Question

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

 

Answer 1

We know that the area of a triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is\[\frac{1}{2}\left| x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right|\]

Now,
Area of ∆PBC

\[= \frac{1}{2}\left| x\left[ 5 - \left( - 2 \right) \right] + \left( - 3 \right)\left( - 2 - y \right) + 4\left( y - 5 \right) \right|\]
\[ = \frac{1}{2}\left| 7x + 7y - 14 \right|\]
\[ = \frac{7}{2}\left| x + y - 2 \right| \text{square units}\]

Area of ∆ABC

\[= \frac{1}{2}\left| 6\left[ 5 - \left( - 2 \right) \right] + \left( - 3 \right)\left( - 2 - 3 \right) + 4\left( 3 - 5 \right) \right|\]
\[ = \frac{1}{2}\left| 42 + 15 - 8 \right|\]
\[ = \frac{49}{2} \text{square units}\]

\[\therefore \frac{ar\left( ∆ PBC \right)}{ar\left( ∆ ABC \right)} = \frac{\frac{7}{2}\left| x + y - 2 \right|}{\frac{49}{2}}\]
\[ \Rightarrow \frac{ar\left( ∆ PBC \right)}{ar\left( ∆ ABC \right)} = \frac{\left| x + y - 2 \right|}{7} = \left| \frac{x + y - 2}{7} \right|\]