equation of a line | ICSE Class 10 Maths Notes

This note covers coordinates, the meaning of a line equation, slope and inclination, horizontal and vertical lines, slope-intercept form, equations through given points, parallel and perpendicular lines, and checking solutions.

What does the equation of a straight line tell us?

How do coordinates describe a point?

The coordinate plane contains two perpendicular number lines: the horizontal x-axis and the vertical y-axis. Perpendicular lines meet at a right angle of ninety degrees. Their intersection is the origin, written as the point (0, 0).

A point has an ordered pair of coordinates, written (x, y). Here x is the horizontal coordinate, or abscissa, and y is the vertical coordinate, or ordinate. The order matters: the first entry locates the point horizontally and the second locates it vertically.

For the point (6, −4), the horizontal position is 6 units in the positive x-direction, and the vertical position is 4 units in the negative y-direction. The point (3, 0) lies on the x-axis because its ordinate is zero.

How does an equation select points?

An equation of a line is an algebraic condition satisfied by the coordinates of every point on that line and by no other point in the plane. An equation is an equality between two expressions. Its variables x and y represent the coordinates of a general point.

To test whether a given point lies on a line, substitute its coordinates into the equation. Substitution means replacing each variable by its specified value. If the equality holds, the point lies on the line; if it fails, the point does not.

A line contains infinitely many points, so its equation describes infinitely many coordinate pairs. Two distinct points, meaning two different points, determine one straight line. This is why two given points provide enough information to find a line equation.

Definition: A point satisfies a line equation when replacing x and y with that point's coordinates makes the two sides equal.

How do slope and inclination describe the direction of a line?

What is inclination?

The inclination of a line is the angle measured anticlockwise from the positive direction of the x-axis to the line. Write this angle as θ, pronounced theta. The symbol ° means degrees, the unit used here for angles. A horizontal line has inclination 0°, while a vertical line has inclination 90°.

An acute angle is greater than 0° and less than 90°. An obtuse angle is greater than 90° and less than 180°. These describe the two possible sloping positions considered here.

Result: Slope from inclination

The slope, also called the gradient, is denoted by m. For a non-vertical line, m = tan θ, where tan means the tangent function. In a right-angled triangle, tangent is the opposite side divided by the adjacent side for the chosen acute angle.

An acute inclination gives a positive slope, while an obtuse inclination gives a negative slope. Read from left to right, a line with positive slope rises and a line with negative slope falls. A horizontal line has zero slope. A vertical line has undefined slope.

Worked example 1. Find the slope of a line whose inclination is 60°.

Answer: Use m = tan θ. With θ = 60°, m = tan 60° = √3. The symbol √ denotes the positive square root. The positive result agrees with the acute inclination.

The axes meet at O, the origin. The sloping line l crosses the x-axis to the right of O and the y-axis below O. The angle θ is marked between the positive x-axis and the line; the adjacent angle is labelled 180° − θ.

See Fig. 9.2 in your NCERT textbook

The letter l in the diagram is simply the name of the line. Measure inclination from the stated positive x-direction. Using the angle with the y-axis directly would generally give a different tangent and therefore the wrong slope.

How is slope calculated from two given points?

Result: Difference of ordinates divided by difference of abscissae

Let P and Q name two distinct points with coordinates (x₁, y₁) and (x₂, y₂), respectively. The small labels 1 and 2 identify the first and second points. Thus x₁ and y₁ belong together, as do x₂ and y₂.

Provided x₂ ≠ x₁, the formula is m = (y₂ − y₁)/(x₂ − x₁). Here ≠ means “is not equal to”, and the slash indicates division. The numerator, above the division, is the change in ordinate; the denominator, below it, is the change in abscissa.

Subtract coordinates in the same point order in both parts. Reversing both differences leaves the quotient unchanged because both are multiplied by −1. Reversing just one difference gives −m instead of m, which is incorrect when m is non-zero; a zero slope remains zero.

