A) \[\frac{1}{8a}({{y}_{1}}+{{y}_{2}})({{y}_{2}}+{{y}_{3}})({{y}_{3}}+{{y}_{1}})\]
B) \[\frac{1}{4a}({{y}_{1}}+{{y}_{2}})({{y}_{2}}+{{y}_{3}})({{y}_{3}}+{{y}_{1}})\]
C) \[\frac{1}{8a}({{y}_{1}}-{{y}_{2}})({{y}_{2}}-{{y}_{3}})({{y}_{3}}-{{y}_{1}})\]
D) \[\frac{1}{4a}({{y}_{1}}-{{y}_{2}})({{y}_{2}}-{{y}_{3}})({{y}_{3}}-{{y}_{1}})\]
Correct Answer: B
Solution :
Points \[\left( \frac{y_{1}^{2}}{4a},{{y}_{1}} \right),\,\left( \frac{y_{2}^{2}}{4a},{{y}_{2}} \right),\,\,\left( \frac{y_{3}^{2}}{4a},{{y}_{3}} \right)\] Use area formula and get \[\Delta =\frac{1}{8a}({{y}_{1}}-{{y}_{2}})({{y}_{2}}-{{y}_{3}})({{y}_{3}}-{{y}_{1}})\].
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