## Coordinate Geometry | Introduction

## Coordinate Geometry | Coordinates of a Point

Location of the position of a point on a plane requires a pair of co-ordinate axes.

The distance of a point from the *x*-axis is called its *y*-coordinate, or ordinate.

The distance of a point from the *y*-axis is called its *x*-co-ordinate or abscissa.

The co-ordinates 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 | Distance Formula>

**The distance between two points P(x_{1, }y_{1}) and Q(x_{2}, y_{2}) is given by the formula**

**Proof:**

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* = *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 D*PTQ*,

we get *PQ*^{2} = *PT*^{2} + *QT*^{2}

= (*x*_{2} – *x*_{1})^{2} + (*y*_{2} – *y*_{1})^{2}

Therefore,

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

which is called the distance formula.

Note: *In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by
*

*OP* =

## Coordinate Geometry | Collinear Points

Three points *A*, *B*, *C* are said to be collinear if they lie on the same straight line.

### Test for Collinearity of Three Points

In order to show that three given points *A*, *B*, *C* are collinear, we find distances AB, BC and AC. If the sum of any two of these distances is equal to the third distance then the given points are collinear.

### Question:

#### Find the distance between the points

#### (i)* P*(–6, 7) and* Q*(–1, –5)

#### (ii) *R*(*a* + *b*, *a* – *b*) and *S*(*a* – *b*, – *a* – *b*)

### Solution:

**(i)**

Here, *x*_{1} = –6, *y*_{1} = 7 and *x*_{2} = –1, *y*_{2} = –5

\

Þ

Þ

**(ii)**

We have,

Þ

### Question:

#### Show that the points (1, –1), (5, 2) and (9, 5) are collinear.

### Solution:

Let *A*(1, – 1), *B*(5, 2) and (9, 5) be the given points. Then, we have

and

Clearly, *AC* = *AB *+ *BC*

Hence, *A*, *B*, *C* are collinear points.

### Question:

#### Do the points* A*(3, 2)*, B*(–2, –3) and *C*(2, 3) form a triangle? If so, name the type of triangle formed.

### Solution:

We have,

and,

As *AB* + *BC* > *AC*, *AC* + *BC* > *AB* and *AB* + *AC* > *BC*. Therefore, points *A*, *B* and *C* form a triangle.

and

\

Therefore, D*ABC* is a right triangle, right angled at *A*.

### Question:

#### Find the value of* x, *if the distance between the points (*x, –*1) and (3, 2) is 5.

### Solution:

Let *P*(*x*, –1) and *Q*(3, 2) be the given points. Then, *PQ* = 5

Þ

or

Þ

Þ

Þ

Þ

Þ (*x* – 7) (*x* + 1) = 0

Þ *x* = 7 or, *x* = –1

### Question:

#### Find a point on x-axis which is equidistant from* A*(2, –5) and* B*(–2, 9).

### Solution:

Let *P*(*x*, 0) be the point equidistant from *A*(2, –5) and *B*(–2, 9). Then, *PA* = *PB*

Þ

Þ

Þ

Þ

Þ – 8*x* = 56

Þ x = –7

Hence, the required point is (–7, 0).

### Question:

#### Find a point on the y-axis which is equidistant from the point* A*(6, 5) and* B*(–4, 3).

### Solution:

Let the required point be *P*(0, *y*). Then, *PA* = *PB*

Þ

Þ

Þ

Þ 4*y* = 36

Þ *y* = 9

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

### Question:

#### Prove that the points (–3, 0), (1, –3) and (4, 1) are the vertices of an isosceles right-angled triangle.

### Solution:

Let *A*(–3, 0), *B*(1, –3) and *C*(4, 1) be the given points. Then,

= = 5 units

= = 5 units

and,

units

Þ *AB* = *BC*.

Therefore, D*ABC* is isosceles.

Also, = 25 + 25 =

D*ABC* is right-angled at *B*. Thus, D*ABC* is a right-angled isosceles triangle.

### Question:

#### Show that the points (*a, a*)*, *(*–a, –a*) and are the vertices of an equilateral triangle.

### Solution:

Let *A*(*a*, *a*), *B*(–*a*, –*a*) and *C* be the given points. Then, we have

Þ

Þ

Þ =

and,

Þ

Þ

Þ

As we have *AB* = *BC* = *AC*

Hence, the triangle *ABC* formed by the given points is an equilateral triangle.

