What is the area of triangle which is enclosed by three lines 4x-3y=12,2x+y=6 and x=0? (Ans:15)
First solve the given equations by substitution :-> Get the value of x and y :-> plot them :-> Find area
yeah, I substituted and got (0, -4) and (0,6) … after that what?
Graph it on a Cartesian Plane
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Solve equations 4x - 3y = 12 and 2x + y = 6 to get the point (3,0)
When you plot these lines on the graph, you can see the height is 3 and the base is 10 (distance from [0,-4] to [0,6]).
Substitute these values into the area of a triangle formula = \frac{1}2\times b \times h
= \frac{1}2\times 3 \times 10 = 15
If you’re not actively working on the solution and just want the answer then its not a very good way to learn, you need to force yourself to critically think (that’s why I didn’t provided you the solution).
I worked on it a lot actually before asking. I constructed the graph too and the problem is I don’t understand how the lines intersected with points (0, -4) and (0,6), and how height is calculated. It may be an easy problem, but I cannot visualize it accurately.
This is what the graph look like
And this is how you are able to get this plot :
4x-3y =12 \tag{1}
2x+y = 6 \tag{2}
x = 0 \tag{3}
Now , substituting the value of eq(3): x = 0 in eq(1) and (2), will give you where equation (1) and (2) meet the Y-axis :
4(0) -3y =12 , y= -4 \\
2(0) + y = 6, y=6
Thus, equation(1) meets Y-axis at coordinate(0,-4) and Eq (2) meets Y-axis at coordinate (0,6).
Now, we need to find the common intersection point b/w Eq(1) and Eq(2):
\text{Substituting, } y=6-2x \text{ from Eq(2) in Eq(1)} \\
4x-3(6-2x) =12 \text{ or } x=3
Now putting x=3 in our Eq y=6-2x or y =0
Thus, both the lines meet at (3,0)
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Thank you very much for the effort. Now, I get it.