Question

1. Draw the graphs of the equations 4x – 3y+ 4=0 and 4x+ 3y-20 =0. shade the triangular region bounded by these lines and x axis and find the area of the triangle bounded by these lines and x axis. Plz do it step by step and send a pic of graph if you are doing it . I will mark u the brainliest. No spam.

Answer 1

Given :

equation of lines

1) 4x – 3y+ 4=0

2)4x+ 3y-20 =0

To Find :

Area of triangular region formed by lines with x axis & shade the triangular region .

Solution :

•Finding , points through whic lines passes

1) 4x – 3y+ 4=0

For x = 2 , y = 4

For x = -1 , y = 0

4x – 3y+ 4=0 passes through points (-1 , 0 ) , ( 2 ,4 )

2)4x+ 3y-20 =0

For x = 2 , y = 4

For x = 5 , y = 0

4x+ 3y-20 =0 passes through points

( 2 ,4 ) , ( 5 , 0 )

•Hence points (-1 , 0 ) , ( 2 ,4 ) ,

( 5 , 0 ) forms the required triangle triangle .

•Also , Area of triangle = 1/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2) ]

Area = 1/2[ (-1)(4-0) + 2(0-0) + 5(0-4)]

Area = 1/2 [ -4 -20 ]

Area = 1/2[-24]

Area = -12

•Since, Area can't be negative

•So , Area of triangle is 12 sq units