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♦ Introduction

♦ Distance Formula

♦ Distance Formula

As we know that to locate the position of a point on a plane, we require a pair 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)`.

Draw a set of a pair of perpendicular axes on a graph paper. Now plot the following points and join them as directed: Join the point` A(4, 8)` to` B(3, 9)` to `C(3, 8) `to `D(1, 6)` to `E(1, 5)` to `F(3, 3)` to `G(6, 3) `to `H(8, 5) `to `I(8, 6) `to `J(6, 8)` to` K(6, 9) `to `L(5, 8) `to A.

Then join the points `P(3.5, 7), Q (3, 6)` and `R(4, 6)` to form a triangle. Also join the points `X(5.5, 7), Y(5, 6)` and `Z(6, 6) `to form a triangle. Now join `S(4, 5), T(4.5, 4)` and `U(5, 5)` to form a triangle. Lastly join `S` to the points `(0, 5)` and `(0, 6) `and join `U` to the points `(9, 5)` and `(9, 6)`. What picture have you got?

Also, you have seen that a linear equation in two variables of the form `ax + by + c = 0`, (`a, b` are not simultaneously zero), when represented graphically, gives a straight line.

Further, you have seen the graph of `y = ax^2 + bx + c (a ≠ 0)`, is a parabola. In fact, 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!

Here, you will learn how to find the distance between the two points whose coordinates are given, and to find the area of the triangle formed by three given points. You will also study how to find the coordinates of the point which divides a line segment joining two given points in a given ratio.

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)`.

Draw a set of a pair of perpendicular axes on a graph paper. Now plot the following points and join them as directed: Join the point` A(4, 8)` to` B(3, 9)` to `C(3, 8) `to `D(1, 6)` to `E(1, 5)` to `F(3, 3)` to `G(6, 3) `to `H(8, 5) `to `I(8, 6) `to `J(6, 8)` to` K(6, 9) `to `L(5, 8) `to A.

Then join the points `P(3.5, 7), Q (3, 6)` and `R(4, 6)` to form a triangle. Also join the points `X(5.5, 7), Y(5, 6)` and `Z(6, 6) `to form a triangle. Now join `S(4, 5), T(4.5, 4)` and `U(5, 5)` to form a triangle. Lastly join `S` to the points `(0, 5)` and `(0, 6) `and join `U` to the points `(9, 5)` and `(9, 6)`. What picture have you got?

Also, you have seen that a linear equation in two variables of the form `ax + by + c = 0`, (`a, b` are not simultaneously zero), when represented graphically, gives a straight line.

Further, you have seen the graph of `y = ax^2 + bx + c (a ≠ 0)`, is a parabola. In fact, 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!

Here, you will learn how to find the distance between the two points whose coordinates are given, and to find the area of the triangle formed by three given points. You will also study how to find the coordinates of the point which divides a line segment joining two given points in a given ratio.

Let us consider the following situation:

A town B is located `36` km east and 15 km north of the town A. How would you find the distance from town A to town B without actually measuring it. Let us see. This situation can be represented graphically as shown in Fig. 7.1.

You may use the Pythagoras Theorem to calculate this distance.

Now, suppose two points lie on the x-axis. Can we find the distance between them? For instance, consider two points `A(4, 0) `and` B(6, 0) `in Fig. 7.2. The points A and B lie on the x-axis.

From the figure you can see that` OA = 4` units and `OB = 6` units.

Therefore, the distance of B from A, i.e., `AB = OB – OA = 6 – 4 = 2` units.

So, if two points lie on the x-axis, we can easily find the distance between them.

Now, suppose we take two points lying on the y-axis. Can you find the distance between them. If the points` C(0, 3) `and `D(0, 8)` lie on the y-axis, similarly we find that` CD = 8 – 3 = 5` units (see Fig. 7.2).

Next, can you find the distance of A from C (in Fig. 7.2)? Since` OA = 4` units and

`OC = 3` units, the distance of `A` from `C`, i.e., `AC = sqrt (3^2 + 4^2 ) = 5` units. Similarly, you can

find the distance of B from` D = BD = 10` units.

Now, if we consider two points not lying on coordinate axis, can we find the distance between them? Yes! We shall use Pythagoras theorem to do so. Let us see an example.

