01 · Explore
Basics of Coordinate Geometry
Analytical geometry, introduced by René Descartes, combines algebra and geometry to represent geometric figures through equations.
Coordinate geometry uses a system of axes to locate points in a plane. The distance between any two points P(x₁, y₁) and Q(x₂, y₂) is calculated using the formula PQ = √((x₂ - x₁)² + (y₂ - y₁)²). This formula is derived from the Pythagorean theorem by considering the horizontal and vertical differences between the points.
The section formula helps find the coordinates of a point dividing a line segment joining (x₁, y₁) and (x₂, y₂) in a ratio m:n. For internal division, the coordinates are ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)). If m = n, the point is the midpoint, with coordinates ((x₁ + x₂)/2, (y₁ + y₂)/2).
The area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by 1/2 |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|. If this area is zero, the three points are collinear, meaning they lie on the same straight line.
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Question
Find the distance between points (6, -4) and (3, 0).
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02 · Explore
Slope of a Line
The slope or gradient of a line measures its steepness and direction relative to the x-axis.
The inclination (θ) of a line is the angle it makes with the positive direction of the x-axis, measured anticlockwise, where 0° ≤ θ < 180°. The slope (m) is defined as m = tan θ. For a horizontal line, θ = 0°, so m = 0. For a vertical line, θ = 90°, and the slope is undefined.
When two points (x₁, y₁) and (x₂, y₂) on a non-vertical line are known, the slope is calculated as the ratio of the change in y-coordinates to the change in x-coordinates: m = (y₂ - y₁) / (x₂ - x₁). This remains consistent whether the inclination is acute or obtuse.
Slope is a fundamental property used to define the relationship between lines. Parallel lines have equal slopes (m₁ = m₂), while perpendicular lines have slopes that are negative reciprocals of each other (m₁ × m₂ = -1).
| Line Type | Inclination (θ) | Slope (m) |
|---|---|---|
| Horizontal | 0° | 0 |
| Vertical | 90° | Undefined |
| Rising (Left to Right) | 0° < θ < 90° | Positive |
| Falling (Left to Right) | 90° < θ < 180° | Negative |
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Question
What is the slope of a line passing through (3, -2) and (-1, 4)?
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03 · Explore
Angle Between Two Lines
When two lines intersect, they form angles that can be determined using their respective slopes.
If two lines L₁ and L₂ have slopes m₁ and m₂, the acute angle θ between them is given by the formula tan θ = |(m₂ - m₁) / (1 + m₁m₂)|, provided that 1 + m₁m₂ ≠ 0. This formula is derived from the trigonometric identity for tan(α₂ - α₁).
The absolute value ensures we find the acute angle. The obtuse angle φ can be found using the supplementary relationship φ = 180° - θ. If 1 + m₁m₂ = 0, the lines are perpendicular (θ = 90°).
This relationship is useful for finding the slope of an unknown line when the angle it makes with a known line is given. In such cases, there are often two possible slopes for the unknown line, representing two different lines that satisfy the angular condition.
Finding Slope via Angle
tan(π/4) = |(m - 1/2) / (1 + m/2)|
Given θ = 45° (π/4) and m₁ = 1/2. Since tan(45°) = 1, we have 1 = |(2m - 1) / (2 + m)|. Case 1: (2m - 1)/(2 + m) = 1 leads to 2m - 1 = 2 + m, so m = 3. Case 2: (2m - 1)/(2 + m) = -1 leads to 2m - 1 = -2 - m, so 3m = -1, m = -1/3. The possible slopes are 3 or -1/3.
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Question
If two lines are parallel, what is the angle between them?
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04 · Explore
Point-Slope and Two-Point Forms
These forms are used to find the equation of a line when specific points or slopes are known.
The point-slope form is used when we know a fixed point (x₀, y₀) on the line and its slope m. The equation is y - y₀ = m(x - x₀). Every point (x, y) on the line must satisfy this algebraic condition.
The two-point form is used when the line passes through two given points (x₁, y₁) and (x₂, y₂). Since the slope m can be expressed as (y₂ - y₁) / (x₂ - x₁), we substitute this into the point-slope form to get: y - y₁ = [(y₂ - y₁) / (x₂ - x₁)] (x - x₁).
Horizontal and vertical lines have simpler equations. A horizontal line at distance 'a' from the x-axis is y = a or y = -a. A vertical line at distance 'b' from the y-axis is x = b or x = -b.
