Reduction of Ax + By + C = 0 in the Standard Forms

Learn how Ax+By+C=0 is reduced’to standard forms, with each step explained clearly.

Reduction of $Ax + By + C = 0$ into Standard Forms

The equation

$Ax + By + C = 0$

is called the general form of the equation of a straight line, where $A$, $B$, and $C$ are constants and $x$, $y$ are variables.

The same straight line can be written in different forms depending on what information we want to show.
The three important forms are:

Slope-intercept form
Intercept form
Perpendicular (normal) form

Section Formula – Coordinates of a Point Dividing a Line Segment

Step 1 of 7

Section Formula

Let, in the figure, A(x₁, y₁) and B(x₂, y₂) are two points. P(x, y) divides the line segment AB internally in the ratio m₁:m₂. Calculate the coordinate of P(x, y).

Now, draw AD⊥OX, BE⊥OX, PF⊥OX, AM⊥PF and PN⊥BE

In the figure, AM = DF = OF − OD = x − x₁,

PN = FE = OE − OF = x₂ − x

PM = PF − MF = PF − AD = y − y₁

BN = BE − NE = BE − PF = y₂ − y

And APBP = m₁m₂

Right angled triangles ΔPMA and ΔBNP are similar.

AMPN = APBP [ ∵ The ratio of corresponding sides of similar triangle]

Or, x − x₁x₂ − x = m₁m₂

Or, m₂x − m₂x₁ = m₁x₂ − m₁x

Or, m₂x + m₁x = m₁x₂ + m₂x₁

Or, x(m₁ + m₂) = m₁x₂ + m₂x₁

Or, x = m₁x₂ + m₂ x₁m₁ + m₂

Again, PMBN = APBP [ ∵ The ratio of corresponding sides of similar triangle]

Or, y − y₁y₂ − y = m₁m₂

Or, m₂y − m₂y₁ = m₁y₂ − m₁y

Or, m₂y + m₁y = m₁y₂ + m₂y₁

Or, y(m₁ + m₂) = m₁y₂ + m₂y₁

Or, y = m₁y₂ + m₂y₁m₁ + m₂

∴ (x, y) = (m₁x₂ + m₂ x₁m₁ + m₂ , m₁y₂ + m₂ y₁m₁ + m₂)

Thus, the coordinates of the point dividing line segment AB are

P(x, y) = (m₁x₂ + m₂ x₁m₁ + m₂ , m₁y₂ + m₂ y₁m₁ + m₂)

Reduction in intercept form

Given the general equation of a line: $Ax + By + C = 0$ (with $A \neq 0,\ B \neq 0,\ C \neq 0$)

1

Move the constant to the right side

Move $C$ to the right side:

$Ax + By = -C$

2

Make the right side equal to 1

Divide both sides by $-C$:

$\dfrac{Ax}{-C} + \dfrac{By}{-C} = 1$

Write each term with $x$ or $y$ in the numerator:

$$\frac{x}{-\dfrac{C}{A}} + \frac{y}{-\dfrac{C}{B}} = 1$$

3

Compare with the intercept form

The intercept form is $\dfrac{x}{a} + \dfrac{y}{b} = 1$

Comparing, we get: $a = -\dfrac{C}{A}$ (x-intercept)

$b = -\dfrac{C}{B}$ (y-intercept)

Therefore, $Ax + By + C = 0$ ⇒ $\dfrac{x}{-C/A} + \dfrac{y}{-C/B} = 1$

where $a = -\dfrac{C}{A}$ and $b = -\dfrac{C}{B}$.

Reduction in perpendicular (normal) form

Given the general equation of a line: $Ax + By + C = 0$

1

Rewrite the equation

Move $C$ to the right side: $Ax + By = -C$ Moving a term across the equals sign changes its sign.

2

Compare with the normal form

The normal form is $x\cos\alpha + y\sin\alpha = p$ $\alpha$: angle of the normal with x-axis
$p$: distance of the line from origin

Multiply by a constant $k$: $kAx + kBy = -kC$ $k$ is chosen so that both equations are identical.

Comparing, we get: $\cos\alpha = kA$ $\sin\alpha = kB$ $p = -kC$

3

Find k

Squaring and adding the first two: $k^2(A^2+B^2) = 1$ Using the identity $\cos^2\alpha+\sin^2\alpha=1$

Solve for $k$: $k = \pm\dfrac{1}{\sqrt{A^2+B^2}}$ Choose the sign of $k$ opposite to the sign of $C$, so that $p>0$.

Therefore, $Ax + By + C = 0$ ⇒ $\dfrac{A}{\pm\sqrt{A^2+B^2}}\,x+\dfrac{B}{\pm\sqrt{A^2+B^2}}\,y=-\dfrac{C}{\pm\sqrt{A^2+B^2}}$

where $\cos\alpha=\dfrac{A}{\pm\sqrt{A^2+B^2}}$, $\sin\alpha=\dfrac{B}{\pm\sqrt{A^2+B^2}}$, and $p=-\dfrac{C}{\pm\sqrt{A^2+B^2}}$.

(The sign of ± is taken opposite to the sign of $C$, so that $p>0$.)

Reduction in slope-intercept form

Given the general equation of a line: $Ax + By + C = 0$

1

Solve for y

Move $Ax + C$ to the right side:

$By = -Ax - C$

Divide by $B$ (assuming $B \neq 0$): $y = -\dfrac{A}{B}x - \dfrac{C}{B}$

$$y = -\frac{A}{B}x - \frac{C}{B}$$

2

Compare with the slope-intercept form

The slope-intercept form is $y = mx + c$

Comparing, we get: $m = -\dfrac{A}{B}$ (slope)

$c = -\dfrac{C}{B}$ (y-intercept)

Therefore, $Ax + By + C = 0$ ⇒ $y = -\dfrac{A}{B}x - \dfrac{C}{B}$

where $m = -\dfrac{A}{B}$ and $c = -\dfrac{C}{B}$.

Example 1 of 2

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