Straight Lines revision notes

Coordinate geometry integrates algebra and geometry to study geometric figures systematically. Before proceeding to straight lines, the foundational formulas for points in a 2D Cartesian plane are essential.

• Distance Formula: The distance between points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
• Section Formula (Internal): The coordinates of a point dividing the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m:n$ are $\left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)$.
• Section Formula (External): If the division is external in the ratio $m:n$, the coordinates are $\left( \frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n} \right)$. JEE Tip Always verify if the ratio is internal or external. External ratios can be treated as internal with a negative fractional ratio.
• Mid-point Formula: For $m=n$, the mid-point is $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.
• Area of a Triangle: For vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, the area is $\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
• Collinearity Condition: Three points A, B, and C lie on a single line (collinear) if and only if the area of the triangle formed by them is zero.

The slope (or gradient) represents the steepness and direction of a line, acting as the fundamental parameter for algebraic representation.

• Slope given two points: For a non-vertical line passing through $P(x_1, y_1)$ and $Q(x_2, y_2)$, the slope is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

The relative orientation of two lines can be determined purely by their slopes.

• Parallel Lines: Two non-vertical lines $l_1$ and $l_2$ with slopes $m_1$ and $m_2$ are parallel if and only if their inclinations are equal ($\alpha = \beta$), thus $m_1 = m_2$.
• Perpendicular Lines: Two non-vertical lines are perpendicular if and only if their slopes are negative reciprocals, meaning $m_1 m_2 = -1$. JEE Tip This rule fails if one line is vertical and the other is horizontal. Always check for zero and undefined slopes first.

A line is defined algebraically by a condition satisfied only by the points $(x,y)$ lying on it.

• Horizontal and Vertical Lines: A line parallel to the x-axis at distance $a$ is $y = \pm a$. A line parallel to the y-axis at distance $b$ is $x = \pm b$.
• Point-Slope Form: The line through fixed point $(x_0, y_0)$ with slope $m$ is $y - y_0 = m(x - x_0)$.
• Two-Point Form: The line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.

• Intercept Form: A line making x-intercept $a$ and y-intercept $b$ is $\frac{x}{a} + \frac{y}{b} = 1$. JEE Tip The area of the triangle formed by this line with the coordinate axes is $\frac{1}{2}|ab|$.

• Parametric / Distance Form: JEE Tip The equation of a line passing through $(x_1, y_1)$ at an angle $\theta$ is $\frac{x - x_1}{\cos \theta} = \frac{y - y_1}{\sin \theta} = r$, where $r$ is the directed distance from $(x_1, y_1)$ to any point $(x,y)$ on the line. Highly useful for finding points at a specific distance along a line.
• Normal Form: $x \cos \alpha + y \sin \alpha = p$, where $p$ is the perpendicular distance from the origin to the line, and $\alpha$ is the angle this perpendicular makes with the positive x-axis.

• Distance of a Point from a Line: The perpendicular distance $d$ of a point $(x_1, y_1)$ from the line $Ax + By + C = 0$ is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
• Distance Between Parallel Lines: For lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$, the distance $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. JEE Tip Before applying this, strictly ensure the coefficients $A$ and $B$ are identical in both equations.
• Relative Position of Two Points: JEE Tip Points $(x_1, y_1)$ and $(x_2, y_2)$ lie on the same side of the line $Ax + By + C = 0$ if $(Ax_1 + By_1 + C)(Ax_2 + By_2 + C) > 0$, and on opposite sides if the product is $< 0$.
• Foot of Perpendicular: JEE Tip The foot $(h, k)$ of the perpendicular from $(x_1, y_1)$ to the line $Ax + By + C = 0$ is found using: $\frac{h - x_1}{A} = \frac{k - y_1}{B} = -\frac{Ax_1 + By_1 + C}{A^2 + B^2}$.
• Image of a Point: JEE Tip The reflection $(h, k)$ of $(x_1, y_1)$ across the line mirror $Ax + By + C = 0$ is found using: $\frac{h - x_1}{A} = \frac{k - y_1}{B} = -2\frac{Ax_1 + By_1 + C}{A^2 + B^2}$.

