Given, line is 3x + 4y - 12 = 0

The point of intersection on the graph can be find out as

x      0          4

y      12        0  

Let the point through which the given line passes is A (4, 0) and B (0, 12)

Let P and Q be the point of intersection of AB

Now, coordinate of P = {(1 * 0 + 2 * 4)/(1 + 2), (1 * 12 + 2 * 0)/(1 + 2)}

                             = (8/3, 4)

Given point O (0, 0) is the origin

So, the equation of OP is

y - 0 = {(4 - 0)/(8/3 - 0)} * (x - 0)

=> y = (3 *4x)/8

=> y = 12x/8

=> y = 3x/2

=> 2y = 3x  ................1

Again,

the equation of OQ is

y - 0 = {(8 - 0)/(4/3 - 0)} * (x - 0)

=> y = (3 *8x)/4

=> y = 3 * 2x

=> y = 6x .......2

So, the equation 1 and 2 are the required equation of the line.