In Fig. 7.3, the points `P(4, 6)` and` Q(6, 8)` lie in the first quadrant. How do we use Pythagoras theorem to find the distance between them?

Let us draw `PR `and `QS` perpendicular to the x-axis from `P` and `Q` respectively. Also, draw a perpendicular from `P` on `QS `to meet `QS` at `T`. Then the coordinates of `R` and `S` are `(4, 0)` and `(6, 0)`, respectively. So, `RS = 2` units. Also, `QS = 8 `units and `TS = PR = 6` units.

Therefore, `QT = 2` units and `PT = RS = 2` units.

Now, using the Pythagoras theorem, we have

`PQ^2 = PT^2 + QT^2`

`=2^2 + 2^2 = 8`

So, `PQ = 2 sqrt 2` units

How will we find the distance between two points in two different quadrants?

Consider the points `P(6, 4)` and `Q(–5, –3)` (see Fig. 7.4). Draw `QS` perpendicular to the x-axis. Also draw a perpendicular `PT` from the point `P` on `QS` (extended) to meet y-axis at the point `R`.

Then `PT = 11` units and `QT = 7 `units. (Why?)

Using the Pythagoras Theorem to the right triangle` PTQ`, we get

`PQ = sqrt(11^2 + 7^2) = sqrt (700` units.

Let us now find the distance between any two points `P(x_1, y_1)` and `Q(x_2, y_2)`. 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` (see Fig. 7.5).

Then, `OR = x_1, OS = x_2`. So, `RS = x_2 – x_1 = PT`.

Also, `SQ = y_2, ST = PR = y_1`. So, `QT = y_2 – y_1`.

Now, applying the Pythagoras theorem in `Delta PTQ`, we get

`PQ^2 = PT^2 +QT^2`

` = (x_2 -x_1)^2 + (y_2 -y_1)^2`

Therefore, ` PQ = sqrt ( (x_2- x_1)^2 + (y_2 -y_1)^2)`

Note that since distance is always non-negative, we take only the positive square root. So, the distance between the points `P(x_1, y_1)` and `Q(x_2, y_2)` is

`color{red}{PQ = d = sqrt ( (x_2 -x_1 )^2 + (y_2 -y_1 )^2 )}` ,

which is called the `"distance formula"`.

`"Remarks :"`

1. In particular, the distance of a point `P(x, y)` from the origin `O(0, 0)` is given by

`OP = sqrt(x^2 +y^2 )`

2. We can also write, `PQ = sqrt( (x_1 +x_2)^2 +( y_1 -y_2)^2 ) `

A town B is located `36` km east and 15 km north of the town A. How would you find the distance from town A to town B without actually measuring it. Let us see. This situation can be represented graphically as shown in Fig. 7.1.

You may use the Pythagoras Theorem to calculate this distance.

Now, suppose two points lie on the x-axis. Can we find the distance between them? For instance, consider two points `A(4, 0) `and` B(6, 0) `in Fig. 7.2. The points A and B lie on the x-axis.

From the figure you can see that` OA = 4` units and `OB = 6` units.

Therefore, the distance of B from A, i.e., `AB = OB – OA = 6 – 4 = 2` units.

So, if two points lie on the x-axis, we can easily find the distance between them.

Now, suppose we take two points lying on the y-axis. Can you find the distance between them. If the points` C(0, 3) `and `D(0, 8)` lie on the y-axis, similarly we find that` CD = 8 – 3 = 5` units (see Fig. 7.2).

Next, can you find the distance of A from C (in Fig. 7.2)? Since` OA = 4` units and

`OC = 3` units, the distance of `A` from `C`, i.e., `AC = sqrt (3^2 + 4^2 ) = 5` units. Similarly, you can

find the distance of B from` D = BD = 10` units.

Now, if we consider two points not lying on coordinate axis, can we find the distance between them? Yes! We shall use Pythagoras theorem to do so. Let us see an example.

In Fig. 7.3, the points `P(4, 6)` and` Q(6, 8)` lie in the first quadrant. How do we use Pythagoras theorem to find the distance between them?

Let us draw `PR `and `QS` perpendicular to the x-axis from `P` and `Q` respectively. Also, draw a perpendicular from `P` on `QS `to meet `QS` at `T`. Then the coordinates of `R` and `S` are `(4, 0)` and `(6, 0)`, respectively. So, `RS = 2` units. Also, `QS = 8 `units and `TS = PR = 6` units.

