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Find the area of the triangle PQR using the method of integration, coordinates of whose vertices are P (3, 0), Q (5, 5) and R (6, 3).

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1 Answer

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by (36.7k points)

Equation of line PQ is

y - 0 = \(\frac{5-0}{5-3}(x-3)\)

⇒ y = \(\frac52\)x - \(\frac{15}2\)

Equation of line QR is

y - 3 = \(\frac{5-3}{5-6}\)(x - 6)

⇒ y = -2x + 15

Equation of line PR is

 y - 0 = \(\frac{3-0}{6-3}\)(x - 3)

⇒ y = x - 3

Area of ΔPQA = \(\int\limits_3^5\left[(\frac52x-\frac{15}2)-(x-3)\right]dx\)

\(\frac34[x^2]_3^5 - \frac92[x]_3^5\)

 = \(\frac34(25-9)-\frac92(5-3)\)

\(\frac34\times16-\frac92\times2\)

= 12 - 9 = 3 square units.

Area of Δ QRA = \(\int\limits_5^6((-2x + 15)-(x-3))dx\)

\(=\int\limits_5^6(-3x + 18)dx\)

\(=-\frac32(x^2)_5^6+18(x)_5^6\)

\(=-\frac32\times11+18\)

\(=\frac{36-33}2=+\frac32\) square units (\(\because\) Area name be negative)

\(\therefore\) area of ΔABC = ar(ΔPQA) + ar(ΔQRA)

= 3 + \(\frac32=\frac92\) square units

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