CBSE Class 12 · 2012 · Area Under CurvesLong Answer

Using the method of integration, find the area of the region bounded by the lines

3 x - 2 y + 1 = 0

2 x + 3 y - 21 = 0

and

x - 5 y + 9 = 0

Answer: $\frac{13}{2}$ square units

Solution

Solving the lines in pairs gives the vertices

(1, 2)

(3, 5)

and

(6, 3)

.
Writing

y

for each line (line 1:

y = \frac{3 x + 1}{2}

, line 2:

y = \frac{21 - 2 x}{3}

, line 3:

y = \frac{x + 9}{5}

) and integrating with the lowest line

x - 5 y + 9 = 0

as the lower edge throughout:

A = \int_{1}^{3} \frac{3 x + 1}{2} d x + \int_{3}^{6} \frac{21 - 2 x}{3} d x - \int_{1}^{6} \frac{x + 9}{5} d x .

= 6 + 1 - 9 + 21 - \frac{7}{2} - 9 = 10 - \frac{7}{2} = \frac{13}{2} .

Hence the area of the triangular region is

\frac{13}{2}

square units.

CBSE Class 12 Board · 2012 · Subjective