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) \[\frac{{{({{c}_{1}}-{{c}_{2}})}^{2}}}{({{m}_{1}}-{{m}_{2}})}\]
Correct Answer: C
Solution :
On solving the equation of lines, we get the vertices of triangle \[(0,\,\,{{c}_{1}}),\,\,(0,\,\,{{c}_{2}})\], and \[\left( \frac{{{c}_{2}}-{{c}_{1}}}{{{m}_{1}}-{{m}_{2}}},\,\,\frac{{{m}_{1}}{{c}_{2}}-{{m}_{2}}{{c}_{1}}}{{{m}_{1}}-{{m}_{2}}} \right)\] Hence, the area \[=\frac{1}{2}\,\left| \,\begin{matrix} 0 & {{c}_{1}} & 1 \\ 0 & {{c}_{2}} & 1 \\ \frac{{{c}_{2}}-{{c}_{1}}}{{{m}_{1}}-{{m}_{2}}} & \frac{{{m}_{1}}{{c}_{2}}-{{m}_{2}}{{c}_{1}}}{{{m}_{1}}-{{m}_{2}}} & 1 \\ \end{matrix}\, \right|\] \[=\frac{1}{2}\left[ 0+{{c}_{1}}\left( \frac{{{c}_{2}}-{{c}_{1}}}{{{m}_{1}}-{{m}_{2}}} \right)-{{c}_{2\,}}\left( \frac{{{c}_{2}}-{{c}_{1}}}{{{m}_{1}}-{{m}_{2}}} \right) \right]\] \[=\frac{1}{2}\frac{({{c}_{2}}-{{c}_{1}})\,\,({{c}_{1}}-{{c}_{2}})}{{{m}_{1}}-{{m}_{2}}}=\frac{1}{2}\frac{{{({{c}_{1}}-{{c}_{2}})}^{2}}}{({{m}_{1}}-{{m}_{2}})}\]. (sign is not considered) Note : Students should remember this question as a formula.
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