Straight Lines

The Two Sides of a Line

Let $P(x_1,y_1)$ and $Q(x_2,y_2)$ be two given points and $Ax+By+C=0$ be the given line. Join $PQ$. Let $PQ$ (produced if necessary) meet the given line at $R$. Let $PR:RQ=m:n$. Then, if $m:n$ is positive, $R$ divides $PQ$ internally i.e. $P$ and $Q$ are on the opposite sides of the line (Figure (a)). If $m:n$ is negative, $R$ divides $PQ$ externally i.e. $P$ and $Q$ are on the same sides of the line (Figure (b)).

From the section formula, coordinates of $R$ are given by,

$\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)$

Since $R$ lies on the line $Ax+By+C=0$, the coordinates of $R$ satisfy the equation of the line. $\text{i.e. }A\left(\frac{mx_2+nx_1}{m+n}\right)+B\left(\frac{my_2+ny_1}{m+n}\right)+C=0$

$m(Ax_2+By_2+C)+n(Ax_1+By_1+C)=0$

$\frac{Ax_1+By_1+C}{Ax_2+By_2+C}=-\frac{m}{n}$

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 $2x+3y-4=0$.