The coordinates of $A, B, C$ are $(6, 3), (- 3, 5), (4, - 2)$ , respectively, and $P$ is any point $(x, y)$ . Show that the ratio of the area of $△ P B C$ to that of $△ A B C$ is $| x + y - 2 | / 7$

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Solution

$\begin{matrix} A r e a Δ A B C = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{3}) | \Rightarrow A r e a o f Δ A B C = \frac{1}{2} | 6 (7) - 3 (- 5) + 4 (- 2) | = \frac{1}{2} | 42 + 15 - 8 | = \frac{1}{2} | 49 | A r e a o f Δ A B C = \frac{49}{2} u n i t s q u a r e \to (i) \Rightarrow A r e a o f Δ A B C = \frac{1}{2} | x (7) - 3 (- 2 - y) + 4 (y - 5) | = \frac{1}{2} | 7 x + 6 + 3 y + 4 y - 20 | = \frac{1}{2} | 7 x + 7 y - 14 | a r (Δ P B C) = \frac{7}{2} | x + y - 2 | \to (i i) \end{matrix}$
Hence, ratio of $a r (Δ P B C)$ to $a r (Δ A B C) = \frac{\frac{7}{2} | x + y - 2 |}{\frac{49}{2}}$
$= \frac{| x + y - 2 |}{7} .$

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P (x, y),

then the ratio of the areas of

△ P B C

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Q. The coordinate of points

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The coordinates of A,B,C are (6,3),(−3,5),(4,−2), respectively, and P is any point (x,y). Show that the ratio of the area of △PBC to that of △ABC is |x+y−2|/7

The coordinates of $A, B, C$ are $(6, 3), (- 3, 5), (4, - 2)$ , respectively, and $P$ is any point $(x, y)$ . Show that the ratio of the area of $△ P B C$ to that of $△ A B C$ is $| x + y - 2 | / 7$