Worked example 2. Find the slope of the line through (3, −2) and (−1, 4).

Answer: Take the second point minus the first in both differences. Then m = [4 − (−2)]/(−1 − 3) = 6/(−4) = −3/2. Square brackets group the numerator. The negative result describes a line falling from left to right.

What happens when a difference is zero?

Worked example 3. Find the slope of the line through (3, −2) and (7, −2).

Answer: m = [−2 − (−2)]/(7 − 3) = 0/4 = 0. The equal ordinates show that this line is horizontal.

Worked example 4. Find the slope of the line through (3, −2) and (3, 4).

Answer: The attempted calculation gives [4 − (−2)]/(3 − 3) = 6/0. Division by zero is undefined, so the line is vertical and its slope is not defined.

Check the denominator before dividing. Zero slope and undefined slope describe different directions. Equal ordinates give the first case; equal abscissae give the second, provided the points are distinct.

How does slope-intercept form connect an equation with its graph?

What do m and c mean?

The slope-intercept form is y = mx + c. Here m is the slope, c is the y-intercept, and (x, y) represents any point on the line. The expression mx means m multiplied by x. The numbers m and c are fixed for that line.

The y-intercept is the signed ordinate of the point where the line meets the y-axis. Since every point on the y-axis has x = 0, substitution gives y = c. The intercept point is therefore (0, c), whereas the intercept value is c.

The sign of c matters. A positive c places the crossing above the origin and a negative c places it below. When c = 0, the equation becomes y = mx and the line passes through the origin.

How is the form used?

Worked example 5. Find the equation of a line with tan θ = 1/2 and y-intercept −3/2, where θ is its inclination.

Answer: Its slope is m = 1/2 and its y-intercept is c = −3/2. Substitute in y = mx + c to obtain y = x/2 − 3/2. Multiplying by 2 and rearranging gives 2y − x + 3 = 0.

Rearranging means carrying out equal operations on both sides to write the same equality in another form. The two equations in the example describe the same points. The slope-intercept version displays the slope and y-intercept directly.

A rising line L crosses the y-axis at the labelled point (0, c). The line is labelled “Slope m”, and the origin O is marked where the coordinate axes meet.

See Fig. 9.12 in your NCERT textbook

The letter L names the line in this diagram. A line equation should be read in its actual form: a constant appearing in a rearranged equation is not automatically the y-intercept. First isolate y with coefficient 1, meaning one times y, then identify c.

How do we find a line through a point with a known slope?

Result: Point-slope form

Suppose a non-vertical line has slope m and passes through the fixed point (x₁, y₁). Its equation is y − y₁ = m(x − x₁). This is the point-slope form: it combines one known point with one known slope.

The coordinates x₁ and y₁ are fixed values, while x and y remain variables for a general point on the line. Do not replace every coordinate symbol by a number. The aim is to retain an equation relating the coordinates of all the line's points.

Why does this equation work?

1. Choose another point (x, y) on the non-vertical line, different from the given point.
2. The slope formula gives m = (y − y₁)/(x − x₁).
3. Multiply by x − x₁ to obtain y − y₁ = m(x − x₁).
4. The given point also satisfies this final equation, because substitution makes both sides zero.

Worked example 6. Find the equation of the line through (−2, 3) with slope −4.

Answer: Substitute x₁ = −2, y₁ = 3 and m = −4. Then y − 3 = −4[x − (−2)] = −4(x + 2). Expanding gives y − 3 = −4x − 8, so 4x + y + 5 = 0.

Expanding means multiplying the factor outside brackets by each term inside. The negative sign applies to both terms: −4(x + 2) becomes −4x − 8. Also, subtracting the negative coordinate −2 produces x + 2.

A useful method is to write the unsimplified substitution before expanding. This preserves the point and slope visibly and makes sign errors easier to locate. The point-slope equation already specifies the line; simplification changes its appearance, not which points satisfy it.

