Introduction

The equation of a straight line is an algebraic rule that is true for every point on the line. Different forms of the equation are used depending on the geometric information you are given.

1. Point-Slope Form

When to use: When you know one point (x1,y1)(x_1, y_1) on the line and its slope 'm'. This form comes directly from the definition of slope.

y−y1=m(x−x1)y - y_1 = m(x - x_1)

2. Two-Point Form

When to use: When you know two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) on the line. This is just the point-slope form where the slope mm has been replaced by its formula y2−y1x2−x1\frac{y_2 - y_1}{x_2 - x_1}.

y−y1=y2−y1x2−x1(x−x1)y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)

3. Slope-Intercept Form

When to use: When you know the slope 'm' and the y-intercept 'c'. This form is the most common for analyzing line properties.

y=mx+cy = mx + c

4. Intercept Form

When to use: When you know the x-intercept 'a' (where the line crosses the x-axis) and the y-intercept 'b' (where it crosses the y-axis).

xa+yb=1\frac{x}{a} + \frac{y}{b} = 1

5. Normal (or Perpendicular) Form

When to use: This form is defined by two parameters: 'p', the length of the perpendicular from the origin to the line, and 'ω\omega', the angle this perpendicular makes with the positive x-axis.

xcos⁡ω+ysin⁡ω=px\cos\omega + y\sin\omega = p

6. General Form

Any straight line can be written in the form Ax+By+C=0Ax + By + C = 0, where A, B, and C are constants. This is the general form.

  • Slope: m=−ABm = -\frac{A}{B}

  • x-intercept: −CA-\frac{C}{A}

  • y-intercept: −CB-\frac{C}{B}

Example 1: Point-Slope Form

Question: Find the equation of the line passing through (-4, 3) with a slope of 1/2.

Solution:

We are given a point (x1,y1)=(−4,3)(x_1, y_1) = (-4, 3) and a slope m=1/2m=1/2. We use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1).

y−3=12(x−(−4))y - 3 = \frac{1}{2}(x - (-4))

y−3=12(x+4)y - 3 = \frac{1}{2}(x + 4)

To write it in the general form, we can multiply by 2 and rearrange the terms:

2(y−3)=x+4  ⟹  2y−6=x+4  ⟹  x−2y+10=02(y-3) = x+4 \implies 2y-6 = x+4 \implies \mathbf{x - 2y + 10 = 0}

Example 2: Two-Point Form

Question: Find the equation of the line passing through (-1, 1) and (2, -4).

Solution:

We are given two points (x1,y1)=(−1,1)(x_1, y_1) = (-1, 1) and (x2,y2)=(2,−4)(x_2, y_2) = (2, -4).

Step 1: Find the slope.

m=−4−12−(−1)=−53m = \frac{-4 - 1}{2 - (-1)} = \frac{-5}{3}.

Step 2: Use the point-slope form with the first point (-1,1).

y−1=−53(x−(−1))y - 1 = -\frac{5}{3}(x - (-1))

3(y−1)=−5(x+1)3(y-1) = -5(x+1)

3y−3=−5x−5  ⟹  5x+3y+2=03y-3 = -5x-5 \implies \mathbf{5x + 3y + 2 = 0}

Example 3: Intercept Form

Question: Find the equation of a line that makes intercepts -3 and 2 on the x- and y-axes respectively.

Solution:

We are given the x-intercept a=−3a=-3 and the y-intercept b=2b=2. We use the intercept form xa+yb=1\frac{x}{a} + \frac{y}{b} = 1.

x−3+y2=1\frac{x}{-3} + \frac{y}{2} = 1

To convert to the general form, we find a common denominator (which is 6) and multiply the entire equation by it:

6(x−3)+6(y2)=6(1)6(\frac{x}{-3}) + 6(\frac{y}{2}) = 6(1)

−2x+3y=6  ⟹  2x−3y+6=0-2x + 3y = 6 \implies \mathbf{2x - 3y + 6 = 0}.

Example 4: Converting from General to Normal Form

Question: Reduce the equation 3x+y−8=0\sqrt{3}x + y - 8 = 0 to normal form.

Solution:

Step 1: Rewrite the equation in the form Ax+By=CAx+By=C.

3x+y=8\sqrt{3}x+y=8. Here A=3,B=1A=\sqrt{3}, B=1.

Step 2: Find the normalizing factor, which is A2+B2\sqrt{A^2+B^2}.

(3)2+12=3+1=4=2\sqrt{(\sqrt{3})^2+1^2} = \sqrt{3+1} = \sqrt{4}=2.

Step 3: Divide the entire equation by this factor.

32x+12y=82  ⟹  32x+12y=4\frac{\sqrt{3}}{2}x + \frac{1}{2}y = \frac{8}{2} \implies \frac{\sqrt{3}}{2}x + \frac{1}{2}y = 4

Step 4: Compare with the normal form xcos⁡ω+ysin⁡ω=px\cos\omega + y\sin\omega = p.

  • cos⁡ω=3/2\cos\omega = \sqrt{3}/2 and sin⁡ω=1/2\sin\omega = 1/2. This means the angle ω\omega is 30∘30^\circ or π/6\pi/6.

  • The perpendicular distance from the origin, p, is 4.