Haryana State Board HBSE 10th Class Maths Notes Chapter 7 Coordinate Geometry Notes.

## Haryana Board 10th Class Maths Notes Chapter 7 Coordinate Geometry

**Introduction**

We have studied in previous classes that how to locate the position of a point in co-ordinate plane of two dimensions in terms of two co-ordinates. The distance of a point from the y-axis is called its x-co-ordinate or abscissa. The distance of a point from the x-axis is called its y co-ordinate or ordinate. The coordiantes 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).

In is chapter, we shall extend our knowledge by learning how to find the distance between the two points whose co-ordinates are given, and to find the area of the triangle formed by three given points. We shall also learn about how to find the co-ordinates of the point which divides a line segment joining two given points in a given ratio.

1. Plane-It is a flat surface, which extends indefinitely in all directions. The surface of a sheet of paper, the surface of a smooth green board, the surface of a smooth table top are some examples of a plane.

2. Co-ordinate axes-To locate the position of a point in a plane, two mutually perpendicular lines are drawn. One of them is horizontal known as x-axis and other is vertical known as y-axis. These axes are combindely called co-ordinate axes.

3. Origin-The point of intersection of two Axes is called origin.

4. Quadrants-The axes divide a plane into four parts called quadrants.

5. Collinear points-If three or more pointa lie on the same line, they are called collinear points. Otherwise, they are called non-collinear points.

**Distance Formula**

a_{1} = a = 0

Let P (x_{1}, y_{1}) and Q(x_{2}, y_{2}) be the given points in the plane. Draw PS ⊥ x-axis, QT ⊥ x-axis and PR ⊥ QT from P, Q and P respectively. Then

OS = x_{1}, OT = x_{2}, PS = y_{1}, QT = y_{2}

PR = ST = OT – OS = x_{2} – x_{1} and QR = QT – RT = y_{2} – y_{1}

ΔQRP is a right angled triangle, right angled at R.

∴ PQ^{2} = PR^{2} + QR^{2} [By Pythagoras theorem]

PQ^{2} = (x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

PQ = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Hence, distance between any two points

\(=\sqrt{\begin{array}{r}

(\text { difference of abscissa })^2+ \\

\text { (difference of ordinates) }^2

\end{array}}\)

It is called the distance formula.

Remark: (i) The distance of a point P(x, y) from the origin O(0, 0) is given by

OP = \(\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}\)

(ii) If three points A, B and C are collinear then AB + BC = AC.

(iii) Properties of different types of quadrilaterals

(a) Square : In a square, all sides and diagonals are equal

(b) Rectangle : In a rectangle opposite sides and diagoals are equal.

(c) Parallelogram : In a parallelogram opposite sides are equal.

(d) Paralleogram but not a rectangle : In this case opposite sides are equal but diagonals are not equal

(e) Rhombus but not a square : In this case all sides are equal, but diagonals are not equal.

**Section Formula**

By section formula we find out the coordinates of the point which divides the line segment joining the two given points in a given ratio internally.

Let P(x, y) be the point dividing the line segment joining the points. A(x_{1}, y_{1}) and B(x_{2}, y_{2}) internally in the ratio m_{1} : m_{2}.

Draw AL ⊥ OX; PM ⊥ OX, BN ⊥ OX, Also draw AK ⊥ PM and PQ ⊥ BN. Then, OL = x_{1}, OM = x, AL = y_{1}, PM = y and BN = y_{2}. ON = x_{2}.

AK = LM = OM – OL = x – x_{1}

PQ = MN = ON – OM

= x_{2} – x

PK = PM – KM = y – y_{1}

BQ = BN – QN = y_{2} – y

In right ΔAKP and ΔPQB.

∠AKP = ∠PQB (Each is 90°)

∠AKP = ∠BPQ (Corresponding ∠s)

ΔAKP ~ ΔPQB [By AA similarity criterion]

Remarks: (1) If P is the mid point of AB then m_{1} : m_{2} = 1 : 1.

Then co-ordinates of P are

\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)

(2) If the ratio in which P divides AB is k : 1. Then the coordiantes of the point P are

\(\left(\frac{k x_2+x_1}{k+1}, \frac{k y_2+y_1}{k+1}\right)\)

**Co-ordinates of the Centroid of a Triangle**

Let A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) be the vertices of a given ΔABC. Let D be the mid point of BC. Let G(x, y) be centroid of the triangle as shown in figure.

Co-ordinates of the point D are \(\left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right)\)

Since centroid divides each median in the ratio 2 : 1.

∴ G divides AD in the ratio 2 : 1

Hence, the coordinates of the centroid of the triangle are \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)

**Area of Triangle**

Let A(x_{1}, y_{1}) B(x_{2}, y_{2}) and C(x_{3}, y_{3}) be the vertices of the given ΔABC. Draw BL, AM and CN perpendicular to the x-axis.

Then, LM = (x_{1} – x_{2}), LN = (x_{3} – x_{2})

MN = (x_{3} – x_{1})

Now, Area of ΔABC = ar(Trap BLMA) + ar(Trap AMNC) – ar(Trap BLNC)

Since, the area is never negative, we have

Area (ΔABC) = \(\frac{1}{2}\)[(x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

**Condition for the collinearity of three points**

If three points A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C(x_{3}, y_{3}) are collinear.

Then Area of ΔABC = 0

⇒ \(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})

⇒ x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})

= 0

Remark : To find the area of a polygon we divide it in triangles and take numerical value of the area of each of the triangles.