This form requires a defined slope. If the given line is vertical, describe its constant x-coordinate directly instead of trying to supply a numerical value for m.

How do two points determine the equation of a line?

How is the two-point form obtained?

Given distinct points (x₁, y₁) and (x₂, y₂), first calculate their slope when x₂ ≠ x₁. Then insert this slope and either point into point-slope form. This gives the two-point form: y − y₁ = [(y₂ − y₁)/(x₂ − x₁)](x − x₁).

This expression is the point-slope equation with the slope written using the two given points. You can therefore solve a two-point problem in two clear stages instead of memorising an unrelated rule. First find the direction; then place the line through a known point.

Worked example 7. Find the equation of the line through (1, −1) and (3, 5).

Answer: The slope is m = [5 − (−1)]/(3 − 1) = 6/2 = 3. Using (1, −1), the equation is y + 1 = 3(x − 1). Hence y = 3x − 4, or −3x + y + 4 = 0.

How can the answer be checked?

For this example, substitute both original points into −3x + y + 4 = 0. At (1, −1), the left side is −3 − 1 + 4 = 0. At (3, 5), it is −9 + 5 + 4 = 0.

Both checks succeed. Because the two distinct given points lie on the resulting straight line, the equation represents their joining line. Checking only the slope would leave the position uncertain: different parallel lines, meaning distinct lines in a plane that do not meet, can have the same slope.

Before starting the fraction, compare the abscissae. If the two points share an x-coordinate, their joining line is vertical. Its equation states that shared x-value directly. If they share a y-coordinate instead, the line is horizontal and states that shared y-value.

Collinear points are points lying on one straight line. Substitution into a line equation is therefore also a way to test whether another point lies on the line determined by the original pair.

What equations describe horizontal and vertical lines?

Which coordinate stays fixed?

A horizontal line runs in the direction of the x-axis. Its points have the same y-coordinate even though their x-coordinates vary. A vertical line runs in the direction of the y-axis. Its points have the same x-coordinate while their y-coordinates vary.

For a positive distance a from the x-axis, the horizontal lines are y = a above it and y = −a below it. Here a denotes that positive distance. For a positive distance b from the y-axis, the vertical lines are x = b to its right and x = −b to its left.

Worked example 8. Find the equations of the lines through (−2, 3) parallel to the coordinate axes. Give one equation for each direction.

Answer: The line parallel to the x-axis has constant ordinate 3, so y = 3. The line parallel to the y-axis has constant abscissa −2, so x = −2.

The horizontal line y = 3 and vertical line x = −2 meet at the labelled point (−2, 3), above and to the left of the origin O.

See Fig. 9.9 in your NCERT textbook

A horizontal line fits y = mx + c with m = 0. A vertical line cannot be written in that form with a defined numerical slope. Its constant-coordinate equation is complete even though it does not contain y.

How do we recognise and construct parallel lines?

Result: Parallel lines have equal slopes

For two distinct non-vertical lines, the condition for being parallel is m₁ = m₂. Here m₁ is the slope of the first line and m₂ is the slope of the second. Their equal inclinations give equal tangents and hence equal slopes.

The condition also works in reverse for distinct non-vertical lines: equal slopes give parallel lines. Their directions agree, even though they occupy different positions in the coordinate plane. The y-intercepts distinguish their positions.

Write the equations as y = mx + c₁ and y = mx + c₂, where c₁ and c₂ are their respective y-intercepts. If c₁ ≠ c₂, these lines are distinct and parallel. If the intercepts are also equal, the equations describe the same line.

How is a parallel line placed through a given point?

1. Identify the given line's slope by writing its equation with y isolated.
2. Use that same slope for the required parallel line.
3. Insert the coordinates of the required point into point-slope form.
4. Simplify the equation and check both the point and the slope.

