2}x + c_{2} and x=0x = 0is.

A

12(c1+c2)2(m1−m2)\frac{1}{2}\frac{(c_{1} + c_{2})^{2}}{(m_{1} - m_{2})}

B

12(c1−c2)2(m1+m2)\frac{1}{2}\frac{(c_{1} - c_{2})^{2}}{(m_{1} + m_{2})}

C

12(c1−c2)2(m1−m2)\frac{1}{2}\frac{(c_{1} - c_{2})^{2}}{(m_{1} - m_{2})}

D

(c1−c2)2(m1−m2)\frac{(c_{1} - c_{2})^{2}}{(m_{1} - m_{2})}

Answer

12(c1−c2)2(m1−m2)\frac{1}{2}\frac{(c_{1} - c_{2})^{2}}{(m_{1} - m_{2})}

Explanation

Solution

On solving the equation of lines, we get the vertices of triangle (0,c1),(0,c2)\left( 0 , c _ { 1 } \right) , \left( 0 , c _ { 2 } \right), and (c2−c1m1−m2,m1c2−m2c1m1−m2)\left( \frac { c _ { 2 } - c _ { 1 } } { m _ { 1 } - m _ { 2 } } , \frac { m _ { 1 } c _ { 2 } - m _ { 2 } c _ { 1 } } { m _ { 1 } - m _ { 2 } } \right)

Hence, the area =12∣0c110c21c2−c1m1−m2m1c2−m2c1m1−m21∣= \frac { 1 } { 2 } \left| \begin{array} { c c c } 0 & c _ { 1 } & 1 \\ 0 & c _ { 2 } & 1 \\ \frac { c _ { 2 } - c _ { 1 } } { m _ { 1 } - m _ { 2 } } & \frac { m _ { 1 } c _ { 2 } - m _ { 2 } c _ { 1 } } { m _ { 1 } - m _ { 2 } } & 1 \end{array} \right|

=12[0+c1(c2−c1m1−m2)−c2(c2−c1m1−m2)]= \frac { 1 } { 2 } \left[ 0 + c _ { 1 } \left( \frac { c _ { 2 } - c _ { 1 } } { m _ { 1 } - m _ { 2 } } \right) - c _ { 2 } \left( \frac { c _ { 2 } - c _ { 1 } } { m _ { 1 } - m _ { 2 } } \right) \right] =12(c2−c1)(c1−c2)m1−m2=12(c1−c2)2(m1−m2)= \frac { 1 } { 2 } \frac { \left( c _ { 2 } - c _ { 1 } \right) \left( c _ { 1 } - c _ { 2 } \right) } { m _ { 1 } - m _ { 2 } } = \frac { 1 } { 2 } \frac { \left( c _ { 1 } - c _ { 2 } \right) ^ { 2 } } { \left( m _ { 1 } - m _ { 2 } \right) } .

(sign is not considered)

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