Class 10 Maths Notes Chapter 7 — Coordinate Geometry#

Distance formula#

The distance between $P(x_1, y_1)$ and $Q(x_2, y_2)$ is

$$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

The distance of $P(x, y)$ from the origin is

$$OP = \sqrt{x^2 + y^2}$$

The distance $PQ$ is the hypotenuse of a right triangle with legs $x_2 - x_1$ and $y_2 - y_1$, so Pythagoras gives the formula.

• A point on the x-axis is $(x, 0)$. A point on the y-axis is $(0, y)$.
• “Equidistant from A and B” means $PA = PB$. Square both sides, $PA^2 = PB^2$, to remove the square roots.

Uses of the distance formula

• Collinear points: A, B, C lie on one line if the sum of the two shorter distances equals the longest, for example $AB + BC = AC$.
• Triangles: two sides equal → isosceles; all three equal → equilateral; $AB^2 + BC^2 = AC^2$ → right-angled.
• Quadrilaterals (find all four sides and both diagonals):

Section formula#

The point $P(x, y)$ dividing the join of $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m_1 : m_2$ is

$$P = \left(\frac{m_1x_2 + m_2x_1}{m_1 + m_2},\ \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\right)$$

• $m_1$ goes with the second point $B$, and $m_2$ with the first point $A$.
• Unknown ratio: take the ratio as $k : 1$. Then $$P = \left(\frac{kx_2 + x_1}{k + 1},\ \frac{ky_2 + y_1}{k + 1}\right)$$ Equate one coordinate to the given value and solve for $k$.
• Divided by the x-axis: put the $y$-coordinate equal to 0. By the y-axis: put the $x$-coordinate equal to 0.

Mid-point and trisection#

Mid-point of the join of $(x_1, y_1)$ and $(x_2, y_2)$ (ratio $1 : 1$):

$$\left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right)$$

• Points of trisection divide a segment into three equal parts. Use the ratios $1 : 2$ and $2 : 1$.
• Four equal parts: ratios $1 : 3$, $1 : 1$ and $3 : 1$.
• Parallelogram: its diagonals bisect each other, so the mid-points of both diagonals are the same point. Use this to find a missing vertex.
• Circle: the centre is the mid-point of any diameter.

Common mistakes#

• Swapping $m_1$ and $m_2$ in the section formula. $m_1$ multiplies the coordinates of the second point.
• Forgetting brackets with negatives: $(x_2 - x_1)^2$ with $x_1 = -5$ is $(x_2 + 5)^2$.
• Deciding a quadrilateral from the sides alone. A rhombus and a square both have four equal sides, so check the diagonals.
• Using “a point on the x-axis” as $(0, y)$. On the x-axis, $y = 0$.