Question

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Answer 1

$\huge\red{Question}$

Solve graphically the pair of linear equations: 4x−3y+4=0,4x+3y−20=0. Find the area of the region bounded by these lines and x-axis. ❓

$\huge\red{Question}$

4x - 3y + 4 = 0 - - - - - - -(i)

4x + 3y - 20 = 0 - - - - - - -(ii)

Putting x = 0 in equation (i) we get

★ 4 xx 0 - 3 y = - 4

x => y = $\frac{4}{3}$

x = 0, y = $\frac{4}{3}$

Putting y = 0 in equation (i)we get

= > 4x - 3xx 0 = -4

=> x = - 1, y = 0

Using the graph table to draw the graph ⇑

The graph of (1) can be obtained by plotting the points ( 0, $\frac{4}{3}$, (-1, 0)

4x + 3 y = 20..............(ii)

Putting x = 0 in equation (ii) we get

=> 4 × 0 + 3y = 20

=> y = $\frac{20}{3}$

x = 0, y = $\frac{20}{3}$

Putting y = 0 in equation (ii) we get

=> 4x + 3 × 0 = 20

=> x = 5

x = 5, y = 0

Use the following table to draw the graph. ⇑

Draw the graph by plotting the two points from table. ⇑

The two lines intersect at p (2,4)

Hence x = 2, y = 2 is the solution of given equation.

Now,

Required area = Area of PBD

Required area = $\frac{1}{2}$ ${ \ base \ × \ Height}$

Required area = $\frac{1}{2}$ ( BD × PM)

Required area = $\frac{1}{2}$ ( 6 × 4)

Hence the area = 12 sq. Units

$\huge\red{Thanks}$