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Using the method of integration find the area of the triangle

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Using the method of integration find the area of the triangle $A BC$ , coordinates of whose vertices are $A (2, 0), B (4, 5)$ and $C (6, 3)$

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Using the method of integration find the area of the triangle

A BC

, coordinates of whose vertices are

A (2, 0), B (4, 5)

and

C (6, 3)

9 mins ago

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Text solutionVerified

The vertices of

Δ A BC

are

A (2, 0), B (4, 5)

, and

C (6, 3)

Equation of line segment

A B

y - 0 = \frac{5 - 0}{4 - 2} (x - 2)

2 y = 5 x - 10

y = \frac{5}{2} (x - 2) .......... (1)

Equation of line segment

BC

y - 5 = \frac{3 - 5}{6 - 4} (x - 4)

2 y - 10 = - 2 x + 8

2 y = - 2 x + 18

y = - x + 9......... (2)

Equation of line segment

C A

y - 3 = \frac{0 - 3}{2 - 6} (x - 6)

- 4 y + 12 = - 3 x + 18

4 y = 3 x - 6

y = \frac{3}{4} (x - 2) ........... (3)

∴ A re a (Δ A BC) = A re a (A B L A) + A re a (B L MCB) - A re a (A CM A)

= \int_{2}^{4} \frac{5}{2} (x - 2) d x + \int_{4}^{6} (- x + 9) d x - \int_{2}^{6} \frac{3}{4} (x - 2) d x

= \frac{5}{2} [\frac{x ^{2}}{2} - 2 x]_{2}^{4} + [\frac{- x ^{2}}{2} + 9 x]_{4}^{6} - \frac{3}{4} [\frac{x ^{2}}{2} - 2 x]_{2}^{6}

= \frac{5}{2} [8 - 8 - 2 + 4] + [- 18 + 54 + 8 - 36] - \frac{3}{4} [18 - 12 - 2 + 4]

= 5 + 8 - \frac{3}{4} (8)

= 13 - 6 = 7

sq. units

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Question Text	Using the method of integration find the area of the triangle $A BC$ , coordinates of whose vertices are $A (2, 0), B (4, 5)$ and $C (6, 3)$
Answer Type	Text solution:1
Upvotes	150