1. DISTANCE FORMULA
(i) The distance between two points P (x1, y2) and Q (x2, y2) is given by
PQ = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
Ex. Distance between the points A (3, 4) and B (6, –3) is
AB = \(\sqrt{(6-3)^2+)(-3-4)^2}\)
= \(\sqrt{3^2+7^2} = \sqrt{9+49}=\sqrt{58}\)
(ii) Distance of a point P (x, y) from the origin is
\(\sqrt{(x-0)^2+)(y-0)^2}\) = \(\sqrt{x^2+y^2}\)
2. MID-POINT FORMULA
If R is the mid-point of P (x1 , y1) and Q (x2, y2), then the co-ordinates of R are
\(\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\)
3. CENTROID OF A TRIANGLE
The point of concurrence of the medians of a triangle is called the centroid of the triangle. It divides the median in the ratio 2 : 1.
The co-ordinates of the centroid of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) are given by
\(\bigg(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\bigg)\)
4. AREA OF A TRIANGLE
Area of ΔABC, whose vertices are A (x1, y1), B (x2, y2) and C is (x3, y3)
\(\frac12\) | x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) |
If \(\frac12\) | x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) | = 0, then the points A, B and C are collinear.
