Introduction
Coordinate Geometry connects algebra and geometry by using coordinates to describe geometric shapes. In this chapter, you will learn the distance formula to find the distance between two points, the section formula to find the point that divides a line segment in a given ratio, and the mid-point formula. These tools are fundamental for solving problems involving triangles, quadrilaterals, and other shapes on the coordinate plane.
Distance Formula
The distance between two points P(x1, y1) and Q(x2, y2) in the coordinate plane is given by PQ = sqrt[(x2-x1)^2 + (y2-y1)^2]. This formula is derived from the Pythagoras theorem by forming a right triangle with PQ as the hypotenuse. It is used to find lengths of sides, check collinearity, and verify types of triangles and quadrilaterals.
Key Points
- •Distance = sqrt[(x2-x1)^2 + (y2-y1)^2]
- •Distance from origin to (x, y) = sqrt(x^2 + y^2)
- •Distance is always non-negative
- •If distance = 0, the two points coincide
- •Used to verify if a triangle is isosceles, equilateral, or right-angled
Worked Example
Find the distance between A(3, 4) and B(-1, 7). AB = sqrt[(-1-3)^2 + (7-4)^2] = sqrt[16 + 9] = sqrt(25) = 5 units
Watch Out
To check if points form a specific type of triangle or quadrilateral, find ALL side lengths using the distance formula, then compare.
Section Formula
The section formula gives the coordinates of a point that divides the line segment joining two given points in a given ratio. If point P divides the line segment joining A(x1, y1) and B(x2, y2) in the ratio m:n internally, then P = [(mx2 + nx1)/(m+n), (my2 + ny1)/(m+n)]. For external division, the formula uses subtraction instead of addition.
Key Points
- •Internal division in ratio m:n: P = [(mx2+nx1)/(m+n), (my2+ny1)/(m+n)]
- •Mid-point formula (special case m:n = 1:1): M = [(x1+x2)/2, (y1+y2)/2]
- •To find the ratio in which a point divides a line, assume ratio k:1 and solve
- •The section formula can be used to find centroids, incentres, etc.
- •Centroid of a triangle = [(x1+x2+x3)/3, (y1+y2+y3)/3]
Worked Example
Find the mid-point of the segment joining A(2, -5) and B(-4, 9). Mid-point = [(2 + (-4))/2, (-5 + 9)/2] = [-2/2, 4/2] = (-1, 2)
Watch Out
When finding the ratio of division, assume the ratio is k:1 (not m:n). This reduces the number of unknowns to just one.
Applications in Geometry
Coordinate geometry formulas can be used to solve various geometric problems: checking if three points are collinear, finding the type of triangle or quadrilateral formed by given points, finding the centroid of a triangle, and verifying geometric properties algebraically. Three points are collinear if the sum of any two distances equals the third.
Key Points
- •Collinearity check: AB + BC = AC (sum of two = third)
- •Equilateral triangle: all three sides equal
- •Isosceles triangle: two sides equal
- •Right triangle: verify Pythagoras theorem on the three sides
- •Parallelogram: opposite sides equal, diagonals bisect each other
Worked Example
Show that A(1, 7), B(4, 2), C(-1, -1) form an isosceles right triangle. AB = sqrt[(4-1)^2 + (2-7)^2] = sqrt[9+25] = sqrt(34) BC = sqrt[(-1-4)^2 + (-1-2)^2] = sqrt[25+9] = sqrt(34) AC = sqrt[(-1-1)^2 + (-1-7)^2] = sqrt[4+64] = sqrt(68) AB = BC = sqrt(34), so it is isosceles. AB^2 + BC^2 = 34 + 34 = 68 = AC^2, so it is right-angled. Hence, it is an isosceles right triangle.
Quick Summary
- ✓Distance formula: sqrt[(x2-x1)^2 + (y2-y1)^2]
- ✓Section formula (m:n): [(mx2+nx1)/(m+n), (my2+ny1)/(m+n)]
- ✓Mid-point: [(x1+x2)/2, (y1+y2)/2]
- ✓Centroid: [(x1+x2+x3)/3, (y1+y2+y3)/3]
- ✓Collinearity: AB + BC = AC
- ✓Use distance formula to verify types of triangles and quadrilaterals
Key Formulas
Distance: PQ = sqrt[(x2-x1)^2 + (y2-y1)^2]
Section formula: P = [(mx2+nx1)/(m+n), (my2+ny1)/(m+n)]
Mid-point: M = [(x1+x2)/2, (y1+y2)/2]
Centroid: G = [(x1+x2+x3)/3, (y1+y2+y3)/3]
Distance from origin: sqrt(x^2 + y^2)
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