# CBSE 10th Mathematics | 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(x1, y1) and Q(x2, y2) is given by the formula

$\displaystyle PQ=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}$

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 = x1, OS = x2, So, RS = x2x1 = PT.

Also, SQ = y2, ST = PR = y1. So, QT = y2y1.

Now, applying the Pythagoras theorem in DPTQ,

we get         PQ2 = PT2 + QT2

= (x2x1)2 + (y2y1)2

Therefore,        $\displaystyle PQ=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}$

Since distance is always non-negative, we take only the positive square root. So, the distance between the points P(x1, y1) and Q(x2, y2) is

$\displaystyle PQ=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}$

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 = $\displaystyle \sqrt{{{x}^{2}}+{{y}^{2}}}$

## 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.

### Solution:

(i)

Here, x1 = –6, y1 = 7 and x2 = –1, y2 = –5

\    $\displaystyle PQ=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}$

Þ    $\displaystyle PQ=\sqrt{{{(-1+6)}^{2}}+{{(-5-7)}^{2}}}$

Þ    $\displaystyle PQ=\sqrt{25+144}=\sqrt{169}=13$

(ii)

We have,

$\displaystyle RS=\sqrt{{{(a-b-a-b)}^{2}}+{{(-a-b-a+b)}^{2}}}$

Þ    $\displaystyle RS=\sqrt{4{{b}^{2}}+4{{a}^{2}}}=2\sqrt{{{a}^{2}}+{{b}^{2}}}$

### Solution:

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

$\displaystyle AB=\sqrt{{{(5-1)}^{2}}+{{(2+1)}^{2}}}=\sqrt{16+9}=5$

$\displaystyle BC=\sqrt{{{(5-9)}^{2}}+{{(2-5)}^{2}}}=\sqrt{16+9}=5$

and        $\displaystyle AC=\sqrt{{{(1-9)}^{2}}+{{(-1-5)}^{2}}}=\sqrt{64+36}=10$

Clearly,    AC = AB + BC

Hence, A, B, C are collinear points.

### Solution:

We have,

$\displaystyle AB=\sqrt{{{(-2-3)}^{2}}+{{(-3-2)}^{2}}}=\sqrt{25+25}=\sqrt{50}$

$\displaystyle BC=\sqrt{{{(2+2)}^{2}}+{{(3+3)}^{2}}}=\sqrt{16+36}=\sqrt{52}$

and,        $\displaystyle AC=\sqrt{{{(2-3)}^{2}}+{{(3-2)}^{2}}}=\sqrt{1+1}=\sqrt{2}$

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

$\displaystyle A{{B}^{2}}=50,\ B{{C}^{2}}=52$ and $\displaystyle A{{C}^{2}}=2$

\        $\displaystyle B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}$

Therefore,    DABC is a right triangle, right angled at A.

### Solution:

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

Þ        $\displaystyle \sqrt{{{(x-3)}^{2}}+{{(-1-2)}^{2}}}=5$

or        $\displaystyle {{(x-3)}^{2}}+9={{5}^{2}}$

Þ        $\displaystyle {{x}^{2}}+9-6x+9=25$

Þ        $\displaystyle {{x}^{2}}-6x+18=25$

Þ        $\displaystyle {{x}^{2}}-6x-7=0$

Þ        $\displaystyle {{x}^{2}}-7x+x-7=0$

Þ        (x – 7) (x + 1) = 0

Þ        x = 7 or, x = –1

### Solution:

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

Þ    $\displaystyle \sqrt{{{(x-2)}^{2}}+{{(0+5)}^{2}}}=\sqrt{{{(x+2)}^{2}}+{{(0-9)}^{2}}}$

Þ    $\displaystyle {{(x-2)}^{2}}+25={{(x+2)}^{2}}+81$

Þ    $\displaystyle {{x}^{2}}-4x+4+25={{x}^{2}}+4x+4+81$

Þ    $\displaystyle {{x}^{2}}-4x+29={{x}^{2}}+4x+85$

Þ    – 8x = 56

Þ    x = –7

Hence, the required point is (–7, 0).