The point condition fixes which line with that slope is required. Slope alone does not provide the intercept. If the specified point is already on the original line, the equation obtained is the original line, so it is not a distinct parallel.

Vertical lines need a separate geometric check because their slopes are undefined. Distinct vertical lines are parallel to one another. A horizontal line through a given point likewise retains the direction of the x-axis, and its equation uses the point's ordinate.

Do not compare just the visible x-coefficients in differently arranged equations. A coefficient is the number multiplying a variable. The multiplier of x gives the slope directly only after the equation is in y = mx + c form.

How do we use slopes to identify perpendicular lines?

Result: Perpendicular slopes have product −1

For two non-vertical lines, the condition for being perpendicular is m₁m₂ = −1. The expression m₁m₂ means the product of their slopes; the symbol × also denotes multiplication. Equivalently, m₂ = −1/m₁: one slope is the negative reciprocal of the other.

For a non-zero number, its reciprocal is one divided by that number. Taking the negative reciprocal also changes the sign. Merely changing a slope's sign is not the perpendicularity rule; the reciprocal operation is essential.

Worked example 9. The line through (−2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (x, 24). Find x, the unknown abscissa of the last point.

Answer: The first slope is m₁ = (8 − 6)/[4 − (−2)] = 1/3. The second is m₂ = (24 − 12)/(x − 8) = 12/(x − 8). Thus (1/3) × [12/(x − 8)] = −1, giving x − 8 = −4 and x = 4.

When does the numerical test need care?

A horizontal line and a vertical line are perpendicular. The product rule cannot be applied numerically to that pair because the vertical slope is undefined. If one slope is zero, the perpendicular line is vertical, rather than a line with a finite negative reciprocal.

In the worked example, x cannot be 8 for the second line to have a defined slope. The first line is neither horizontal nor vertical, so a perpendicular to it has a defined, non-zero slope. The final value x = 4 satisfies that restriction.

To construct a perpendicular through a specified point, first find the given line's slope. Take its negative reciprocal when appropriate, then use the new slope and the point in point-slope form. Use constant-coordinate equations for the horizontal and vertical cases.

How can we choose a method and verify the complete equation?

Which information should be used first?

Begin by identifying exactly what is given: a slope and an intercept, a slope and a point, two points, or a line with a parallel or perpendicular condition. Each combination supplies the direction and position of the required line in a different way.

How can a perpendicular-line problem be completed?

Worked example 10. Find the equation of the line through (−3, 5) perpendicular to the line through (2, 5) and (−3, 6).

Answer: The given line's slope is (6 − 5)/(−3 − 2) = −1/5. The perpendicular slope is 5. Through (−3, 5), the required equation is y − 5 = 5(x + 3), or y = 5x + 20.

For this answer, substitution of (−3, 5) gives 5 = −15 + 20, which is true. The product of the slopes is (−1/5) × 5 = −1. These checks verify both the specified point and the required perpendicular direction.

What should the final check include?

A general linear equation has the form Ax + By + C = 0, where A, B and C are fixed numbers and A and B are not both zero. Here A and B are coefficients; C is the constant term, which contains no variable.

When B ≠ 0, rearranging gives y = (−A/B)x − C/B. Thus the slope is −A/B and the y-intercept is −C/B. This explains why neither the x-coefficient nor the constant can be read as slope or intercept without considering the equation's form.

Finally, check every supplied point, the required slope relationship, and any zero denominator. Equivalent equations describe the same points even when their terms are arranged differently. A correct final equation must satisfy all the stated conditions together.