### Question:

#### If the point (*x, y*) is equidistant from the points (*a + b, b – a*) and (*a – b, a + b*)*, *prove that* bx = ay.*

### Solution:

Let *P*(*x*, *y*), *Q*(*a* + *b*, *b* – *a*) and *R* (*a* – *b*, *a* + *b*) be the given points.

Then, *PQ* = *PR*

Þ

or

Þ

Þ

Þ – 2*x*(*a* + *b*) – 2*y*(*b* – *a*) = –2*x*(*a* – *b*) – 2*y*(*a* + *b*)

or – *ax *– *bx* – *by* + *ay* = – *ax* + *bx* – *ay* – *by*

Þ – *bx* + *ay* = *bx* – *ay*

Þ – *bx* – *bx* = – 2*ay*

Þ 2*bx* = 2*ay*

Þ *bx* = *ay*

### Question:

#### Show that four points (0, –1), (6, 7), (–2, 3) and (8, 3) are the vertices of a rectangle.

### Solution:

Let *A*(0 –1), *B*(6, 7), *C*(–2, 3) and *D*(8, 3) be the given points. Then,

and,

\ *AD* = *BC* and *AC* = *BD*

So, *ADBC* is a parallelogram.

Now,

and,

\ *AB*^{2} = *AD*^{2} + *DB*^{2} and *CD*^{2} = *CB*^{2} + *BD*^{2}

Hence, *ADBC* is a parallelogram.

### Question:

#### Find the coordinates of the point equidistant from three given points *A*(5, 1)* B*(–3, –7) and* C*(7, –1).

### Solution:

Let the required point be *P*(*x*, *y*). Then,

*PA* = *PB* = *PC*

Þ

Þ

Þ

and

Þ

and

Now,

Þ 16*x* + 16*y* = – 32

Þ *x* + *y* = – 2 … (i)

Also,

Þ 20*x* + 12*y* = – 8 Þ 5*x* + 3*y* = – 2 … (ii)

Multiplying (i) by 5 and subtracting (ii) from the result, we get

2*y* = – 8 Þ *y* = – 4

Putting *y* = – 4 in (i), we get *x* = 2

\ *x* = 2 and *y* = – 4

Hence, the required point is *P*(2, – 4).

### Question:

#### Find the coordinates of the circumcentre of a triangle whose vertices are *A*(4, 6),* B*(0, 4) and* C*(6, 2). Also, find its circumradius.

### Solution:

Let *A*(4, 6), *B*(0, 4) and *C*(6, 2) be the vertices of the given D*ABC*.

Let *P*(*x*, *y*) be the circumcentre of D*ABC*.

Then, *PA* = *PB* = *PC*

Þ

Þ

Now,

**Þ **

Þ

Þ 8*x* + 4*y* = 36

Þ 2*x* + *y* = 9 ….. (i)

Again,

Þ

Þ

Þ 12*x* – 4*y* = 24

Þ 3*x* – *y* = 6 ….. (ii)

On solving (i) and (ii), we get x = 3 and y = 3.

\ Coordinates of the circumcentre of D*ABC* are *P*(3, 3)

Circumradius = = units.

### Question:

#### Find the coordinates of the centre of a circle passing through the points *A*(2, 1),* B*(5, –8) and* C*(2, –9). Also, find the radius of this circle.

### Solution:

Let *P*(*x*, *y*) be the centre of the circle passing through the points *A*(2, 1), *B*(5, –8) and *C*(2, –9). Then,

PA = PB = PC

Þ

Þ

Now,

Þ

Þ

Þ 6x – 18y – 84 = 0

Þ x – 3y = 14 ….. (i)

And,

Þ

Þ

Þ 6*x* + 2*y* – 4 = 0

Þ 3*x* + *y* = 2 ….. (ii)

On solving (i) and (ii), we get x = 2 and y = – 4.

Hence, the coordinates of the centre of the circle are *P*(2, –4).

Radius of the circle = PA

=

= = 5 units.

### Question:

#### If two vertices of an equilateral triangle be* O*(0, 0) and* A(3,**), *find the coordinates of its third vertex.

### Solution:

Let *O*(0, 0) and *A*(3,) be two vertices of the equilateral D*OAB*. Let *B*(*x, y*) be its third vertex. Then, *OA* = *OB* = *AB *Þ

Þ

Now,

Þ

Þ … (i)

Þ

Þ

Þ

or

Þ

Þ … (ii)

Putting from (ii) in (i), we get

Þ

Þ

Þ 4*x*(*x* – 3) = 0

Þ *x* = 0 or *x* = 3

Now, x = 0 Þ y =

And, x = 3 Þ y =

Hence, the coordinates of B are or