CBSE Class 10 Mathematics Coordinate Geometry Notes Set 05

Introduction to Coordinate Geometry

Coordinate geometry or analytical geometry is the study of geometry using a coordinate system. Coordinate geometry is used in Physics and Engineering and also in space science, rocketry and spaceflight etc.

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.

Take two lines X'OX and Y'OY which are perpendicular to each other and intersect at the point O. X'OX and Y'OY are known as the co-ordinate axes. Here O is the origin, X'OX is known as the x-axis and Y'OY is known as the y-axis.

Now suppose A is a point as shown in the figure. Draw AN and AM perpendicular on the y-axis and x-axis, respectively. The length NA (or OM) is known as the abscissa of point A or the x-coordinate and MA (or ON) is known as the ordinate of point A or the y-coordinate. Here, point A having abscissa x and ordinate y is named as A(x, y).

Quadrants in Co-ordinate Geometry

A plane is divided into four parts known as quadrants by the axes of a two-dimensional Cartesian system. These quadrants are denoted by Roman Numerals (with the signs of the (x, y) coordinates).

Note.
1. Co-ordinates of the origin O are (0, 0).
2. The abscissa of any point on the y-axis is zero. It is positive to the right of the y-axis and is negative to the left of the y-axis.
3. The ordinate of any point of x-axis is zero. It is positive above the x-axis and is negative below the x-axis.

Distance Formula in Coordinate Geometry

The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) in a coordinate system is given by the formula \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). This is known as the distance formula.

Proof. \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) are the given points. Draw PR and QS perpendicular to the x-axis. A perpendicular is drawn from point P on QS to meet at point M.Let \( OR = x_1 \), \( OS = x_2 \). Then \( RS = x_2 - x_1 = PM \). Let \( SQ = y_2 \), \( SM = PR = y_1 \). Then \( QM = y_2 - y_1 \).
On applying the Pythagoras theorem in \( \triangle PMQ \), we get\[ PQ^2 = PM^2 + QM^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \]\[ \Rightarrow PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]Since distance can never be negative, only the positive square root is taken.

Note. From the origin, the distance of a point \( (x, y) \) is given by \( \sqrt{x^2 + y^2} \).

Test of Geometrical Figures

Solved Examples

Example 1. Write down (i) abscissa (ii) ordinate (iii) Co-ordinates of A, B, C and D as given in the figure.

Answer:
For point A: abscissa of \( A = 2 \); ordinate of \( A = 3 \); co-ordinates of \( A = (2, 3) \).
For point B: abscissa of \( B = -2 \); ordinate of \( B = 4 \); co-ordinates of \( B = (-2, 4) \).
For point C: abscissa of \( C = -5 \); ordinate of \( C = -3 \); co-ordinates of \( C = (-5, -3) \).
For point D: abscissa of \( D = 5 \); ordinate of \( D = -1 \); co-ordinates of \( D = (5, -1) \).

Example 2. Draw a triangle PQR where vertices P, Q and R are \( (0, 2),\ (2, -2) \) and \( (-2, 2) \) respectively.

Answer:
By taking abscissa = 0 and ordinate = 2, plot the point P. In the similar way, plot points Q and R taking abscissa 2 and \( -2 \) and ordinates \( -2 \) and 2 respectively. Now join P, Q and R.
Hence, \( \triangle PQR \) is the required triangle.

Example 3. Find the distance between the points \( P(7, 5) \) and \( Q(2, 5) \).
Answer:
From the distance formula, \( PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Here \( x_1 = 7,\ y_1 = 5,\ x_2 = 2 \) and \( y_2 = 5 \).
\( \therefore \) Required distance \( = \sqrt{(2 - 7)^2 + (5 - 5)^2} = \sqrt{(-5)^2 + 0^2} = \sqrt{25} = 5 \) units.

Example 4. Find the points on the x-axis which are at a distance of 5 units from the point \( Q(-1, 4) \).
Answer:
Let the point on the x-axis be \( P(x, 0) \). Distance \( PQ = 5 \) units.
\( \sqrt{(x - (-1))^2 + (0 - 4)^2} = 5 \). Squaring both sides: \( (x + 1)^2 + 16 = 25 \Rightarrow x^2 + 2x + 1 + 16 = 25 \Rightarrow x^2 + 2x - 8 = 0 \).
By the quadratic formula, \( x = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm 6}{2} = 2 \) or \( -4 \).
Hence, the required points on the x-axis are \( (2, 0) \) and \( (-4, 0) \).