• Concurrency Condition: Three lines $A_1x + B_1y + C_1 = 0$, $A_2x + B_2y + C_2 = 0$, and $A_3x + B_3y + C_3 = 0$ are concurrent (pass through a common point) if the determinant of their coefficients is zero: $\begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix} = 0$. Alternatively, solve two equations and check if the intersection satisfies the third.
• Family of Lines: JEE Tip The equation of any line passing through the intersection of two lines $L_1 = 0$ and $L_2 = 0$ is $L_1 + \lambda L_2 = 0$, where $\lambda$ is a parameter ($\lambda \ne \text{coefficient constraint}$).

JEE Tip For two intersecting lines $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$, the equations of their angle bisectors are: $\frac{A_1x + B_1y + C_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{A_2x + B_2y + C_2}{\sqrt{A_2^2 + B_2^2}}$

• Origin Containing Bisector: Ensure $C_1 > 0$ and $C_2 > 0$ by multiplying the equations by $-1$ if necessary. Then, the positive sign ($+$) corresponds to the bisector containing the origin.
• Acute/Obtuse Bisector: With $C_1, C_2 > 0$, evaluate the sign of $(A_1A_2 + B_1B_2)$. If $(A_1A_2 + B_1B_2) > 0$, the $+$ sign gives the obtuse angle bisector, and the $-$ sign gives the acute angle bisector. If $(A_1A_2 + B_1B_2) < 0$, the roles reverse.

• Slope limitation: The slope $m$ is undefined for vertical lines (inclination $90^\circ$). Equations involving slope directly (like $y=mx+c$) cannot represent vertical lines.
• Angle limit: Angle formula between lines uses tangent. If the lines are perpendicular, $1+m_1m_2 = 0$, making $\tan \theta$ undefined (i.e., $\theta = 90^\circ$).
• Parallel Lines Distance: The formula $d = |C_1-C_2|/\sqrt{A^2+B^2}$ CANNOT be used if the coefficients of $x$ and $y$ are proportional but not exactly identical. (e.g., $2x+3y+5=0$ and $4x+6y+7=0$. You must divide the second equation by 2 first).

• Inclination Range vs Standard Angles: Inclination $\theta$ is restricted strictly to $[0, 180^\circ]$. Never use negative angles or angles $>180^\circ$ for inclination.
• Modulus in Angle Formula: Students often forget the modulus in $\tan \theta = |\frac{m_2-m_1}{1+m_1m_2}|$. Without it, you might calculate the obtuse angle instead of the acute angle.
• Intercepts vs Lengths: An intercept ($a$ or $b$) is a coordinate, meaning it can be negative. It is NOT merely a length (which is strictly positive). A negative intercept implies it cuts the negative axis.
• Modulus in Distance: The distance formulas $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$ MUST include the absolute value operator. Distance cannot be negative.
• Perpendicular Slopes Trap: The rule $m_1 m_2 = -1$ breaks if one line is the x-axis ($m=0$) and the other is the y-axis (m = undefined). The product $0 \times \text{undefined}$ is an indeterminate form, but the lines are still geometrically perpendicular.

• Image and foot of perpendicular of a point with respect to a straight line.
• Equations of angle bisectors and identifying the region containing the origin.
• Family of lines (proving lines pass through a fixed point regardless of a parameter).
• Locus of a point satisfying given geometric distance constraints.
• Triangle centers (Incenter, Orthocenter) constructed via intersections of specifically defined lines.

Derivation of Distance of a Point from a Line

1. Let line $L: Ax + By + C = 0$ and point $P(x_1, y_1)$.
2. The line meets axes at $Q(-C/A, 0)$ and $R(0, -C/B)$.
3. Calculate Area of $\Delta PQR$ using vertex coordinates: Area = $\frac{1}{2} |x_1(0 - (-C/B)) + (-C/A)(-C/B - y_1) + 0(y_1 - 0)|$.
4. Simplify to Area = $\frac{1}{2} |\frac{C}{AB}(Ax_1 + By_1 + C)|$.
5. Also, Area of $\Delta PQR$ = $\frac{1}{2} \times \text{base (QR)} \times \text{height (PM, which is } d)$.
6. Distance $QR = \sqrt{(-C/A - 0)^2 + (0 - (-C/B))^2} = \frac{|C|}{|AB|} \sqrt{A^2 + B^2}$.
7. Equating the areas: $\frac{1}{2} d \times \frac{|C|}{|AB|} \sqrt{A^2 + B^2} = \frac{1}{2} |\frac{C}{AB}(Ax_1 + By_1 + C)|$.
8. Solving for $d$: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.