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Solve graphically the system of linear equations $4 x - 3 y + 4 = 0 and 4 x + 3 y − 20 = 0$ .

Find the area bounded by these lines and x -axis.

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x=2, y=4 and area= 12 sq unit

x=2, y=4 and area= 10 sq unit

x=2, y=4 and area= 15 sq unit

x=2, y=4 and area= 20 sq unit

answer is B.

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Detailed Solution

Given equations are

4 x - 3 y + 4 = 0 and 4 x + 3 y − 20 = 0

.
Put several values for one variable and get the value of its corresponding variable.
Put different 'x' in 4x - 3y + 4 = 0 values to find different 'y' values.
Put, x = -1.

⇒ 4 − 1 − 3 y + 4 = 0

= > 3 y = 0

Put, x = 2.

= > 42 − 3 y + 4 = 0

= > 3 y = 12

⇒ y = 4

x	-1	2
y	0	4

Put different 'x' values in 4x + 3y - 20 = 0 to find different 'y' values.
Put, x= 2.

⇒ 42 + 3 y − 20 = 0

= > 3 y = 12

⇒ y = 4

Put, x = 5.

⇒ 45 + 3 y − 20 = 0

= > 3 y = 0

⇒ y = 0

x	2	5
y	4	0

Draw the graph with above obtained values.
Question Image

Triangle ABC is formed with the graphs of given two equations and x-axis.
Coordinates of triangle are A(-1,0), B(2,4), C(5,0).
We know that,

Area of Δ A B C = 12 × Base × H eight

Area of Δ A B C = 12 × 6 × 4

= 12 sq unit
The correct option is (2).

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