### Solution:

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

Þ        $\displaystyle \sqrt{{{(0-6)}^{2}}+{{(y-5)}^{2}}}=\sqrt{{{(0+4)}^{2}}+{{(y-3)}^{2}}}$

Þ        $\displaystyle 36+{{(y-5)}^{2}}=16+{{(y-3)}^{2}}$

Þ        $\displaystyle 36+{{y}^{2}}-10y+25=16+{{y}^{2}}-6y+9$

Þ        4y = 36

Þ        y = 9

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

### Solution:

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

$\displaystyle AB=\sqrt{{{[1-(-3)]}^{2}}+{{(-3-0)}^{2}}}$

$\displaystyle =\sqrt{{{4}^{2}}+{{(-3)}^{2}}}$ = $\displaystyle \sqrt{16+9}$ = 5 units

$\displaystyle BC=\sqrt{{{(4-1)}^{2}}+{{(1+3)}^{2}}}$

= $\displaystyle \sqrt{9+16}$ = 5 units

and, $\displaystyle CA=\sqrt{{{(4+3)}^{2}}+{{(1-0)}^{2}}}$

$\displaystyle =\sqrt{49+1}$$\displaystyle =5\sqrt{2}$ units

Þ AB = BC.

Therefore, DABC is isosceles.

Also,    $\displaystyle A{{B}^{2}}+B{{C}^{2}}$ = 25 + 25 = $\displaystyle {{(5\sqrt{2})}^{2}}=C{{A}^{2}}$

DABC is right-angled at B. Thus, DABC is a right-angled isosceles triangle.

### Solution:

Let A(a, a), B(–a, –a) and C$\displaystyle \left( -\sqrt{3}a,\ \sqrt{3}a \right)$ be the given points. Then, we have

$\displaystyle AB=\sqrt{{{(-a-a)}^{2}}+{{(-a-a)}^{2}}}=\sqrt{4{{a}^{2}}+4{{a}^{2}}}=2\sqrt{2}a$

$\displaystyle BC=\sqrt{{{(-\sqrt{3}a+a)}^{2}}+{{(\sqrt{3}a+a)}^{2}}}$

Þ    $\displaystyle BC=\sqrt{{{a}^{2}}{{(1-\sqrt{3})}^{2}}+{{a}^{2}}{{(\sqrt{3}+1)}^{2}}}$

Þ    $\displaystyle BC=a\sqrt{{{(1-\sqrt{3})}^{2}}+{{(1+\sqrt{3})}^{2}}}$

Þ    $\displaystyle BC=a\sqrt{1+3-2\sqrt{3}+1+3+2\sqrt{3}}$ = $\displaystyle a\sqrt{8}=2\sqrt{2}a$

and,    $\displaystyle AC=\sqrt{{{(-\sqrt{3}a-a)}^{2}}+{{(\sqrt{3}a-a)}^{2}}}$

Þ    $\displaystyle AC=\sqrt{{{a}^{2}}{{(\sqrt{3}+1)}^{2}}+{{a}^{2}}{{(\sqrt{3}-1)}^{2}}}$

Þ    $\displaystyle AC=a\sqrt{{{(\sqrt{3}+1)}^{2}}+{{(\sqrt{3}-1)}^{2}}}$

Þ    $\displaystyle AC=a\sqrt{3+1+2\sqrt{3}+3+1-2\sqrt{3}}=a\sqrt{8}=2\sqrt{2}a$

As we have AB = BC = AC

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

### Solution:

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

Then, PQ = PR

Þ    $\displaystyle \sqrt{{{[x-(a+b)]}^{2}}+{{[y-(b-a)]}^{2}}}=\sqrt{{{[x-(a-b)]}^{2}}+{{[y-(a+b)]}^{2}}}$

or    $\displaystyle {{[x-(a+b)]}^{2}}+{{[y-(b-a)]}^{2}}={{[x-(a-b)]}^{2}}+{{[y-(a+b)]}^{2}}$

Þ    $\displaystyle {{x}^{2}}-2x(a+b)+{{(a+b)}^{2}}+{{y}^{2}}-2y(b-a)+{{(b-a)}^{2}}$

Þ    $\displaystyle {{x}^{2}}+{{(a-b)}^{2}}-2x(a-b)+{{y}^{2}}-2y(a+b)+{{(a+b)}^{2}}$

