The coordinates of A, B, C are (6, 3), (–3, 5) and (4, – 2) respectively and P is any point (x, y). Show that the ratio of the areas of triangle PBC and ABC is .
| $\frac{x + y - 2}{7}$ | [4 MARKS]

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Solution

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

We have,

$∴ A r e a o f Δ P B C = \frac{1}{2} | (5 x + 6 + 4 y) - (- 3 y + 20 - 2 x) |$

$\Rightarrow A r e a o f Δ P B C = \frac{1}{2} | 5 x + 6 + 4 y + 3 y - 20 + 2 x |$

$\Rightarrow A r e a o f Δ P B C = \frac{1}{2} | 7 x + 7 y - 14 |$

$\Rightarrow A r e a o f Δ P B C = \frac{7}{2} | x + y - 2 |$

$\Rightarrow A r e a o f Δ A B C = \frac{7}{2} | 6 + 3 - 2 |$

[ Replacing x by 6 and y = 3 in Area of $Δ P B C$ ]

$\Rightarrow A r e a o f Δ A B C = \frac{49}{2}$

$∴ \frac{A r e a o f Δ P B C}{A r e a o f Δ A B C} = \frac{\frac{7}{2} | x + y - 2 |}{\frac{49}{2}}$

$\Rightarrow \frac{A r e a o f Δ P B C}{A r e a o f Δ A B C} = \frac{| x + y - 2 |}{7}$

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A, B, C

(6, 3), (- 3, 5), (4, - 2)

P

(x, y)

△ P B C

△ A B C

| x + y - 2 | / 7

A (6, 3), B (- 3, 5), C (4, - 2)

P (x, y),

△ P B C

△ A B C

a r (P B C) \div a r (t r i a n g l e A B C) = | (x + y - 2) \div (7) |

Δ A B C, A \equiv (6, 3), B \equiv (- 3, 5), C \equiv (4, - 2)

P

(x, y)

\frac{Δ (P B C)}{Δ (A B C)} = ∣ ∣ ∣ \frac{x + y - 2}{7} ∣ ∣ ∣

(6, 3), (- 3, 5), (4, - 2)

(x, y),

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