Therefore, `QT = 2` units and `PT = RS = 2` units.

Now, using the Pythagoras theorem, we have

`PQ^2 = PT^2 + QT^2`

`=2^2 + 2^2 = 8`

So, `PQ = 2 sqrt 2` units

How will we find the distance between two points in two different quadrants?

Consider the points `P(6, 4)` and `Q(–5, –3)` (see Fig. 7.4). Draw `QS` perpendicular to the x-axis. Also draw a perpendicular `PT` from the point `P` on `QS` (extended) to meet y-axis at the point `R`.

Then `PT = 11` units and `QT = 7 `units. (Why?)

Using the Pythagoras Theorem to the right triangle` PTQ`, we get

`PQ = sqrt(11^2 + 7^2) = sqrt (700` units.

Let us now find the distance between any two points `P(x_1, y_1)` and `Q(x_2, y_2)`. 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` (see Fig. 7.5).

Then, `OR = x_1, OS = x_2`. So, `RS = x_2 – x_1 = PT`.

Also, `SQ = y_2, ST = PR = y_1`. So, `QT = y_2 – y_1`.

Now, applying the Pythagoras theorem in `Delta PTQ`, we get

`PQ^2 = PT^2 +QT^2`

` = (x_2 -x_1)^2 + (y_2 -y_1)^2`

Therefore, ` PQ = sqrt ( (x_2- x_1)^2 + (y_2 -y_1)^2)`

Note that since distance is always non-negative, we take only the positive square root. So, the distance between the points `P(x_1, y_1)` and `Q(x_2, y_2)` is

`color{red}{PQ = d = sqrt ( (x_2 -x_1 )^2 + (y_2 -y_1 )^2 )}` ,

which is called the `"distance formula"`.

`"Remarks :"`

1. In particular, the distance of a point `P(x, y)` from the origin `O(0, 0)` is given by

`OP = sqrt(x^2 +y^2 )`

2. We can also write, `PQ = sqrt( (x_1 +x_2)^2 +( y_1 -y_2)^2 ) `

Q 3189480317

Do the points` (3, 2), (–2, –3)` and `(2, 3) ` form a triangle? If so, name the

type of triangle formed.

Class 10 Chapter 7 Example 1

type of triangle formed.

Class 10 Chapter 7 Example 1

Let us apply the distance formula to find the distances` PQ, QR` and `PR`,

where` P(3, 2), Q(–2, –3)` and `R(2, 3)` are the given points. We have

`PQ = sqrt ( (3+2)^2 +(2+3)^2 ) = sqrt( 5^2 + 5^2) = sqrt (50) = 7.07` (approx)

`QR = sqrt ( (-2-2)^2 +(-3-3)^2 ) = sqrt ( (-4)^2 +(-6)^2 ) = sqrt (52) =7.21 ` (approx)

`PR = sqrt ( (3-2)^2 +( 2-3)^2 ) = sqrt (1^2 + (-1)^2 ) = sqrt 2 = 1.41` (approx )

Since the sum of any two of these distances is greater than the third distance, therefore, the points P, Q and R form a triangle.

Also,` PQ^2 + PR^2 = QR^2`, by the converse of Pythagoras theorem, we have `∠ P = 90°`.

Therefore, `PQR` is a right triangle.

Q 3109480318

Show that the points` (1, 7), (4, 2), (–1, –1)` and `(– 4, 4) `are the vertices of a square.

Class 10 Chapter 7 Example 2

Class 10 Chapter 7 Example 2

Let `A(1, 7), B(4, 2), C(–1, –1)` and `D(– 4, 4)` be the given points. One way

of showing that `ABCD ` is a square is to use the property that all its sides should be

equal and both its digonals should also be equal. Now,

`AB = sqrt( (1-4)^2 + (7-2)^2 ) = sqrt(9+25) = sqrt(34)`

`BC = sqrt ( (4+1)^2 + (2+1)^2 ) = sqrt (25 + 9) = sqrt (34)`

`CD = sqrt ( (-1+4)^2 + (-1-4)^2 ) = sqrt(9+25) = sqrt (34)`

`DA = sqrt ( (1+4)^2 + (7-4)^2 ) = sqrt (25 + 9) = sqrt (34)`

`AC = sqrt ( ( 1+1 )^2 + (7+1)^2 ) = sqrt(4+64) = sqrt (68)`

`BD = sqrt ( (4+4)^2 + ( 2-4)^2 ) = sqrt (64 +4) = sqrt (68 )`

Since, `AB = BC = CD = DA` and `AC = BD`, all the four sides of the quadrilateral

ABCD are equal and its diagonals AC and BD are also equal. Thereore, `ABCD` is a

square.