Þ    – 2x(a + b) – 2y(ba) = –2x(ab) – 2y(a + b)

or    – ax bxby + ay = – ax + bxayby

Þ    – bx + ay = bxay

Þ    – bxbx = – 2ay

Þ    2bx = 2ay

Þ    bx = ay

### Solution:

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

$\displaystyle AD=\sqrt{{{(8-0)}^{2}}+{{(3+1)}^{2}}}=\sqrt{64+16}=4\sqrt{5}$

$\displaystyle BC=\sqrt{{{(6+2)}^{2}}+{{(7-3)}^{2}}}=\sqrt{64+16}=4\sqrt{5}$

$\displaystyle AC=\sqrt{{{(-2-0)}^{2}}+{{(3+1)}^{2}}}=\sqrt{4+16}=2\sqrt{5}$

and,    $\displaystyle BD=\sqrt{{{(8-6)}^{2}}+{{(3-7)}^{2}}}=\sqrt{4+16}=2\sqrt{5}$

\    AD = BC and AC = BD

Now,    $\displaystyle AB=\sqrt{{{(6-0)}^{2}}+{{(7+1)}^{2}}}=\sqrt{36+64}=10$

and,    $\displaystyle CD=\sqrt{{{(8+2)}^{2}}+{{(3-3)}^{2}}}=10$

\    AB2 = AD2 + DB2 and CD2 = CB2 + BD2

### Solution:

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

PA = PB = PC

Þ    $\displaystyle P{{A}^{2}}=P{{B}^{2}}=P{{C}^{2}}$

Þ    $\displaystyle P{{A}^{2}}=P{{B}^{2}}\ \text{and}\ P{{B}^{2}}=P{{C}^{2}}$

Þ    $\displaystyle {{(x-5)}^{2}}+{{(y-1)}^{2}}={{(x+3)}^{2}}+{{(y+7)}^{2}}$

and    $\displaystyle {{(x+3)}^{2}}+{{(y+7)}^{2}}={{(x-7)}^{2}}+{{(y+1)}^{2}}$

Þ    $\displaystyle {{x}^{2}}+{{y}^{2}}-10x-2y+26={{x}^{2}}+{{y}^{2}}+6x+14y+58$

and    $\displaystyle {{x}^{2}}+{{y}^{2}}+6x+14y+58={{x}^{2}}+{{y}^{2}}-14x+2y+50$

Now,    $\displaystyle {{x}^{2}}+{{y}^{2}}-10x-2y+26={{x}^{2}}+{{y}^{2}}+6x+14y+58$

Þ    16x + 16y = – 32

Þ    x + y = – 2                    … (i)

Also,    $\displaystyle {{x}^{2}}+{{y}^{2}}+6x+14y+58={{x}^{2}}+{{y}^{2}}-14x+2y+50$

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

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

2y = – 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).

### Solution:

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

Let P(x, y) be the circumcentre of DABC.

Then, PA = PB = PC

Þ    $\displaystyle P{{A}^{2}}=P{{B}^{2}}=P{{C}^{2}}$

Þ    $\displaystyle P{{A}^{2}}=P{{B}^{2}}\ \text{and}\ P{{B}^{2}}=P{{C}^{2}}$

Now,    $\displaystyle P{{A}^{2}}=P{{B}^{2}}$

Þ    $\displaystyle {{(x-4)}^{2}}+{{(y-6)}^{2}}={{(x-0)}^{2}}+{{(y-4)}^{2}}$

Þ    $\displaystyle {{x}^{2}}+{{y}^{2}}-8x-12y+52={{x}^{2}}+{{y}^{2}}-8y+16$

Þ    8x + 4y = 36

Þ    2x + y = 9            ….. (i)

Again,    $\displaystyle P{{B}^{2}}=P{{C}^{2}}$

Þ    $\displaystyle {{(x-0)}^{2}}+{{(y-4)}^{2}}={{(x-6)}^{2}}+{{(y-2)}^{2}}$

Þ    $\displaystyle {{x}^{2}}+{{y}^{2}}-8y+16={{x}^{2}}+{{y}^{2}}-12x-4y+40$

Þ    12x – 4y = 24

Þ    3xy = 6            ….. (ii)

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

\    Coordinates of the circumcentre of DABC are P(3, 3)

Circumradius = $\displaystyle PA=\sqrt{{{(4-3)}^{2}}+{{(6-3)}^{2}}}$ = $\displaystyle \sqrt{{{1}^{2}}+{{3}^{2}}}=\sqrt{10}$ units.