Glossary

• Coordinate plane — A plane in which perpendicular horizontal and vertical axes locate points using ordered pairs.
• Origin — The point with coordinates (0, 0), where the two coordinate axes intersect.
• Abscissa — The first coordinate of a point, specifying its horizontal position relative to the origin.
• Ordinate — The second coordinate of a point, specifying its vertical position relative to the origin.
• Line equation — An algebraic condition satisfied by the coordinates of exactly the points on a straight line.
• Inclination — The angle measured anticlockwise from the positive x-direction to a line.
• Slope — The tangent of a non-vertical line's inclination, also called its gradient.
• Y-intercept — The signed ordinate of the point where a line crosses the y-axis.
• Point-slope form — An equation combining a line's defined slope with the coordinates of one fixed point.
• Parallel lines — Distinct lines in the same plane that do not meet each other.
• Perpendicular lines — Lines that intersect at a right angle, such as the two coordinate axes.
• Negative reciprocal — The negative of one divided by a given non-zero number.
• Collinear points — Points that lie together on the same straight line in the coordinate plane.

Common errors and misconceptions

• Misconception: Slope is the change in x divided by the change in y. Correct: Divide the change in ordinate by the change in abscissa, provided the denominator is non-zero.
• Misconception: Either subtraction order can be used independently. Correct: Use the same point order in the numerator and denominator; reversing just one gives −m instead of m, which changes the result unless m = 0.
• Misconception: A vertical line has zero slope. Correct: A vertical line has undefined slope. Zero slope belongs to a horizontal line.
• Misconception: The y-intercept is the point (c, 0). Correct: The y-intercept value is c and its point is (0, c), because x is zero on the y-axis.
• Misconception: A perpendicular slope is obtained just by changing the sign. Correct: Take the negative reciprocal of a non-zero slope; treat horizontal and vertical lines separately.
• Misconception: Equal slopes prove two equations represent distinct parallel lines. Correct: They may represent the same line. Compare their positions, for example their y-intercepts.
• Misconception: Any constant in a line equation is its y-intercept. Correct: Rearrange to y = mx + c before identifying c, or set x = 0 and solve for y.

Exam-style questions with model answers

Q1. Find the slope of the line through (3, −2) and (−1, 4). [2 marks]

1. Use m = (y₂ − y₁)/(x₂ − x₁), keeping the same point order in both differences.
2. Substitution gives m = [4 − (−2)]/(−1 − 3) = 6/(−4) = −3/2.

Q2. Find the slope of the line through (3, −2) and (3, 4), and explain its direction. [2 marks]

1. The slope calculation gives [4 − (−2)]/(3 − 3) = 6/0, so the slope is undefined.
2. The abscissa is the same at both distinct points. Their joining line is vertical, with equation x = 3.

Q3. Find the equation of the line through (−2, 3) with slope −4. Show the point-slope substitution and simplify. [3 marks]

1. Use point-slope form y − y₁ = m(x − x₁). Here the given fixed point has x₁ = −2 and y₁ = 3, and the slope is m = −4.
2. Substitution gives y − 3 = −4[x − (−2)] = −4(x + 2). Subtracting the negative abscissa produces a plus sign inside the brackets.
3. Expand to obtain y − 3 = −4x − 8. Rearranging gives the required line equation 4x + y + 5 = 0.

Q4. A line has inclination θ with tan θ = 1/2 and y-intercept −3/2. Find its equation and give the coordinates where it meets the y-axis. [3 marks]

1. The slope equals the tangent of the inclination. Therefore m = 1/2, while the given signed y-intercept is c = −3/2.
2. Insert these values into y = mx + c. The required equation is y = x/2 − 3/2, equivalently 2y − x + 3 = 0.
3. At the y-axis, x = 0. The ordinate is consequently −3/2, so the point of intersection is (0, −3/2).

Q5. Find the equation of the line through (1, −1) and (3, 5). Verify both given points and state its y-intercept. [5 marks]

1. The abscissae are different, so the line has a defined slope. Calculate m = [5 − (−1)]/(3 − 1) = 6/2 = 3.
2. Use the first point in point-slope form: y − (−1) = 3(x − 1). Thus y + 1 = 3x − 3, giving y = 3x − 4.
3. Check the first point by setting x = 1. The equation gives y = 3 − 4 = −1, agreeing with its given ordinate.
4. Check the second point by setting x = 3. The equation gives y = 9 − 4 = 5, agreeing with its given ordinate.
5. In slope-intercept form, the constant term is −4. Therefore the y-intercept is −4 and the crossing point is (0, −4).

