Question

Draw the graphs of the 5x – 2y = 10 and 3x + 4y = 12. Shade the region bounded by these lines and x-axis. Also find the area of shaded region.

Answer 1

Answer: 15/13 sq. units

Step-by-step explanation:

We have drawn two lines  5x-2y=10, 3x+4y=12 on the graph.

We got two three points which forms triangle A, B,  C.

A is at y= 0 on line 5x-2y=10,

= > 5x=10

=> x = 2

=> A(2,0)

Now,  B is on line  3x+4y=12 , at y=0 ,

=> x = 4

=>B (4,0)

Then, C is at point of intersection of both lines 3x+4y=12 and 5x-2y=10.

That is,  Using Elimination method,

2*(5x-2y)=2*10

=>10x-4y=20

=>(3x+4y)= 12

+

= >13x=32

=>x=32/13

Substituting in any line out of two,  we get y = 15/13 .

C(32/13,15/13)

Area of triangle =

$\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\ \\=\frac{1}{2}[2(0-15/13)+4(15/13-0)+32/13(0-0)]\\ \\=15/13$

Hence,  Area of triangle is 15/13 sq. units.

Answer 2

Step-by-step explanation:

We have drawn two lines  5x-2y=10, 3x+4y=12 on the graph.

We got two three points which forms triangle A, B,  C.

A is at y= 0 on line 5x-2y=10,

= > 5x=10

=> x = 2

=> A(2,0)

Now,  B is on line  3x+4y=12 , at y=0 ,

=> x = 4

=>B (4,0)

Then, C is at point of intersection of both lines 3x+4y=12 and 5x-2y=10.

That is,  Using Elimination method,

2*(5x-2y)=2*10

=>10x-4y=20

=>(3x+4y)= 12

+

= >13x=32

=>x=32/13

Substituting in any line out of two,  we get y = 15/13 .

C(32/13,15/13)

Area of triangle =

Hence,  Area of triangle is 15/13 sq. units.