Example 5. The vertices of a triangle are \( (-2, 0),\ (2, 3) \) and \( (1, -3) \). Is the triangle equilateral, isosceles or scalene?
Answer:
Let \( P = (-2, 0),\ Q = (2, 3),\ R = (1, -3) \). By the distance formula:
\( PQ = \sqrt{(2 + 2)^2 + (3 - 0)^2} = \sqrt{16 + 9} = 5 \) units.
\( QR = \sqrt{(1 - 2)^2 + (-3 - 3)^2} = \sqrt{1 + 36} = \sqrt{37} \) units.
\( RP = \sqrt{(-2 - 1)^2 + (0 + 3)^2} = \sqrt{18} = 3\sqrt{2} \) units.
As \( PQ \neq QR \neq RP \), \( \triangle PQR \) is a scalene triangle.

Example 6. The two opposite vertices of a square are \( (-1, 2) \) and \( (3, 2) \). Find the co-ordinates of the other two vertices.

Answer:
Let \( PQRS \) be a square with the opposite vertices \( P(-1, 2) \) and \( R(3, 2) \), and let \( Q = (x, y) \).
As \( PQ = QR \), \( PQ^2 = QR^2 \Rightarrow (x + 1)^2 + (y - 2)^2 = (x - 3)^2 + (y - 2)^2 \Rightarrow 8x = 8 \Rightarrow x = 1 \).
\( \angle Q = 90^\circ \), so \( PR^2 = PQ^2 + QR^2 \): \( 16 = 2\left[(1 + 1)^2 + (y - 2)^2\right] \Rightarrow (y - 2)^2 = 4 \Rightarrow y = 4 \) or \( 0 \).
Hence the other two vertices are \( (1, 4) \) and \( (1, 0) \).

Example 7. Use analytical geometry (co-ordinate geometry) to prove that the mid-point of the hypotenuse of a right-angled triangle is equidistant from its vertices.

Answer:
Let \( POQ \) be a right-angled triangle with the right angle at O, taken as the origin. Let \( OP = a \) and \( OQ = b \). Then the coordinates of P and Q are \( (a, 0) \) and \( (0, b) \).
Let M be the mid-point of the hypotenuse PQ. Then \( M = \left(\frac{a + 0}{2}, \frac{b + 0}{2}\right) = \left(\frac{a}{2}, \frac{b}{2}\right) \).
\( MO = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2} = \frac{1}{2}\sqrt{a^2 + b^2} \).
\( MP = \sqrt{\left(\frac{a}{2} - a\right)^2 + \left(\frac{b}{2}\right)^2} = \frac{1}{2}\sqrt{a^2 + b^2} \) and \( MQ = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{b}{2} - b\right)^2} = \frac{1}{2}\sqrt{a^2 + b^2} \).
Since \( MP = MQ = MO \), M is equidistant from the vertices of \( \triangle POQ \).

Example 8. If \( B(2, 1) \) and \( C(-3, 2) \). Also, \( A(x, y) \) lies on the right bisector of BC, then show that \( 5x - y + 4 = 0 \).

Answer:
A is a point on the right (perpendicular) bisector of BC, so it is equidistant from B and C. Hence \( AB = AC \Rightarrow AB^2 = AC^2 \).
\( (x - 2)^2 + (y - 1)^2 = (x + 3)^2 + (y - 2)^2 \Rightarrow x^2 - 4x + 4 + y^2 - 2y + 1 = x^2 + 6x + 9 + y^2 - 4y + 4 \).
\( \Rightarrow 10x - 2y + 8 = 0 \Rightarrow 2(5x - y + 4) = 0 \Rightarrow 5x - y + 4 = 0 \). Hence proved.

Example 9. Prove that the triangle whose vertices are \( (-3, -4),\ (2, 6) \) and \( (-6, 10) \) is a right triangle.
Answer:
Let \( P = (-3, -4),\ Q = (2, 6),\ R = (-6, 10) \). By the distance formula:
\( PQ^2 = (2 + 3)^2 + (6 + 4)^2 = 25 + 100 = 125 \).
\( QR^2 = (-6 - 2)^2 + (10 - 6)^2 = 64 + 16 = 80 \).
\( PR^2 = (-6 + 3)^2 + (10 + 4)^2 = 9 + 196 = 205 \).
Here \( PQ^2 + QR^2 = 125 + 80 = 205 = PR^2 \). By the converse of the Pythagoras theorem, \( PQR \) is a right triangle with \( \angle Q = 90^\circ \).

Example 10. If the point \( A(x, y) \) is equidistant from the points \( P(a, 0) \) and \( Q(2a, a) \), prove that \( x + y = 2a \).
Answer:
\( AP = \sqrt{(a - x)^2 + y^2} \) and \( AQ = \sqrt{(2a - x)^2 + (a - y)^2} \).
As A is equidistant from P and Q, \( AP^2 = AQ^2 \Rightarrow a^2 - 2ax + x^2 + y^2 = 4a^2 - 4ax + x^2 + a^2 - 2ay + y^2 \).
\( \Rightarrow 2ax + 2ay = 4a^2 \Rightarrow 2a(x + y) = 4a^2 \Rightarrow x + y = 2a \). Hence proved.