CBSE Class 10 Mathematics Coordinate Geometry Notes Set 13

Coordinate Geometry

A point is fixed by its abscissa (distance from the y-axis) and ordinate (distance from the x-axis). Signs by quadrant: (+, +), (-, +), (-, -), (+, -). Points on the x-axis have ordinate 0 and points on the y-axis have abscissa 0.
Distance between \( (x_1,y_1) \) and \( (x_2,y_2) \) is \( \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \). The distance from the origin is \( \sqrt{x^2+y^2} \).

Using the Distance Formula

For a triangle compute the three sides: two equal sides give an isosceles triangle, three equal an equilateral one, and if the sum of the squares of two equals the square of the third it is right-angled. For a quadrilateral check opposite sides and the diagonals: a square has four equal sides and equal diagonals; a rhombus four equal sides; a rectangle equal opposite sides and equal diagonals; a parallelogram equal opposite sides, or diagonals that bisect each other.
To find a point equidistant from two points, set the squares of the distances equal; this gives a linear equation.

Section Formula

Internal division in the ratio m : n gives \( \left(\frac{mx_2+nx_1}{m+n},\ \frac{my_2+ny_1}{m+n}\right) \). External division uses \( m-n \) in the denominator. The mid-point is the case m = n. A ratio that comes out negative means the division is external. To find where a line or axis cuts a segment, take the ratio as k : 1 and put the dividing point on the line.

Centroid and Area

The centroid is \( \left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right) \). Area of a triangle \( =\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| \); three points are collinear when the area is zero. A quadrilateral is split by a diagonal into two triangles.
Example: the vertices (-2, 1), (5, 4), (2, -3) enclose area 20 square units, and (-3, 2), (5, 4), (7, -6), (-5, -4) in order enclose 80 square units.