CBSE Class 10 Mathematics Coordinate Geometry Notes Set 14

Cartesian Plane and Distance Formula

A point is fixed by its abscissa x and ordinate y. The distance between (x₁, y₁) and (x₂, y₂) is √[(x₂ − x₁)² + (y₂ − y₁)²]. Distances decide the type of a triangle (equal sides, Pythagoras) or quadrilateral (equal diagonals for a rectangle, equal sides for a rhombus).

Section and Midpoint Formulae

The point dividing the join of (x₁, y₁) and (x₂, y₂) internally in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)). For external division replace n by −n. The midpoint is the case m = n.
A negative ratio from the formula means external division. The diagonals of a parallelogram bisect each other.

Centres of a Triangle

Centroid: ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3), dividing each median 2 : 1. Incentre: the average of the vertices weighted by the opposite side lengths. Circumcentre: equidistant from the three vertices. Orthocentre: intersection of altitudes. For any triangle, H = 3G − 2S (orthocentre, centroid, circumcentre).

Area and Collinearity

Area of a triangle = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. Three points are collinear when this area is zero. For a polygon, use the shoelace method with the vertices in order.

Straight Lines

Slope-intercept form y = mx + c; point-slope y − y₁ = m(x − x₁); two-point form; intercept form x/a + y/b = 1; general form ax + by + c = 0 with slope −a/b.
Parallel lines have equal slopes; perpendicular lines have m₁m₂ = −1. The distance of (x₁, y₁) from ax + by + c = 0 is |ax₁ + by₁ + c|/√(a² + b²). A locus is the path of a point obeying a condition: equal distances give a perpendicular bisector.

Solved Examples

Example 1. Find the distance between (2, 3) and (6, 6).
Answer:
√(16 + 9) = 5.
5 units

Example 2. Find the ratio in which (−1, 1) divides the line joining (−4, −2) and (2, 4)
Answer:
m(2) + n(−4) = −1(m + n) gives 3m = 3n.
1 : 1, so (−1, 1) is the midpoint.

Example 3. Find the centroid of the triangle with vertices (1, 2), (3, 4) and (5, 9).
Answer:
((1 + 3 + 5)/3, (2 + 4 + 9)/3).
(3, 5)

Example 4. Find the area of the triangle with vertices (0, 0), (4, 0) and (0, 3).
Answer:
Area = ½ × 4 × 3.
6 square units

Example 5. For what value of k are (1, 2), (2, 4) and (3, k) collinear?
Answer:
Slopes: 2 and (k − 4)/1, so k − 4 = 2.
k = 6

Example 6. Find the equation of the line through (1, 2) and (3, 8).
Answer:
Slope = 6/2 = 3, so y − 2 = 3(x − 1).
3x − y − 1 = 0

Example 7. Find the distance of (3, 4) from the line 3x + 4y − 5 = 0.
Answer:
|9 + 16 − 5|/5 = 20/5.
4 units

Example 8. Find the fourth vertex of the parallelogram ABCD with A(1, 1), B(4, 2) and C(5, 5).
Answer:
Diagonals AC and BD bisect each other, so D = A + C − B = (2, 4).
D = (2, 4)

Example 9. Find the equation of the perpendicular bisector of the segment joining (1, 1) and (3, 5).
Answer:
Midpoint (2, 3), slope of the segment 2, so the perpendicular slope is −1/2: y − 3 = −(x − 2)/2.
x + 2y = 8

Example 10. Find the locus of a point equidistant from (0, 0) and (6, 0).
Answer:
x² + y² = (x − 6)² + y² gives 12x = 36.
x = 3