Straight line equations — slope form, intercept form, normal form selection

Question

A line passes through $(3, -2)$ and makes an angle of 45° with the positive x-axis. Write its equation in slope-intercept form, point-slope form, and normal form.

Solution — Step by Step

Slope $m = \tan 45° = 1$ . Point: $(3, -2)$ .

y - (-2) = 1(x - 3) \implies \mathbf{y + 2 = x - 3}

\mathbf{x - y - 5 = 0}

From $y + 2 = x - 3$ : $y = x - 5$

So slope $m = 1$ and y-intercept $c = -5$ : $\mathbf{y = x - 5}$

Normal form: $x\cos\alpha + y\sin\alpha = p$ where $p > 0$ is the perpendicular distance from origin and $\alpha$ is the angle the normal makes with x-axis.

From $x - y - 5 = 0$ : divide by $\sqrt{1^2 + (-1)^2} = \sqrt{2}$ .

But we need the RHS to be positive, so write as $-x + y + 5 = 0 \implies -x + y = -5$ … Actually, let us rewrite: $x - y = 5$ , divide by $\sqrt{2}$ :

\frac{x}{\sqrt{2}} - \frac{y}{\sqrt{2}} = \frac{5}{\sqrt{2}}

$\cos\alpha = 1/\sqrt{2}$ , $\sin\alpha = -1/\sqrt{2}$ → $\alpha = -\pi/4$ (or $7\pi/4$ , or equivalently $315°$ ).

$p = 5/\sqrt{2} = 5\sqrt{2}/2$ . Normal form: $\mathbf{x\cos(315°) + y\sin(315°) = \frac{5\sqrt{2}}{2}}$

Why This Works

graph TD
    A["Which line equation form to use?"] --> B["Given slope and y-intercept?"]
    B -->|Yes| C["Slope-intercept: y = mx + c"]
    A --> D["Given slope and a point?"]
    D -->|Yes| E["Point-slope: y - y₁ = m x - x₁"]
    A --> F["Given two intercepts?"]
    F -->|Yes| G["Intercept form: x/a + y/b = 1"]
    A --> H["Need perpendicular distance from origin?"]
    H -->|Yes| I["Normal form: x cos α + y sin α = p"]
    A --> J["Given two points?"]
    J -->|Yes| K["Two-point: y-y₁ / y₂-y₁ = x-x₁ / x₂-x₁"]

Each form of the line equation is suited to different given information. The point-slope form is the most versatile — it works whenever you have any point on the line and the slope. The intercept form is elegant when both intercepts are known. The normal form is essential for distance calculations.

All forms ultimately represent the same line — converting between them is just algebra. The general form $ax + by + c = 0$ is the most compact and is used for distance formulas: distance from $(x_1, y_1)$ to line $ax + by + c = 0$ is $\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$ .

Alternative Method

For JEE, the parametric form (also called symmetric form) is extremely useful: $\frac{x - x_1}{\cos\theta} = \frac{y - y_1}{\sin\theta} = r$ , where $r$ is the distance from $(x_1, y_1)$ along the line. Any point on the line is $(x_1 + r\cos\theta, y_1 + r\sin\theta)$ .

This form is a JEE Advanced favourite because it lets you express points on a line in terms of a single parameter $r$ (the distance), making optimization problems on lines much easier.

Common Mistake

Using the intercept form when the line passes through the origin. The intercept form $x/a + y/b = 1$ requires non-zero intercepts. If the line passes through the origin, both intercepts are 0, and the form breaks down (division by zero). Use the slope form $y = mx$ instead. Similarly, the intercept form fails for vertical lines ( $x = k$ ) and horizontal lines through origin ( $y = 0$ ). Always check applicability before using a specific form.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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