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Coordinate Geometry

Question

If the area of triangle ABC formed by A(x, y), B(1, 2) and C(2, 1) is 6 square units, then prove that x + y = 15.

Solution

$The given vertises are A (x, y), B (1, 2) C (2, 1) . It is know that the area of a tringle whose vertises are (x_{1}, y_{1}) . (x_{2,} y_{2}), and (x_{3,} y_{3)} is given by \frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})| ∴ Area of ∆ ABC = \frac{1}{2} |x (2 - 1) + 1 (1 - y) + 2 (y - 2)| = \frac{1}{2} |x + 1 - y + 2 y - 4| = \frac{1}{2} |x + y - 3| The area of ∆ ABC is given as 6 sq . units . \Rightarrow \frac{1}{2} |x + y - 3| = 6 \Rightarrow x + y - 3 = 12 ∴ x + y = 15$

Some More Questions From Coordinate Geometry Chapter

Prove that the diagonals of a rectangle ABCD, with vertices A(2, -1), B(5, -1), C(5, 6) and D(2, 6), are equal and bisect each other.

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Coordinate Geometry

If the area of triangle ABC formed by A(x, y), B(1, 2) and C(2, 1) is 6 square units, then prove that x + y = 15.

Some More Questions From Coordinate Geometry Chapter

Prove that the diagonals of a rectangle ABCD, with vertices A(2, -1), B(5, -1), C(5, 6) and D(2, 6), are equal and bisect each other.

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