Coordinate Geometry Class 10 Maths Notes

Coordinate geometry describes position with numbers. Two perpendicular number lines, the x-axis and the y-axis, cross at the origin O(0, 0) and divide the plane into four quadrants. The x-axis measures left or right and the y-axis measures up or down, so every point needs two numbers and always in the order x first, y second.

• Quadrant I has x > 0 and y > 0; Quadrant II has x < 0 and y > 0; Quadrant III has x < 0 and y < 0; Quadrant IV has x > 0 and y < 0.
• A point on the x-axis has y = 0, as in (5, 0); a point on the y-axis has x = 0, as in (0, −3).
• The signs of the coordinates decide the quadrant, and the magnitude of each coordinate decides the distance from each axis — this is the idea the distance formula formalises.
• The plane is infinite in all four directions, so there is no largest or smallest coordinate value.
• Two points coincide if and only if both their x-coordinates and their y-coordinates are equal.

A linear equation in two variables is of the form ax + by + c = 0, and its graph is a single straight line. Drawing that line needs only two correct points, which you get from a small table of values, and the easiest points are the intercepts — where the line meets each axis.

• For y = mx + c the line crosses the y-axis at (0, c), and that single point fixes c immediately.
• For x = a the line is vertical, parallel to the y-axis, so it has no y-intercept and no gradient.
• Substituting a second convenient value of x gives the second point; plot both and join them with a ruler extended in both directions.
• The slope m is the gradient: rise over run, taken along the line, and it is positive for a line rising to the right.
• Points on the same line satisfy the equation, and a point not on the line does not — a common one-mark check.

The reason a single linear equation in two variables has an infinite number of solutions is that its graph is a whole line of points. The way round this is to take two equations at once: a pair of linear equations whose solution must satisfy both. The graph of each is a straight line, and the one point common to both lines is the solution of the pair.

• Draw both lines on the same axes with the same scale.
• If the two lines cross at exactly one point, that point is the unique solution of the pair.
• The point of intersection read off the graph gives only approximate values; the exact solution comes from algebra.
• Because a pair can have 0, 1 or infinitely many solutions depending on the slopes and positions of the lines, the graph is the quickest way to decide which case you are in.

Find the solution of the pair y = 2x + 1 and y = 3x − 2. Draw y = 2x + 1: it crosses the y-axis at (0, 1) and passes through (2, 5). Draw y = 3x − 2: it crosses the y-axis at (0, −2) and passes through (1, 1). Plot these points, extend both lines, and the two lines cross at a single point.

• The graph gives the answer, and setting the two expressions for y equal is the algebraic confirmation of the same point.
• Substitute to check: (3, 7) satisfies 7 = 2(3) + 1 and 7 = 3(3) − 2, both true.
• Write the answer as the ordered pair (3, 7) — never as 3, 7 or as 'x = 3 and y = 7' alone in a question that asks for the solution of the pair.

The distance between two points is the length of the straight line joining them, and the formula for it is built from a right triangle. Drop a perpendicular from one point to a horizontal or vertical line through the other; the two legs of that right triangle are the differences of the x-coordinates and of the y-coordinates.

• Both differences are squared, so it makes no difference which point you call point 1 and which point 2 — subtract in either order.
• Substitute the whole difference in brackets before squaring; squaring x₂ and x₁ separately and subtracting is a common slip.
• Write the two squares under one radical and simplify only at the end, so the square root covers the entire sum.
• If both differences are zero the points coincide and the distance is 0, which the formula confirms.
• Always end with the unit: cm, m or km as the question gives.

Example 1. Find the distance between P(1, 2) and Q(4, 6). Here x₂ − x₁ = 4 − 1 = 3 and y₂ − y₁ = 6 − 2 = 4.

Example 2. Find the distance between A(−1, 4) and B(4, −8). Here x₂ − x₁ = 4 − (−1) = 5 and y₂ − y₁ = −8 − 4 = −12.

To divide the segment joining two points in a given ratio you need one formula, and the whole difficulty is the convention it uses. State the ratio with the endpoints named — AP : PB = m : n — and the formula below follows with no ambiguity. Read the ratio aloud, and say which letter comes first.

• AP : PB = m : n is the statement to say before the formula: m measures the piece from A to P, n measures the piece from P to B.
• The weight m is therefore multiplied by the coordinates of B, the endpoint you travel towards, and the weight n by the coordinates of A.
• m + n is the same as AB in total, so if the problem gives AB and AP you get PB by subtracting before you substitute.
• Internal division means P lies strictly between A and B. If your computed point falls outside the segment, the two weights have been swapped — redo it with m and n exchanged.
• Always confirm at the end that the point you got really does divide AB in the stated ratio.

Example 1. A(1, 1) and B(7, 11), and P divides AB internally so that AP : PB = 2 : 3. Then m = 2 and n = 3, so the weight 2 goes with B(7, 11) and the weight 3 goes with A(1, 1).

Example 2. A(1, 2) and B(4, 6), with AP : PB = 1 : 2. Here m = 1 goes with B(4, 6) and n = 2 goes with A(1, 2).

The retained content of this unit is exactly the graph of a pair of linear equations, the distance formula and the section formula for internal division. Two results that older books place beside them are not in the rationalised syllabus.

• The area of a triangle from its vertices' coordinates is DELETED for 2024-25. No shoelace or determinant area formula will be needed.
• The mid-point formula is not among the retained results either. If a question ever produces equal halves, take the ratio as AP : PB = 1 : 1 and use the section formula, which covers that case exactly.
• External division, the slope-intercept treatment of a family of lines, and the polar or rectangular coordinate forms are outside the retained list.
• So the entire examinable formula list here is short: the distance formula and the internal section formula.

Coordinate geometry carries 6 of the 80 theory marks, so most of it appears as a one-mark or two-mark item and as the arithmetic inside a longer question. The case-study section is the most likely home for a two-part coordinate question.

• Find the distance between two given points. (One or two marks.)
• Find the coordinates of the point that divides a segment in a given ratio. (Two or three marks.)
• Plot a linear equation and state whether a given point lies on it. (One mark.)
• Find the graphical solution of a pair of linear equations by plotting. (Two marks, with the marked intersection.)
• A case-study question on a map or a floor plan typically asks one part using the distance formula and one part using the section formula, and both parts may carry a choice.