Straight Lines
Coordinate Geometry: Straight Lines
Linear Slopes & Core Functional Forms
Slopes, Inclinations, and Perpendicular ConditionsTopic 1
The inclination of a non-vertical straight line is defined as the angle $\theta$ ($0 \le \theta < \pi$) measured in the anti-clockwise direction from the positive direction of the x-axis to the line. The slope or gradient $m$ of the line is the trigonometric tangent of its inclination: \[ m = \tan\theta \] If a line passes through two distinct coordinate points $P(x_1, y_1)$ and $Q(x_2, y_2)$, its slope is uniquely calculated using the vertical change divided by the horizontal change: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \quad (x_1 \neq x_2) \]
Geometric relationships between multiple lines are governed by their respective slopes:- Parallel Lines: Two non-vertical lines are parallel if and only if their slopes are identically equal ($m_1 = m_2$).
- Perpendicular Lines: Two non-vertical lines are perpendicular if and only if the product of their slopes evaluates to negative unity: \[ m_1 \cdot m_2 = -1 \implies m_2 = -\frac{1}{m_1} \]
Find the slope of a straight line that is perpendicular to the line passing through the distinct points $A(2, 5)$ and $B(-4, 3)$.
Show solution
First, determine the slope $m_1$ of the baseline path connecting coordinates $A$ and $B$: \[ m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 5}{-4 - 2} = \frac{-2}{-6} = \frac{1}{3} \] Let $m_2$ represent the slope of the line perpendicular to $AB$. Apply the standard perpendicular slope condition: \[ m_1 \cdot m_2 = -1 \implies \left(\frac{1}{3}\right) \cdot m_2 = -1 \implies m_2 = -3 \] Final Answer: The slope of the perpendicular line is $-3$.
Standard Forms of Equations and Symmetric ParameterizationsTopic 2
- Slope-Intercept Form: $y = mx + c$ (where $c$ is the y-intercept)
- Point-Slope Form: $y - y_1 = m(x - x_1)$ (passes through $P(x_1, y_1)$)
- Two-Point Form: $y - y_1 = \left(\frac{y_2 - y_1}{x_2 - x_1}\right)(x - x_1)$
- Intercept Form: $\frac{x}{a} + \frac{y}{b} = 1$ (where $a$ and $b$ are the x and y intercepts)
- Normal Form: $x \cos\alpha + y \sin\alpha = p$ (where $p$ is the perpendicular distance from the origin and $\alpha$ is the inclination angle of that perpendicular vector)
- Symmetric / Parametric Form: Tracks coordinates step-wise using a directional distance parameter $r$ from a baseline point $P(x_1, y_1)$ along an inclination angle $\theta$: \[ \frac{x - x_1}{\cos\theta} = \frac{y - y_1}{\sin\theta} = r \implies x = x_1 + r\cos\theta, \quad y = y_1 + r\sin\theta \]
Find the coordinate position of a point located at a distance of $r = 5$ units away from the baseline point $P(2, 3)$ along the direction of the line whose inclination angle is $\theta = 30^\circ$.
Show solution
Apply the symmetric parametric structural equations directly to map the shifts along the coordinates: \[ x = x_1 + r\cos\theta \quad \text{and} \quad y = y_1 + r\sin\theta \] Substitute the given parameters ($x_1 = 2$, $y_1 = 3$, $r = 5$, and $\theta = 30^\circ$) into the formulas: \[ x = 2 + 5\cos(30^\circ) = 2 + 5\left(\frac{\sqrt{3}}{2}\right) = 2 + \frac{5\sqrt{3}}{2} \] \[ y = 3 + 5\sin(30^\circ) = 3 + 5\left(\frac{1}{2}\right) = 3 + \frac{5}{2} = \frac{11}{2} \] Final Answer: The coordinate point is $\left(2 + \frac{5\sqrt{3}}{2}, \frac{11}{2}\right)$.
Metric Geometry & Point Transformations
Distances, Angles, and Parallel SplitsTopic 1
- Distance of a Point from a Line: The perpendicular distance $d$ from a point $P(x_1, y_1)$ to the general line $Ax + By + C = 0$ is: \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
- Distance Between Parallel Lines: Two parallel lines share identical leading coefficients when normalized, written as $L_1: Ax + By + C_1 = 0$ and $L_2: Ax + By + C_2 = 0$. The perpendicular distance between them is the absolute difference of their constants over the norm coefficient: \[ d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \]
Calculate the perpendicular distance between the two parallel lines $3x - 4y + 7 = 0$ and $6x - 8y - 1 = 0$.
