CBSE 10th Mathematics | Coordinate Geometry | Introduction

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.

Question:

Find the distance between the points

(i)    P(–6, 7) and Q(–1, –5)

(ii)    R(a + b, ab) and S(ab, – ab)

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}}}

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

\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.

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,

\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.

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        \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

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    \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).

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        \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).

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,

\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.

Question:

Show that the points (a, a), (–a, –a) and \displaystyle \left( -\sqrt{3}a,\ \sqrt{3}a \right) are the vertices of an equilateral 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.

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, 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

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,

\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

So, ADBC is a parallelogram.

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

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    \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).

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 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.

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    \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.

Question:

If two vertices of an equilateral triangle be O(0, 0) and A(3,\displaystyle \sqrt{3}), find the coordinates of its third vertex.

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})

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