### 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    $\displaystyle P{{A}^{2}}=P{{B}^{2}}=P{{C}^{2}}$

Þ    $\displaystyle P{{A}^{2}}=P{{B}^{2}}\ \text{and}\ P{{B}^{2}}=P{{C}^{2}}$

Now,    $\displaystyle P{{A}^{2}}=P{{B}^{2}}$

Þ    $\displaystyle {{(x-2)}^{2}}+{{(y-1)}^{2}}={{(x-5)}^{2}}+{{(y+8)}^{2}}$

Þ    $\displaystyle {{x}^{2}}+{{y}^{2}}-4x-2y+5={{x}^{2}}+{{y}^{2}}-10x+16y+89$

Þ    6x – 18y – 84 = 0

Þ    x – 3y = 14                    ….. (i)

And,    $\displaystyle P{{B}^{2}}=P{{C}^{2}}$

Þ    $\displaystyle {{(x-5)}^{2}}+{{(y+8)}^{2}}={{(x-2)}^{2}}+{{(y+9)}^{2}}$

Þ    $\displaystyle {{x}^{2}}+{{y}^{2}}-10x+16y+89={{x}^{2}}+{{y}^{2}}-4x+18y+85$

Þ    6x + 2y – 4 = 0

Þ    3x + 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

= $\displaystyle \sqrt{{{(2-2)}^{2}}+{{(1+4)}^{2}}}=\sqrt{{{0}^{2}}+{{5}^{2}}}$

= $\displaystyle \sqrt{25}$ = 5 units.

### Solution:

Let O(0, 0) and A(3,$\displaystyle \sqrt{3}$) be two vertices of the equilateral DOAB. Let B(x, y) be its third vertex. Then, OA = OB = AB Þ $\displaystyle O{{A}^{2}}=O{{B}^{2}}=A{{B}^{2}}$

Þ    $\displaystyle O{{A}^{2}}=O{{B}^{2}}\ \text{and}\ O{{B}^{2}}=A{{B}^{2}}$

Now,    $\displaystyle O{{A}^{2}}=O{{B}^{2}}$

Þ    $\displaystyle {{(3-0)}^{2}}+{{(\sqrt{3}-0)}^{2}}={{(x-0)}^{2}}+{{(y-0)}^{2}}$

Þ    $\displaystyle {{x}^{2}}+{{y}^{2}}=12$                … (i)

$\displaystyle O{{B}^{2}}=A{{B}^{2}}$    Þ $\displaystyle {{(x-0)}^{2}}+{{(y-0)}^{2}}={{(x-3)}^{2}}+{{(y-\sqrt{3})}^{2}}$

Þ $\displaystyle {{x}^{2}}+{{y}^{2}}={{x}^{2}}+9-6x+{{y}^{2}}+3-2\sqrt{3}$

Þ $\displaystyle 2\sqrt{3}\,y=12-6x$

or $\displaystyle \sqrt{3}y=6-3x$

Þ $\displaystyle \sqrt{3}y=3\,(2-x)$

Þ $\displaystyle y=\sqrt{3}\,(2-x)$        … (ii)

Putting $\displaystyle y=\sqrt{3}\,(2-x)$ from (ii) in (i), we get

$\displaystyle {{x}^{2}}+3{{(2-x)}^{2}}=12$    Þ $\displaystyle {{x}^{2}}+3(4+{{x}^{2}}-4x)=12$

Þ $\displaystyle 4{{x}^{2}}-12x=0$

Þ 4x(x – 3) = 0

Þ x = 0 or x = 3

Now,    x = 0 Þ y = $\displaystyle 2\sqrt{3}$

And,    x = 3 Þ y = $\displaystyle -\sqrt{3}$

Hence, the coordinates of B are $\displaystyle (0,\ 2\sqrt{3})$ or $\displaystyle (3,\ \sqrt[-]{3})$