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What is the area of the triangle formed by the lines 5x-2y=10,3x+4y=12 and x-axis?

 Jul 27, 2020
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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)

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

 

(By the way this is someone else's answer, not mine. I do not take credit for the detailed explanation.)

 

Hope this helped!

Caffeine :)

 Jul 27, 2020

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