Maharashtra Board Class 10 Maths Part II Chapter 5 Co ordinate Geometry PDF Download

5 Co-ordinate Geometry

Let's Study

Distance formula

Section formula

Slope of a line

Let's Recall

We know how to find the distance between any two points on a number line.

If co-ordinates of points P, Q and R are -1, -5 and 4 respectively then find the length of seg PQ, seg QR.

If x₁ and x₂ are the co-ordinates of points A and B and x₂ > x₁ then length of seg AB = d(A,B) = x₂ - x₁

As shown in the figure, co-ordinates of points P, Q and R are -1, -5 and 4 respectively.

d(P, Q) = (-1)-(-5) = -1 + 5 = 4

and d(Q, R) = 4 - (-5) = 4 + 5 = 9

Using the same concept we can find the distance between two points on the same axis in XY-plane.

Let's Learn

(1) To Find Distance Between Any Two Points On An Axis

Two points on an axis are like two points on the number line. Note that points on the X-axis have co-ordinates such as (2, 0), (-5/2, 0), (8, 0). Similarly points on the Y-axis have co-ordinates such as (0, 1), (0, 17/2), (0, -3). Part of the X-axis which shows negative co-ordinates is OX' and part of the Y-axis which shows negative co-ordinates is OY'.

Teacher's Note

Distance is the space between two points. In real life, we measure distance between two cities using kilometers.

Exam Trick

Remember: Distance formula uses \((x_2 - x_1)\) and \((y_2 - y_1)\). This is like finding how far apart the x-values and y-values are from each other.

Points to Remember

Distance between two points means how far they are from each other.

We use the formula: distance = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

The distance is always a positive number.

Two points on the same line can have zero distance only if they are the same point.

(2) To Find The Distance Between Two Points If The Segment Joining These Points Is Parallel To Any Axis In The XY Plane

i) In the figure, seg AB is parallel to X-axis.

The y co-ordinates of points A and B are equal

Draw seg AL and seg BM perpendicular to X-axis

Therefore ABML is a rectangle.

Therefore AB = LM

But, LM = x₂ - x₁

Therefore d(A,B) = x₂ - x₁

ii) In the figure seg PQ is parallel to Y-axis.

The x co-ordinates of points P and Q are equal

Draw seg PR and seg QS perpendicular to Y-axis.

Therefore PQSR is a rectangle

Therefore PQ = RS

But, RS = y₂ - y₁

Therefore d(P,Q) = y₂ - y₁

In the above figure, points A(x₁, 0) and B(x₂, 0) are on X-axis such that, x₂ > x₁

Therefore d(A, B) = x₂ - x₁

In the above figure, points P(0, y₁) and Q(0, y₂) are on Y-axis such that, y₂ > y₁

Therefore d(P,Q) = y₂ - y₁

Teacher's Note

When two points are on the same axis, finding distance is easy. Just like finding how many steps between two positions on a straight road.

Exam Trick

Remember: If points are on X-axis, use \(x_2 - x_1\). If points are on Y-axis, use \(y_2 - y_1\). The coordinate that changes tells you which formula to use.

Points to Remember

Points on X-axis have y-coordinate = 0.

Points on Y-axis have x-coordinate = 0.

For parallel lines to X-axis, only x-coordinates change.

For parallel lines to Y-axis, only y-coordinates change.

Activity

In the figure, seg AB is parallel to Y-axis and seg CB is parallel to X-axis. Co-ordinates of points A and C are given.

To find AC, fill in the boxes given below.

Triangle ABC is a right angled triangle.

According to Pythagoras theorem, (AB)² + (BC)² = (AC)²

We will find co-ordinates of point B to find the lengths AB and BC,

CB is parallel to X-axis, so y co-ordinate of B = 2

BA is parallel to Y-axis, so x co-ordinate of B = 2

AB = 3 - 2 = 1 and BC = 2 - (-2) = 4

Therefore AC² = 1 + 16 = 17, so AC = \(\sqrt{17}\)

Let's Learn

Distance Formula

In the figure 5.7, A(x₁, y₁) and B(x₂, y₂) are any two points in the XY plane.

From point B draw perpendicular BP on X-axis.

Similarly from point A draw perpendicular AD on seg BP.

seg BP is parallel to Y-axis.

Therefore the x co-ordinate of point D is x₂.

seg AD is parallel to X-axis.

Therefore the y co-ordinate of point D is y₁.

Therefore AD = d(A, D) = x₂ - x₁ and BD = d(B, D) = y₂ - y₁

In right angled triangle ABD, AB² = AD² + BD²

AB² = \((x_2 - x_1)^2 + (y_2 - y_1)^2\)

Therefore AB = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

This is known as distance formula.

Teacher's Note

Distance formula helps us find distance between any two points, just like GPS finds distance between two locations in a city.

Exam Trick

Remember the formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Always square the differences, add them, then take the square root.

Points to Remember

Distance formula works for any two points in the plane.

The distance is always positive or zero.

We always take the positive square root.

If two points are the same, distance = 0.

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