**Case I:** If the points $P$ and $Q$ are on opposite sides of the line, $\frac{m}{n}$ is positive, hence $Ax_1+By_1+C$ and $Ax_2+By_2+C$ are of opposite signs.

**Case II:** If the points $P$ and $Q$ are on the same side of the line, $\frac{m}{n}$ is negative, hence $Ax_1+By_1+C$ and $Ax_2+By_2+C$ are of same sign.

If a given point $(x’,y’)$ and the origin are on the same side of the line $Ax+By+C=0$ then $Ax’+By’+C$ and $C$ have the same sign. They are on opposite sides if they are of opposite signs.

### Are the points $(-1,2)$ and $(3,-2)$ lie on the same side of the line with equation $x+3y=6$?

The given equation of the line is, \[x+3y-6=0\] For $(-1,2)$: \[x+3y-6=-1+3×2-6=-1\text{ (-ve)}\] For $(3,-2)$: \[x+3y-6=3+3×(-2)-6=-9\text{ (-ve)}\]

Since $x+3y-6$ has the same sign for the points $(-1,2)$ and $(3,-2)$, both points lie on the same side of the given line.

### Show that two of the three points $(0,0)$, $(2,3)$ and $(3,4)$ lie on one side and the remaining on the other side of the line $x-3y+3=0$.

Given points are $A(0,0)$, $B(2,3)$ and $C(3,4)$. Substituting these points in the expression \[x-3y+3=0\] For $A(0,0)$: \[0+0+3=3\text{ (+ve)}\] For $B(2,3)$: \[2-9+3=-4\text{ (-ve)}\] For $C(3,4)$: \[3-12+3=-6\text{ (-ve)}\]

Therefore, $B$ and $C$ are on the same side and $A$ is in the opposite side of the line $x-3y+3=0$.

### Prove that two of the vertices of the triangle formed by the lines $y-x=0$, $2y-x=0$ and $y=1$ lie on one side and the third vertex lies on the other side of the line $2x+3y-4=0$.

The equations of the sides of the triangle are $y-x=0$, $2y-x=0$ and $y=1$. By solving these equations, we get the vertices of the triangle; $A(0,0)$, $B(1,1)$ and $C(2,1)$. Substituting the coordinates of these points in the expression, \[2x+3y-4=0\]

For $A(0,0)$: \[0+0-4=-4\text{ (-ve)}\]

For $B(1,1)$: \[2×1+3×1-4=1\text{ (+ve)}\] For $C(2,1)$: \[2×2+3×1-4=3\text{ (+ve)}\]

Hence, $B$ and $C$ lie on one side and $A$ lies on the other side of the line $2x+3y-4=0$.