Show solution
First, scale the first line equation by $2$ so that the leading directional coefficients match the second equation exactly: \[ 2(3x - 4y + 7) = 0 \implies 6x - 8y + 14 = 0 \] Now identify the normalized constant parameters from both parallel lines: \[ A = 6, \quad B = -8, \quad C_1 = 14, \quad C_2 = -1 \] Substitute these parameters into the parallel distance formula: \[ d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} = \frac{|14 - (-1)|}{\sqrt{6^2 + (-8)^2}} = \frac{|14 + 1|}{\sqrt{36 + 64}} = \frac{15}{\sqrt{100}} = \frac{15}{10} = 1.5 \] Final Answer: The distance between the parallel lines is $1.5$ units (or $\frac{3}{2}$).
Foot of a Perpendicular and Mirror Reflection ImagesTopic 2
- Foot of the Perpendicular: Let $H(h, k)$ represent the coordinates of the foot of the perpendicular dropped from a point $P(x_1, y_1)$ onto the line $Ax + By + C = 0$. The coordinate shifts are calculated using the ratio formula: \[ \frac{h - x_1}{A} = \frac{k - y_1}{B} = -\frac{Ax_1 + By_1 + C}{A^2 + B^2} \]
- Reflection (Image) of a Point: Let $I(x_2, y_2)$ represent the coordinates of the mirror reflection image of a point $P(x_1, y_1)$ across the line mirror $Ax + By + C = 0$. Since the line acts as the perpendicular bisector connecting the object $P$ and its image $I$, the transformation ratio doubles its scalar displacement factor: \[ \frac{x_2 - x_1}{A} = \frac{y_2 - y_1}{B} = -2\left(\frac{Ax_1 + By_1 + C}{A^2 + B^2}\right) \]
Find the coordinates of the reflection mirror image of the point $P(1, 2)$ in the line $x - y + 4 = 0$.
Show solution
Identify the line parameters and point coordinates: \[ A = 1, \quad B = -1, \quad C = 4, \quad x_1 = 1, \quad y_1 = 2 \] Evaluate the core scale numerator term ($Ax_1 + By_1 + C$): \[ 1(1) + (-1)(2) + 4 = 1 - 2 + 4 = 3 \] Evaluate the norm coefficient squared ($A^2 + B^2$): \[ 1^2 + (-1)^2 = 1 + 1 = 2 \] Apply the reflection transformation ratio formula to find the coordinates $(x_2, y_2)$: \[ \frac{x_2 - 1}{1} = \frac{y_2 - 2}{-1} = -2\left(\frac{3}{2}\right) = -3 \] Separate the equations to solve for each coordinate independently: \[ x_2 - 1 = -3 \implies x_2 = -2 \] \[ \frac{y_2 - 2}{-1} = -3 \implies y_2 - 2 = 3 \implies y_2 = 5 \] Final Answer: The reflection image coordinates are $I(-2, 5)$.
Linear Systems & Concurrency Operators
Concurrency Determinants and Families of LinesTopic 1
A set of three or more distinct straight lines is said to be concurrent if they all pass through the exact same point of intersection. Three lines given by $L_i: A_i x + B_i y + C_i = 0$ (where $i = 1, 2, 3$) are concurrent if and only if the determinant of their coefficients evaluates to zero: \[ \begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix} = 0 \]
The infinite set of lines passing through the intersection of two baseline lines $L_1 = 0$ and $L_2 = 0$ is defined as a family of straight lines. Any line belonging to this family can be modeled using the linear combination formula: \[ L_1 + \lambda L_2 = 0 \implies (A_1 x + B_1 y + C_1) + \lambda(A_2 x + B_2 y + C_2) = 0 \] where $\lambda \in \mathbb{R}$ is a variable scalar parameter. This formula allows you to find a specific line from the family by applying a given geometric constraint to solve for $\lambda$.
Find the value of the constant parameter $k$ for which the three straight lines $x - y + 1 = 0$, $2x + y - 4 = 0$, and $kx + 2y - 5 = 0$ are concurrent.