Alternative Solution : We find the four sides and one diagonal, say,

`AC `as above. Here `AD^2 + DC^2 = 34 + 34 = 68 = AC^2`. Therefore, by

the converse of Pythagoras theorem,` ∠ D = 90°`. A quadrilateral with all four sides equal and one

angle `90°` is a square. So, `ABCD` is a square.

Q 3129580411

Fig. 7.6 shows the arrangement of desks in a classroom. Ashima, Bharti and

Camella are seated at `A(3, 1), B(6, 4)` and` C(8, 6) `respectively. Do you think they are seated in a

line? Give reasons for your answer.

Class 10 Chapter 7 Example 3

Camella are seated at `A(3, 1), B(6, 4)` and` C(8, 6) `respectively. Do you think they are seated in a

line? Give reasons for your answer.

Class 10 Chapter 7 Example 3

Using the distance formula, we have

`AB = sqrt ( (6-3)^2 + ( 4-1)^2 ) = sqrt (9+9 ) = sqrt (18) = 3 sqrt 2`

`BC = sqrt ( (8-6)^2 + (6-4)^2 ) = sqrt (4+4) = sqrt 8 = 2 sqrt 2`

`AC = sqrt ( (8-3)^2 + ( 6-1)^2 ) = sqrt (25 +25) = sqrt (25) = 5 sqrt 2`

Since, `AB + BC = 3 sqrt 2 + 2 sqrt 2 = 5 sqrt 2= AC` , we can say that the points A, B and C

are collinear. Therefore, they are seated in a line.

Q 3149580413

Find a relation between x and y such that the point `(x , y)` is equidistant

from the points `(7, 1)` and `(3, 5)`.

Class 10 Chapter 7 Example 4

from the points `(7, 1)` and `(3, 5)`.

Class 10 Chapter 7 Example 4

Let `P(x, y)` be equidistant from the points `A(7, 1) `and `B(3, 5)`.

We are given that `AP = BP` . So, `AP^2 = BP^2`

i.e., `(x – 7)^2 + (y – 1)^2 = (x – 3)^2 + (y – 5)^2`

i.e., `x^2 – 14x + 49 + y^2 – 2y + 1 = x^2 – 6x + 9 + y^2 – 10y + 25`

i.e., `x – y = 2`

which is the required relation.

Remark : Note that the graph of the equation `x – y = 2 `is a line. From your earlier studies,

you know that a point which is equidistant from A and B lies on the perpendicular

bisector of AB. Therefore, the graph of `x – y = 2` is the perpendicular bisector of AB

(see Fig. 7.7).

Q 3179580416

Find a point on the y-axis which is equidistant from the points A(6, 5) and

B(– 4, 3).

Class 10 Chapter 7 Example 5

B(– 4, 3).

Class 10 Chapter 7 Example 5

We know that a point on the y-axis is of the form (0, y). So, let the point

P(0, y) be equidistant from A and B. Then

` (6 – 0)^2 + (5 – y)^2 = (– 4 – 0)^2 + (3 – y)^2`

i.e., `36 + 25 + y^2 – 10y = 16 + 9 + y^2 – 6y`

i.e.,` 4y = 36`

i.e., `y = 9`

So, the required point is `(0, 9)`.

Let us check our solution `: AP = sqrt ( (6-0)^2 + ( 5-9)^2 ) = sqrt (36 +16) = sqrt (52)`

`BP = sqrt ( ( -4 -0)^2 + (3-9)^2 ) = sqrt (16 +36 ) = sqrt (52 )`

Note : Using the remark above, we see that (0, 9) is the intersection of the y-axis and

the perpendicular bisector of AB.