We have learnt that to locate the position of a point in a plane of coordinate axes. The distance of a point from the y-axis is called its x-coordinate, or abscissa. The distance of a point from the x-axis is called its y-coordinate, or ordinate. The coordinates of a point on the x-axis are of the form (x, 0), and of a point on the y-axis are of the form (0, y).
Coordinate geometry has been developed as an algebraic tool for studying geometry of figures. It helps us to study geometry using algebra, and understand algebra with the help of geometry. Because of this, coordinate geometry is widely applied in various fields such as physics, engineering, navigation, seismology and art!
Let us now find the distance between any two points P(x1, y1) and Q(x2, y2). Draw PR and QS
perpendicular to the x-axis. A perpendicular from the point P on QS is drawn to meet it at the point T.
Then, OR = x1, OS = x2. So, RS = x2 − x1 = PT.
Also, SQ = y2, ST = PR = y1. So, QT = y2 − y1.
Now, applying the Pythagoras theorem in △ PTQ, we get
PQ2 = PT2 + QT2
= (x2 − x1)2 + (y2 − y1)2
PQ = √(x1 - x2)2 + (y1 - y2)2
Note that since distance is always non-negative, we take only the positive square root. So, the distance between the points P(x1, y1) and Q(x2, y2) is PQ = √(x 1 - x2)2 + (y1 - y2)2 which is called the distance formula.
In the figure any two points A(x1, y1) and B(x2, y2) and assume that P (x, y) divides AB internally in the ratio m1 : m2, i.e.,
by the AA similarity criterion, △ PAQ ≅ △ BPC
PA/BP = AQ/PC = PQ/BG
AQ = RS = OS − OR = x − x1
PC = ST = OT − OS = x2 − x
PQ = PS − QS = PS − AR = y − y1
BC = BT− CT = BT − PS = y2 − y
Substituting these values in (1), we get
m1/m = x - x1/x2 - x = y - y1/y2 - y
So, the coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2), internally, in the ratio m1 : m2 are
(m1x2 + m2x1/m1 + m2, m1y2 + m2y1/m1 + m2) This is known as the section formula.
We have learnt how to calculate the area of a triangle when its base and corresponding height (altitude) are given. You have used the formula :
Area of a triangle = 1/2 x base x altitude