Show solution
Set up the coefficient matrix determinant and equate it to zero to enforce concurrency: \[ \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & -4 \\ k & 2 & -5 \end{vmatrix} = 0 \] Expand the matrix determinant along the first row: \[ 1 \cdot [1(-5) - (-4)(2)] - (-1) \cdot [2(-5) - (-4)(k)] + 1 \cdot [2(2) - 1(k)] = 0 \] Simplify the terms inside each bracket: \[ 1 \cdot [-5 + 8] + 1 \cdot [-10 + 4k] + 1 \cdot [4 - k] = 0 \] \[ 3 - 10 + 4k + 4 - k = 0 \] Combine the constants and the variable terms to solve for $k$: \[ 3k - 3 = 0 \implies 3k = 3 \implies k = 1 \] Final Answer: The lines are concurrent when $k = 1$.
Homogeneous Quadratic Forms & Joint Line Pairs
Homogeneous Second-Degree Equations and Angle BisectorsTopic 1
- Coincident Lines: The lines are coincident (or parallel) if $h^2 - ab = 0 \implies h^2 = ab$.
- Perpendicular Lines: The lines are perpendicular if the sum of the squared coefficients vanishes: $a + b = 0$.
Find the angle between the pair of straight lines represented by the homogeneous equation $3x^2 + 4xy + y^2 = 0$, and find the joint equation of their angle bisectors.
Show solution
Identify the homogeneous coefficients from the given equation: \[ a = 3, \quad 2h = 4 \implies h = 2, \quad b = 1 \] Calculate the angle $\theta$ using the joint tangent formula: \[ \tan\theta = \left| \frac{2\sqrt{2^2 - 3(1)}}{3 + 1} \right| = \left| \frac{2\sqrt{4 - 3}}{4} \right| = \frac{2(1)}{4} = \frac{1}{2} \implies \theta = \tan^{-1}\left(\frac{1}{2}\right) \] Now set up the joint equation of the angle bisectors using the standard formula: \[ \frac{x^2 - y^2}{a - b} = \frac{xy}{h} \implies \frac{x^2 - y^2}{3 - 1} = \frac{xy}{2} \implies \frac{x^2 - y^2}{2} = \frac{xy}{2} \] Cancel the denominators to find the final equation: \[ x^2 - y^2 = xy \implies x^2 - xy - y^2 = 0 \] Final Answer: $\theta = \tan^{-1}\left(\frac{1}{2}\right)$; Bisectors: $x^2 - xy - y^2 = 0$.
General Second-Degree Equations and Locus TrackingTopic 2
The general second-degree equation in terms of $x$ and $y$ incorporates linear and constant terms along with the quadratic terms: \[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \] This general equation represents a pair of straight lines if and only if the determinant of its complete coefficient matrix vanishes. This condition is represented by the discriminant equation $\Delta = 0$: \[ \Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \] Additionally, to represent real lines, the coefficients must satisfy $h^2 \ge ab$.
If $\Delta = 0$, the general equation can be factored into two distinct linear equations representing lines that intersect at an interior point away from the origin. The coordinates of this intersection point can be found by taking the partial derivatives of the general equation with respect to $x$ and $y$ and solving the resulting linear system: \[ \frac{\partial f}{\partial x} = 2ax + 2hy + 2g = 0 \quad \text{and} \quad \frac{\partial f}{\partial y} = 2hx + 2by + 2f = 0 \]
Determine the value of the constant parameter $c$ for which the general second-degree equation $2x^2 + 5xy + 2y^2 + 3x + 3y + c = 0$ represents a pair of straight lines.
Show solution
Identify the complete set of coefficients from the general equation structure: \[ a = 2, \quad b = 2, \quad 2h = 5 \implies h = \frac{5}{2} \] \[ 2g = 3 \implies g = \frac{3}{2}, \quad 2f = 3 \implies f = \frac{3}{2}, \quad c = c \] Set up the discriminant equation $\Delta = 0$ to enforce the line pair condition: \[ abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \] Substitute the extracted coefficients into the formula: \[ (2)(2)(c) + 2\left(\frac{3}{2}\right)\left(\frac{3}{2}\right)\left(\frac{5}{2}\right) - 2\left(\frac{3}{2}\right)^2 - 2\left(\frac{3}{2}\right)^2 - c\left(\frac{5}{2}\right)^2 = 0 \] \[ 4c + \frac{45}{4} - \frac{9}{2} - \frac{9}{2} - \frac{25}{4}c = 0 \] Combine the constant fractions and the variable terms: \[ \left(4 - \frac{25}{4}\right)c + \left(\frac{45}{4} - 9\right) = 0 \] \[ -\frac{9}{4}c + \frac{9}{4} = 0 \implies \frac{9}{4}c = \frac{9}{4} \implies c = 1 \] Final Answer: The equation represents a pair of lines when $c = 1$.
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