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Bihar Board Class 10 · Maths · Matric exam 2027

Chapter 7: Coordinate Geometry — notes

The chapter’s 6 parts in the book’s order — the first idea of each and the points Medha Ma’am writes on the board — and the summary at the end. The full lesson and a check question for each part are in the app.

  1. 1. Locating a point: coordinates

    Okay, so, to locate a point on a plane, we need a pair of coordinate axes.

    • • x-coordinate (abscissa) = distance from y-axis
    • • y-coordinate (ordinate) = distance from x-axis
    • • Point on x-axis: (x, 0)
    • • Point on y-axis: (0, y)
  2. 2. The Distance Formula

    Okay, so imagine town B is 36 km east and 15 km north of town A.

    • • Distance formula: PQ = √[(x2–x1)² + (y2–y1)²]
    • • Distance from origin: OP = √(x² + y²)
    • • Example: P(3,2), Q(–2,–3), R(2,3)
    • • PQ = √50, QR = √52, PR = √2
    • • PQ² + PR² = QR² → angle P = 90°
    • • So PQR is a right triangle
  3. 3. Using the Distance Formula

    Okay, so the distance formula can do more than just find lengths.

    • • Square check: all 4 sides equal, both diagonals equal
    • • Collinear check: AB + BC = AC means points lie in a line
    • • Equidistant point: set AP² = BP², simplify
    • • Example 4: relation found is x – y = 2
    • • Example 5: point on y-axis equidistant from A(6,5), B(–4,3) is (0, 9)
  4. 4. The Section Formula

    Okay, so — think of a telephone company placing a relay tower at P, between A and B.

    • • Section formula (ratio m1 : m2):
    • • x = (m1x2 + m2x1)/(m1+m2)
    • • y = (m1y2 + m2y1)/(m1+m2)
    • • Mid-point formula: ((x1+x2)/2, (y1+y2)/2)
    • • Example 6: ratio 3:1 of (4,–3),(8,5) → point (7, 3)
    • • Example 8: trisection points of A(2,–2),B(–7,4): (–1,0) and (–4,2)
  5. 5. Finding ratio, and the parallelogram idea

    Okay, so sometimes we're given a point, and we need to find in what ratio it divides a segment.

    • • To find ratio: use section formula, solve for m1:m2 or k:1
    • • On y-axis, abscissa (x) = 0
    • • On x-axis, ordinate (y) = 0
    • • Diagonals of a parallelogram bisect each other
    • • Example 10: mid-point AC = mid-point BD → p = 7
  6. 6. Area of a Triangle

    Okay, so you already know area of a triangle = ½ × base × altitude, and you've used Heron's formula too.

    • • Area = ½ |x1(y2–y3) + x2(y3–y1) + x3(y1–y2)|
    • • Area is always taken as a positive (numerical) value
    • • Area = 0 means the points are collinear
    • • Example 11: vertices (1,–1),(–4,6),(–3,–5) → area 24 sq units
    • • Example 12: vertices (5,2),(4,7),(7,–4) → area 2 sq units
    • • Quadrilateral: split into 2 triangles, add their areas

Summary

Okay, so let's look back at everything in this chapter. We started with how to locate a point — using its abscissa and ordinate. A point on the x-axis looks like (x, 0), and a point on the y-axis looks like (0, y). Then we moved to the distance formula, PQ = √[(x2 − x1)² + (y2 − y1)²]. Remember, this came from the Pythagoras theorem. We used it to check if points form a triangle, a square, if they're collinear, or if a point is equidistant from two given points. After that came the section formula — it gives the coordinates of a point dividing a line segment in a ratio m1 : m2. The mid-point formula is just a special case of this, where the ratio is 1 : 1. We used these to find ratios, trisection points, and even to show that a parallelogram's diagonals bisect each other. Finally, we found the area of a triangle using Area = ½|x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)|. Here's something neat — if this area comes out to 0, the three points are collinear. And we even used this to find the area of a quadrilateral, by splitting it into two triangles. So, tell me — out of all these, which formula did you enjoy using the most?

AskMedha · Medha Ma’am — Bihar Board Class 10 Maths, Chapter 7. Written from the textbook, checked twice.

Questions parents and students ask

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