What is the area of the triangle formed by the lines 5x-2y=10,3x+4y=12 and x-axis?
+284 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 :)
