Coordinate Geometry Class 10: Notes and Important Questions

This guide covers Coordinate Geometry Class 10, chapter 7 of NCERT Class 10 Mathematics. You will learn the distance formula and how it proves shapes, the section formula and mid-point formula, and how to find the ratio in which a point divides a segment. Worked examples, important questions and common mistakes follow.

Coordinates: a quick recap

• The x-coordinate (abscissa) of a point is its distance from the y-axis.
• The y-coordinate (ordinate) is its distance from the x-axis.
• Every point on the x-axis is (x, 0). Every point on the y-axis is (0, y).

So a point on the y-axis, 5 units from the x-axis in the positive direction, is (0, 5).

This chapter has two main goals: finding the distance between two points, and finding the point that divides a segment in a given ratio.

The distance formula

Take P(x₁, y₁) and Q(x₂, y₂). Draw a horizontal line from P and a vertical line from Q to make a right triangle PTQ.

• Horizontal side PT = |x₂ − x₁|
• Vertical side QT = |y₂ − y₁|

The bars mean absolute value, because lengths cannot be negative. By Pythagoras, PQ² = PT² + QT², so

PQ = √(x₂ − x₁)² + (y₂ − y₁)²

The order of subtraction does not matter, because (a − b)² = (b − a)².

Distance from the origin: for P(x, y), OP = √(x² + y²).

• (36, 15): OP = √(1296 + 225) = √1521 = 39 units.
• (−6, 8): OP = √(36 + 64) = √100 = 10 units.

Using distances to identify shapes

• Triangle: three points form a triangle only if the sum of any two sides is greater than the third.
• Right triangle: the sum of the squares of two sides equals the square of the third.
• Collinear points: if AB + BC = AC exactly, then A, B, C lie on one straight line.
• Square: all four sides equal and both diagonals equal.
• Rhombus: all four sides equal, but the diagonals are usually not equal. Its area is d₁d₂/2, where d₁ and d₂ are the diagonals.

Example: A(1, 7), B(4, 2), C(−1, −1), D(−4, 4).

• AB² = 3² + (−5)² = 34, BC² = (−5)² + (−3)² = 34, CD² = (−3)² + 5² = 34, DA² = 5² + 3² = 34. So every side is √34.
• AC² = (−2)² + (−8)² = 68 and BD² = (−8)² + 2² = 68. Both diagonals are √68.

All sides equal and diagonals equal, so ABCD is a square.

Points that are equidistant

If P is equidistant from A and B, then AP = BP. Square both sides (AP² = BP²) to remove the roots. All such points lie on the perpendicular bisector of AB.

If the point must lie on an axis, use (x, 0) for the x-axis or (0, y) for the y-axis.

Example: find a point on the y-axis equidistant from A(6, 5) and B(−4, 3). Let P = (0, y).

• AP² = 36 + (5 − y)² and BP² = 16 + (3 − y)².
• 36 + 25 − 10y + y² = 16 + 9 − 6y + y²
• 61 − 10y = 25 − 6y, so 4y = 36 and y = 9.

The point is (0, 9). Check: AP² = 36 + 16 = 52 and BP² = 16 + 36 = 52.

The section formula

If P(x, y) divides the segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ (so AP : PB = m₁ : m₂), then

x = (m₁x₂ + m₂x₁)/(m₁ + m₂), y = (m₁y₂ + m₂y₁)/(m₁ + m₂)

Notice the cross-over: m₁ multiplies B's coordinates and m₂ multiplies A's. The formula is derived using similar triangles.

Example: divide (4, −3) and (8, 5) in the ratio 3 : 1. x = (3 × 8 + 1 × 4)/4 = 28/4 = 7 and y = (3 × 5 + 1 × (−3))/4 = 12/4 = 3. The point is (7, 3), which lies between the endpoints, as it should.

The mid-point formula

The mid-point divides the segment in the ratio 1 : 1, so

Mid-point = ((x₁ + x₂)/2, (y₁ + y₂)/2)

It is simply the average of the coordinates. It is useful for finding the centre of a circle from the ends of a diameter, and for parallelograms, whose diagonals bisect each other.

Example: in parallelogram ABCD with A(6, 1), B(8, 2), C(9, 4) and D(p, 3), the mid-points of AC and BD are the same. So (6 + 9)/2 = (8 + p)/2, giving 8 + p = 15 and p = 7.

Finding the ratio, and trisection

To find an unknown ratio, take it as k : 1. Then P = ((kx₂ + x₁)/(k + 1), (ky₂ + y₁)/(k + 1)). Set one coordinate equal to the known value and solve for k. If k = 2/7, the ratio is 2 : 7.

Example: in what ratio does (−4, 6) divide A(−6, 10) and B(3, −8)? Using x: −4 = (3k − 6)/(k + 1), so −4k − 4 = 3k − 6, giving 7k = 2 and k = 2/7. The ratio is 2 : 7. Check y: (2 × (−8) + 7 × 10)/9 = 54/9 = 6.

Trisection: the two points that cut AB into three equal parts divide it in the ratios 1 : 2 and 2 : 1. The second point is also the mid-point of the first point and B. For four equal parts, find the mid-point of AB and then the mid-points of each half.

Remember this

• Distance: √(x₂ − x₁)² + (y₂ − y₁)²; from origin √(x² + y²).
• Section formula: ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)).
• Mid-point: average the coordinates.
• Unknown ratio: take k : 1. A point on the y-axis has x = 0.
• Square: equal sides and equal diagonals. Rhombus: equal sides only.

Important questions with answers

1. Find the distance between (2, 3) and (5, 7). √(5 − 2)² + (7 − 3)² = √(9 + 16) = √25 = 5 units.

2. Find the distance of (−6, 8) from the origin. √(36 + 64) = 10 units.

3. If AB = 3, BC = 4 and AC = 7, are A, B, C collinear? Yes, since AB + BC = 7 = AC.

4. Find the mid-point of the segment joining (2, −4) and (6, 10). ((2 + 6)/2, (−4 + 10)/2) = (4, 3).

5. Find the points of trisection of the segment joining A(1, 1) and B(7, 4). Ratio 1 : 2: x = (1 × 7 + 2 × 1)/3 = 3, y = (1 × 4 + 2 × 1)/3 = 2, so P(3, 2). Ratio 2 : 1: x = (2 × 7 + 1 × 1)/3 = 5, y = (2 × 4 + 1 × 1)/3 = 3, so Q(5, 3). Check: the mid-point of P and B is (5, 3).

6. In what ratio does the y-axis divide the segment joining A(−2, 3) and B(4, 1)? Find the point. Take k : 1. On the y-axis x = 0: (4k − 2)/(k + 1) = 0, so k = 1/2. The ratio is 1 : 2. y = (1 × 1 + 2 × 3)/3 = 7/3, so the point is (0, 7/3).

7. A segment is divided into four equal parts. In what ratio does the point nearest the start divide it? 1 : 3.

8. What is the first step to find a point on the x-axis equidistant from two points? Take the point as (x, 0), since every point on the x-axis has y-coordinate 0.

Common mistakes to avoid

• Swapping m₁ and m₂ in the section formula. m₁ goes with B, m₂ with A.
• Forgetting to square negative differences properly: (−5)² = 25, not −25.
• Calling a shape a square after checking only the sides. Check the diagonals too.
• Writing a point on the x-axis as (0, x).
• Not checking that an internal division point lies between the endpoints.

Sketch the points on rough axes before you calculate, and study this chapter with Joy to practise more questions.