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Area of a Triangle Given Its Vertices

The area of a...

Question

The area of a triangle whose vertices are A(6,3), B(-3,5) and C(4,-2) is

⁴⁹/₂ sq. units

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⁹/₄ sq. units

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48 sq. units

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49 sq. units

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Solution

The correct option is A
⁴⁹/₂ sq. units

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Similar questions

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

P (x, y),

△ P B C

△ A B C

A (6, 3)

B) - 3, 5)

C (4, - 2)

D (x, 3 x)

\frac{Δ D B C}{Δ A B C} = \frac{1}{2}

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

D (x, 3 x)

\frac{△ D B C}{△ A B C} = \frac{1}{2}

x

\frac{∆ D B C}{∆ A B C} = \frac{1}{2},

Four points A (6, 3), B (-3, 5), C (4, -2) and D (x, 3x) are given in such a way that $\frac{△ D B C}{△ A B C} = \frac{1}{2},$ find x.

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