The general equation of first degree in $x$ and $y$ is $Ax+By+C=0$, where $A$, $B$ and $C$ are constants, and $A$ and $B$ are not simultaneously zero. This is also called the **linear equation** in $x$ and $y$ because it always represents a straight line.

### $Ax+By+C=0$ always represents a straight line.

Let $P(x_1,y_1)$, $Q(x_2,y_2)$ and $R(x_3,y_3)$ be any three points on the locus represented by the equation $Ax+By+C=0$. Then, the coordinates of the points must satisfy the equation. \[\therefore Ax_1+By_1+C=0\] \[Ax_2+By_2+C=0\] \[Ax_3+By_3+C=0\]

From the first two equations, by the rule of cross multiplication, we have \[\frac{A}{y_1-y_2}=\frac{B}{x_2-x_1}=\frac{C}{x_1y_2-x_2y_1}=k\text{ (say)}\]

\[\therefore A=(y_1-y_2)k\] \[B=(x_2-x_1)k\] \[C=(x_1y_2-x_2y_1)k\] Substituting the value of $A$, $B$ and $C$ in the third equation, we have \[k(y_1-y_2)x_3+k(x_2-x_1)y_3+k(x_1y_2-x_2y_1)=0\]

Here, the L.H.S. is twice the area of the triangle $PQR$. Hence, the area of the triangle is zero, which means that $P$, $Q$, $R$ lie in a straight line. Thus the linear equation represents a straight line.

The converse is easily seen to be true. Because equation of any line not parallel to y-axis can be written as $y=mx+c$, and that of any line parallel to y-axis is $x=a$. Both of these equations are linear.

## Reduction of the Linear Equation to Three Standard Forms

The linear equation $Ax+By+C=0$ can be reduced to the three standard forms: slope intercept form, double intercept form and normal form.

### Reduction to the Slope Intercept Form

\[Ax+By+C=0\] \[By=-Ax-C\] \[y=-\frac{A}{B}x-\frac{C}{B}\] which is of the slope intercept form $y=mx+c$ where \[\text{Slope }(m)=-\frac{A}{B}=-\frac{\text{Coefficient of }x}{\text{Coefficient of }y}\] \[\text{Y-intercept }(c)=-\frac{C}{B}\]

### Reduction to the Double Intercept Form

\[Ax+By+C=0\] \[Ax+By=-C\] \[\frac{A}{-C}x+\frac{B}{-C}y=1\] \[\frac{x}{\frac{-C}{A}}+\frac{y}{\frac{-C}{B}}=1\] which is of the double intercept form $\frac{x}{a}+\frac{y}{b}=1$ where \[\text{X-intercept }(a)=-\frac{C}{A}\] \[\text{Y-intercept }(b)=-\frac{C}{B}\]

### Reduction to the Normal Form

The equations \[Ax+By+C=0\] \[\text{and, }x\cos\alpha+y\sin\alpha-p=0\] will represent one and the same straight line if their corresponding coefficients are proportional. \[\text{i.e. }\frac{\cos\alpha}{A}=\frac{\sin\alpha}{B}=\frac{-p}{C}=k\text{ (say)}\] \[\text{so that }\cos\alpha=Ak,\text{ }\sin\alpha=Bk,\text{ }-p=Ck\] \[\therefore A^2k^2+B^2k^2=1\] \[k^2=\frac{1}{A^2+B^2}\] \[\therefore k=\pm\frac{1}{\sqrt{A^2+B^2}}\]

If $C>0$, then $k=-\frac{1}{\sqrt{A^2+B^2}}$ So, \[p=\frac{C}{\sqrt{A^2+B^2}}\] \[\cos\alpha=\frac{-A}{\sqrt{A^2+B^2}}\] \[\sin\alpha=\frac{-B}{\sqrt{A^2+B^2}}\] Hence the equation of the line in the normal form is \[-\frac{A}{\sqrt{A^2+B^2}}x-\frac{B}{\sqrt{A^2+B^2}}y=\frac{C}{\sqrt{A^2+B^2}}\]

If $C<0$, then $k=\frac{1}{\sqrt{A^2+B^2}}$ So, \[p=-\frac{C}{\sqrt{A^2+B^2}}\] \[\cos\alpha=\frac{A}{\sqrt{A^2+B^2}}\] \[\sin\alpha=\frac{B}{\sqrt{A^2+B^2}}\] Hence the equation of the line in the normal form is \[\frac{A}{\sqrt{A^2+B^2}}x+\frac{B}{\sqrt{A^2+B^2}}y=-\frac{C}{\sqrt{A^2+B^2}}\]

## The Point of Intersection of Two Straight Lines

Let $A_1x+B_1y+C_1=0$ and $A_2x+B_2y+C_2=0$ be any two given straight lines. We can find the point of intersection of these two given lines simply by solving for the values of $x$ and $y$. $(x,y)$ is the coordinates of the point of intersection. By solving for the values of $x$ and $y$, we get the coordinates of the point of intersection of two given lines to be, \[\left(\frac{B_1C_2-C_1B_2}{A_1B_2-B_1A_2},\frac{C_1A_2-A_1C_2}{A_1B_2-B_1A_2}\right)\] provided that $A_1B_2-B_1A_2≠0$.

### Obtain the equations of two lines passing through the intersection of the lines $4x-3y-1=0$ and $2x-5y+3=0$ and equally inclined to the axes.

Here, given lines are, \[4x-3y-1=0\] \[2x-5y+3=0\] Solving these equations, we get, $x=1$ and $y=1$. Hence, the coordinates of the point of intersection of the given lines is $(1,1)$.

Since the lines are equally inclined to the axes, so, \[\text{Slope }(m)=\pm 1\] Taking $m=1$, the equation of the given line is \[y-1=1(x-1)\] \[\therefore x-y=0\] Taking $m=-1$, the equation of the given line is \[y-1=-1(x-1)\] \[y-1=-x+1\] \[\therefore x+y-2=0\]

**Previous:** Point Slope Form and Two Points Form