Question

Draw the graphs of the linear equations 4x-3y+4=0 and 4x+3y-20=0. Find the area bounded by these lines and x-axis.

Answer 1

the area bounded by 4x-3y+4=0 and 4x+3y-20=0 lines and x-axis. = 12 sq units

Step-by-step explanation:

4x-3y+4=0

4x+3y-20=0

Adding both

8x - 16 = 0

=> x = 2

=> y = 4

(2 , 4) is intersection point of 4x-3y+4=0 and 4x+3y-20=0

lines and x-axis

at x axis y = 0

=> x = - 1    for 4x-3y+4=0

& x = 5  for 4x+3y-20=0

(-1 , 0)  , (5 , 0) , ( 2, 4)   are the points

Simply Base = 6  & Altitude = 4

Area = (1/2) * 6 * 4= 12 sq units

or

= (1/2) | - 1(0 - 4)  + 5(4 - 0)  + 2(0 - 0) |

= (1/2) | 4 + 20 + 0 |

= (1/2)(24)

= 12

the area bounded by 4x-3y+4=0 and 4x+3y-20=0 lines and x-axis. = 12 sq units

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Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

The given set of linear equations are 4x-3y+4 =0 ......(1) and 4x+3y-20=0.......(2)

first we will plot these straight lines.

points (-1,0) and (0,3/4) satisfy the eq(1) plot the points and join them

points (5,0) and (5/4 ,5) satisfy eq(2) plot the points and join them.

now from the graph it is clear that the given lines intersect at point (2,4)

thus x =2 and y =4 is the solution of the given equation.

the shaded region is the required region.

area of the triangle ABC = 1/2 *base*height of the triangle ABC

= 1/2 *BC*4

=2BC

=2*6=12 sq units