Putting $x=4$ and $y=2$ in each of the given equations, we get the following:
$2x+3y=12$:
LHS $=2\times 4+3\times 2=14\ne$ RHS
$4x+y=19$:
LHS $=4\times 4+2=18\ne$ RHS
$x+y=6$:
LHS $=4+2=6=$ RHS
$3x+2y=15$:
LHS $=3\times 4+2\times 2=16\ne$ RHS
Hence, $x=4,y=2$ is a solution of the equation $x+y=6$.

If x = 0,
0 + 2y = 0
$\Rightarrow$ y = 0
$\therefore$ The line x + 2y = 0 passes through the origin.
We know that for x + 2y = 0 ,
 y = 0 when x = 0.
Therefore, (0,3) does not lie on the line.

y = 4x can be rewritten as -4x + y = 0.
On comparing 4x - y = 0 with
ax + by + c = 0, we get
a = -4
b = 1
c = 0
$\therefore$ 4, -1, 0 are the values of a, b and c respectively.

At the y-axis, value of x = 0
Putting x = 0 in the given equation,
3 x 0 - 2y = 6
$\Rightarrow$ y = -3.

Hence, the line 3x - 2y = 6 cuts y-axis at the point (0,-3).

The solution of the equation remains unchanged if the same numbers are added to both sides of the equation.
Also, when we multiply or divide both the sides of the equation by the same non-zero number, the solution remains unchanged.
For example, in the equation x + 2y = 3, if we add 4 to both sides of the equation, the new equation will be x + 2y + 4 = 7. The values of x and y that satisfies the original equation will satisfy the new equation obtained as well.
​

Let us consider the equation x + y = 3
when x = 0 then y = 3
similarly, when y = 0 then x = 3
Let us consider the equation x + y = 5
when x = 0 then y = 5
similarly, when y = 0 then x = 5
By plotting the graphs of both of the equations, we can easily observe that the line x + y = 3 lies closer to the origin.

If the point lies on the graph then it satisfies the given equation,
Putting x = 3, y = 5 in the given equation,
$3\times 5=a\times 3+9$
$\Rightarrow$ 3a + 9 = 15
$\Rightarrow$ a = 2