Coordinate geometry, also called analytic geometry, is the branch of mathematics that connects algebra and geometry by describing points, lines, and figures using coordinates. Every point in a plane is located by an ordered pair (x, y), where x is the distance from the y-axis and y is the distance from the x-axis. This simple yet revolutionary idea, credited to Rene Descartes, turns geometric problems into algebraic calculations.
This chapter deals with the plane coordinate system, the distance formula between two points, the section formula that divides a line segment in a given ratio, and the area of a triangle from its vertices' coordinates. These tools allow us to compute distances, locate points, and find areas purely by arithmetic, without drawing any figure.
Coordinate geometry is the meeting point of the algebra and geometry that students have been studying separately. It is essential for understanding graphs, including the graphs of linear and quadratic equations studied earlier, and it forms the foundation for the applications of trigonometry and for calculus in higher classes. This chapter is also a reliable source of marks in the board examination.
The Cartesian plane consists of two perpendicular number lines intersecting at their zero points. The horizontal line is the x-axis and the vertical line is the y-axis; their intersection is the origin O with coordinates (0, 0).
The plane is divided into four quadrants: - First quadrant: x > 0, y > 0 - Second quadrant: x < 0, y > 0 - Third quadrant: x < 0, y < 0 - Fourth quadrant: x > 0, y < 0
Every point P in the plane has coordinates (x, y). The x-coordinate is called the abscissa and the y-coordinate is called the ordinate. A point on the x-axis has y = 0, and a point on the y-axis has x = 0. The coordinates are always written in the fixed order (x, y), so the point (3, 5) is different from (5, 3).
The distance between two points P(x1, y1) and Q(x2, y2) is given by: PQ = sqrt[(x2 - x1)^2 + (y2 - y1)^2]
This formula is a direct application of Pythagoras' theorem to the right triangle formed by the horizontal and vertical differences between the points.
Example: Distance between A(2, 3) and B(6, 6). AB = sqrt[(6 - 2)^2 + (6 - 3)^2] = sqrt[16 + 9] = sqrt[25] = 5.
Special case: The distance of a point P(x, y) from the origin is sqrt[x^2 + y^2]. For example, the distance of (3, 4) from the origin is 5.
If point P(x, y) divides the line segment joining A(x1, y1) and B(x2, y2) in the ratio m:n (that is, AP:PB = m:n), then: x = (mx2 + nx1)/(m + n) y = (my2 + ny1)/(m + n)
When P is the midpoint, m:n = 1:1, so: x = (x1 + x2)/2 y = (y1 + y2)/2
The section formula is used to find the point of division, to verify division ratios, and to find coordinates of points dividing a segment into equal parts. A point dividing a segment in the ratio k:1 is also expressed directly as P = ((k x2 + x1)/(k + 1), (k y2 + y1)/(k + 1)).
The area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is given by: Area = 1/2 [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]
The area is always taken as a positive value, so the absolute value is used if the expression comes out negative.
Three points are collinear if and only if the area of the triangle formed by them is zero.
Example: Are the points (1, 1), (2, 3), and (3, 5) collinear? Area = 1/2 [1(3 - 5) + 2(5 - 1) + 3(1 - 3)] = 1/2 [-2 + 8 - 6] = 0. Hence the points are collinear.
The area formula can also find the area of polygons by dividing them into triangles, though this is less common at the Class 10 level.
| Quantity | Formula | Remarks |
|---|---|---|
| Distance between two points | sqrt[(x2 - x1)^2 + (y2 - y1)^2] | From Pythagoras |
| Distance from origin | sqrt[x^2 + y^2] | Origin is (0, 0) |
| Section formula | x = (mx2 + nx1)/(m + n), y = (my2 + ny1)/(m + n) | Ratio m:n |
| Midpoint | ((x1 + x2)/2, (y1 + y2)/2) | m = n = 1 |
| Area of triangle | 1/2 [x1(y2-y3) + x2(y3-y1) + x3(y1-y2)] | Absolute value |
| Quadrant | Sign of x | Sign of y | Example |
|---|---|---|---|
| First | + | + | (3, 5) |
| Second | - | + | (-3, 5) |
| Third | - | - | (-3, -5) |
| Fourth | + | - | (3, -5) |
A reliable method for solving any coordinate geometry problem is to write the coordinates in the form (x, y), identify which formula applies, substitute carefully, and then simplify step by step. Consider the problem of finding the distance between the points A(-2, 3) and B(4, -5). Here x1 = -2, y1 = 3, x2 = 4, and y2 = -5. The horizontal difference x2 - x1 equals 4 - (-2) = 6, and the vertical difference y2 - y1 equals -5 - 3 = -8. Squaring gives 36 and 64, whose sum is 100, so the distance is the square root of 100, which is 10. The most common error in such problems is mishandling the negative signs; it is always safer to write the subtraction in brackets, as we have done here, so that subtracting a negative number is done correctly.
A second problem type asks us to find the point that divides a segment in a given ratio. Suppose P divides the segment joining A(1, 2) and B(7, 8) in the ratio 1 : 2. Taking m = 1 and n = 2, the x-coordinate is (1 x 7 + 2 x 1)/(1 + 2) = (7 + 2)/3 = 3, and the y-coordinate is (1 x 8 + 2 x 2)/3 = (8 + 4)/3 = 4. Hence the point P is (3, 4). We can check the answer by verifying the ratio of the distances: AP is the distance between (1, 2) and (3, 4), and PB is the distance between (3, 4) and (7, 8). The two distances are in the ratio 1 : 2, confirming the computation. When the ratio is 1 : 1, the section formula reduces to the midpoint formula, and the coordinates are simply the averages of the two endpoints.
The area formula is used both to find areas and to test whether three points are collinear. To find the area of a triangle with vertices (0, 0), (5, 0), and (0, 12), we substitute into the formula: the area equals one-half of 0(0 - 12) + 5(12 - 0) + 0(0 - 0), which is one-half of 60, giving 30 square units. This agrees with the familiar base-times-height result, since this is a right triangle with legs 5 and 12. For a collinearity test, we compute the expression and check whether it is exactly zero; if the three points lie on one straight line, the triangle formed by them has zero area. Students should always take the absolute value of the final area, because the formula can produce a negative result depending on the order in which the vertices are named.
A useful strategy for numerical problems is to draw a quick sketch of the points on the Cartesian plane before substituting into any formula. The sketch helps us judge whether an answer is reasonable, such as noticing that a computed distance is far too large or that a point lies in the wrong quadrant. It also helps us choose which point to treat as (x1, y1) and which as (x2, y2), since the distance formula gives the same result either way, while the section formula does not. Combining a rough diagram with careful algebraic substitution is the fastest and safest route to full marks in this chapter, and this habit of visual checking will serve students well in the later chapters on trigonometry and on the applications of coordinate geometry.
Coordinate geometry transforms geometric reasoning into precise numerical computation. The distance formula, section formula, and area of a triangle provide a complete toolbox for solving positional problems, and the concept of collinearity connects the algebraic and geometric views of points lying on a line. This chapter deepens the understanding of the Cartesian plane first encountered in earlier classes and is indispensable for the applications of trigonometry and higher studies. With regular practice of both formula substitution and numerical simplification, students can secure full marks in this predictable and scoring chapter.