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

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 :

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):

Now putting x=3 in our Eq y=6-2x or y =0

Thus, both the lines meet at (3,0)

Thank you very much for the effort. Now, I get it.