Equation from Two Points
y - (-1) = [(5 - (-1)) / (3 - 1)] (x - 1)
For points (1, -1) and (3, 5): Slope m = (5 + 1) / (3 - 1) = 6/2 = 3. Equation: y + 1 = 3(x - 1) => y + 1 = 3x - 3 => 3x - y - 4 = 0.
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Question
Find the equation of a line passing through (-2, 3) with slope -4.
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05 · Explore
Slope-Intercept and Intercept Forms
These forms represent lines based on where they cross the coordinate axes.
The slope-intercept form is y = mx + c, where m is the slope and c is the y-intercept (the y-coordinate where the line crosses the y-axis). If the line makes an x-intercept 'd', the equation can also be written as y = m(x - d).
The intercept form is used when a line makes an x-intercept 'a' and a y-intercept 'b'. The equation is (x/a) + (y/b) = 1. This form is particularly useful for quickly sketching a line by plotting the points (a, 0) and (0, b).
The general equation of any straight line is Ax + By + C = 0, where A and B are not both zero. Any of the specific forms can be reduced to this general form, and vice versa, to identify the line's properties.
Using Intercept Form
x/(-3) + y/2 = 1
For x-intercept -3 and y-intercept 2: Multiply by 6 to clear denominators: -2x + 3y = 6, which can be rearranged as 2x - 3y + 6 = 0.
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Question
Convert 6x + 3y - 5 = 0 into slope-intercept form.
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06 · Explore
Distance of a Point From a Line
The distance from a point to a line is defined as the length of the perpendicular segment from the point to the line.
For a point (x₁, y₁) and a line Ax + By + C = 0, the perpendicular distance 'd' is given by d = |Ax₁ + By₁ + C| / √(A² + B²). The absolute value ensures the distance is always non-negative.
This formula is derived by calculating the area of a triangle formed by the point and the intercepts of the line on the axes, then using the base-height relationship to find the altitude (the perpendicular distance).
A common application is finding the distance from the origin (0, 0) to a line, which simplifies the numerator to |C|. This distance represents the shortest path from the point to the line.
Point to Line Distance
d = |3(3) + (-4)(-5) - 26| / √(3² + (-4)²)
To find the distance of (3, -5) from 3x - 4y - 26 = 0: d = |9 + 20 - 26| / √(9 + 16) = |3| / √25 = 3/5 units.
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Question
What is the distance of the line 3x - 4y + 5 = 0 from the origin?
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07 · Explore
Distance Between Parallel Lines
Parallel lines never intersect and maintain a constant perpendicular distance from each other.
Two parallel lines can be represented as Ax + By + C₁ = 0 and Ax + By + C₂ = 0. Note that the coefficients A and B are identical because the slopes are equal.
The distance 'd' between these lines is calculated as d = |C₁ - C₂| / √(A² + B²). This is equivalent to finding the distance from a point on one line to the other line.
If the lines are given in slope-intercept form as y = mx + c₁ and y = mx + c₂, the formula becomes d = |c₁ - c₂| / √(1 + m²). Both formulas provide the same result for the perpendicular gap between the lines.
Distance Between Parallel Lines
d = |7 - 5| / √(3² + (-4)²)
For lines 3x - 4y + 7 = 0 and 3x - 4y + 5 = 0: d = 2 / √(9 + 16) = 2/5 units.
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Question
Find the distance between y = 2x + 3 and y = 2x - 7.
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08 · Explore
Concurrency and Image of a Point
Advanced concepts in straight lines include the intersection of multiple lines and the reflection of points across a line.
Three lines are said to be concurrent if they all pass through a single common point. To check for concurrency, find the intersection point of any two lines and verify if it satisfies the equation of the third line.
The image of a point P(x₁, y₁) with respect to a line (acting as a mirror) is a point Q(h, k) such that the line is the perpendicular bisector of the segment PQ. This implies two conditions: the slope of PQ is the negative reciprocal of the line's slope, and the midpoint of PQ lies on the line.
Solving these two conditions simultaneously allows for the determination of the coordinates (h, k) of the image point. This technique is widely used in optics and coordinate transformations.
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Question
If lines 2x + y - 3 = 0, 3x - y - 2 = 0, and 5x + ky - 3 = 0 are concurrent, find k.
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