A) \[\frac{({{d}_{1}}-{{c}_{1}})({{d}_{2}}-{{c}_{2}})}{{{[(a_{1}^{2}+b_{1}^{2})(a_{2}^{2}+b_{2}^{2})]}^{1/2}}}\]
B) \[\frac{({{d}_{1}}-{{c}_{1}})({{d}_{2}}-{{c}_{2}})}{{{a}_{1}}{{a}_{2}}-{{b}_{1}}{{b}_{2}}}\]
C) \[\frac{({{d}_{1}}+{{c}_{1}})({{d}_{2}}+{{c}_{2}})}{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}}\]
D) \[\frac{({{d}_{1}}-{{c}_{1}})({{d}_{2}}-{{c}_{2}})}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\]
Correct Answer: D
Solution :
Area \[=\frac{{{p}_{1}}{{p}_{2}}}{\sin \theta }={{p}_{1}}{{p}_{2}}\text{cosec}\theta \] , where \[{{p}_{1}}=\frac{{{d}_{1}}-{{c}_{1}}}{\sqrt{(a_{1}^{2}+b_{1}^{2})}},{{p}_{2}}=\frac{{{d}_{2}}-{{c}_{2}}}{\sqrt{(a_{2}^{2}+b_{2}^{2})}}\] Also \[\tan \theta =\frac{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}}\]. Since \[\text{cose}{{\text{c}}^{2}}\theta =1+{{\cot }^{2}}\theta \] or \[\text{cose}{{\text{c}}^{2}}\theta =\frac{{{({{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}})}^{2}}+{{({{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}})}^{2}}}{{{({{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}})}^{2}}}\] \[=\frac{(a_{1}^{2}+b_{1}^{2})(a_{2}^{2}+b_{2}^{2})}{{{({{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}})}^{2}}}\] Putting for \[{{p}_{1}},{{p}_{2}}\] and \[\text{cosec}\theta \], we get Area =\[\frac{({{d}_{1}}-{{c}_{1}})({{d}_{2}}-{{c}_{2}})}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\].
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