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The co-ordinates of the points A, B and C are (6, 3), (−3, 5) and (4, −2) respectively. P(*x*, *y*) is any point in the plane. Show that \[\frac{ar\left( ∆ PBC \right)}{ar\left( ∆ ABC \right)} = \left| \frac{x + y - 2}{7} \right|\]

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#### Solution

We know that the area of a triangle having vertices (*x*_{1}, *y*_{1}), (*x*_{2}, *y*_{2}) and (*x*_{3}, *y*_{3}) is\[\frac{1}{2}\left| x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right|\]

Area of ∆PBC

\[ = \frac{1}{2}\left| 7x + 7y - 14 \right|\]

\[ = \frac{7}{2}\left| x + y - 2 \right| \text{square units}\]

\[ = \frac{1}{2}\left| 42 + 15 - 8 \right|\]

\[ = \frac{49}{2} \text{square units}\]

\[ \Rightarrow \frac{ar\left( ∆ PBC \right)}{ar\left( ∆ ABC \right)} = \frac{\left| x + y - 2 \right|}{7} = \left| \frac{x + y - 2}{7} \right|\]

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