Q6. The line through (−2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (x, 24). Find the unknown abscissa x and check perpendicularity. [5 marks]

1. Calculate the first line's slope: m₁ = (8 − 6)/[4 − (−2)] = 2/6 = 1/3. This is a defined, non-zero slope.
2. For the second line, the slope is m₂ = (24 − 12)/(x − 8) = 12/(x − 8), with x ≠ 8.
3. Perpendicular lines with defined slopes satisfy m₁m₂ = −1. Hence (1/3) × [12/(x − 8)] = −1, or 4/(x − 8) = −1.
4. Multiply by the non-zero denominator to obtain 4 = −(x − 8). Therefore x − 8 = −4 and the required value is x = 4.
5. With x = 4, the second slope is 12/(−4) = −3. Its product with 1/3 is −1, confirming perpendicularity.

Q7. Find the equation of the line through (−3, 5) perpendicular to the line through (2, 5) and (−3, 6). Check both conditions. [5 marks]

1. The given line has slope (6 − 5)/(−3 − 2) = −1/5. The unequal abscissae allow the slope formula to be used.
2. The required perpendicular slope is the negative reciprocal, namely 5. The slopes satisfy (−1/5) × 5 = −1, which checks the direction condition.
3. Use the specified point (−3, 5) and slope 5 in point-slope form. This gives y − 5 = 5[x − (−3)] = 5(x + 3).
4. Expand and simplify: y − 5 = 5x + 15, so the required equation is y = 5x + 20.
5. Substitute the required point to check its position: with x = −3, the equation gives y = −15 + 20 = 5. Both conditions hold.

Key takeaways

• A line equation is satisfied by every point on that line and by no other point in the plane.
• Slope equals the tangent of inclination for a non-vertical line; inclination is measured anticlockwise from the positive x-direction.
• Calculate slope as change in ordinate divided by change in abscissa, retaining the same subtraction order.
• In y = mx + c, m gives the slope and c gives the signed y-intercept.
• Point-slope form uses a known point and a defined slope; two-point problems first supply that slope.
• Distinct non-vertical parallel lines have equal slopes, while non-vertical perpendicular lines have slopes whose product is −1.
• Horizontal lines have constant ordinates and zero slope; vertical lines have constant abscissae and undefined slope.
• Verify an equation using the supplied points and the required direction, checking that no denominator is zero.

Test yourself

Why must the subtraction order match in both parts of the slope formula?

Both differences must describe the same direction between the points. Reversing just one difference gives −m instead of m, producing an incorrect slope when m is non-zero; a zero slope remains zero.

What distinguishes zero slope from undefined slope?

Zero slope belongs to a horizontal line with equal ordinates. Undefined slope belongs to a vertical line with equal abscissae.

Where does y = mx + c meet the y-axis?

It meets the y-axis at (0, c), because putting x = 0 gives y = c.

What equation describes a non-vertical line through the origin with slope m?

The equation is y = mx, because passing through the origin makes its y-intercept zero.

What are the equations of the lines through (−2, 3) parallel to the axes?

The horizontal line is y = 3, and the vertical line is x = −2.

Why is equal slope insufficient to prove that two equations represent distinct parallel lines?

The equations might describe the same line. For non-vertical lines, equal slopes with different y-intercepts establish distinct parallels.

What is the slope of a perpendicular to a line with slope 1/3?

The negative reciprocal is −3, since multiplying 1/3 by −3 gives the required product −1.

How can you check whether (3, 5) lies on y = 3x − 4?

Substitute x = 3: the right side is 9 − 4 = 5, matching the point's ordinate.