A) \[\frac{x+y-2}{7}\]
B) \[\frac{x-y+2}{2}\]
C) \[\frac{x-y-2}{7}\]
D) none of these
Correct Answer: A
Solution :
Key Idea: If \[A({{x}_{1}},{{y}_{1}}).B({{x}_{2}}{{y}_{2}})\]and \[C({{x}_{3}},{{y}_{3}})\]are the vertices of \[a\]triangle, then Area of \[\Delta =\frac{1}{2}\{{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})\] \[+\,{{x}_{3}}({{y}_{1}}-{{y}_{2}})\}\] Given points are \[A(6,3),B(-3,5),C(4-,2)\] and \[p(x,y)\] \[\therefore \] \[\frac{\Delta PBC}{\Delta ABC}\] \[=\frac{\frac{1}{2}}{\frac{1}{2}}\left[ \frac{x(5+2)-3(-2-y)+4(y-5)}{6(5+2)-3(-2-3)+4(3-5)} \right]\] \[=\left[ \frac{7x+7y-14}{42+15-8} \right]\] \[=\frac{7x+7y-14}{49}=\frac{x+y-2}{7}\]
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