**TOPIC 4: COORDINATE GEOMETRY I ~ ADV MATHEMATICS FORM 5**

# COORDINATE GEOMETRY

-Is the conic section whose eccentricity is equal to zero.

-Coordinate geometry is the study of representation of geometry figure either on two or three dimension under the one of the following steps.

1. DISTANCE BETWEEN TWO POINTS

-Let’s consider point A and on XY- plane now we need to find the distance from A to B.

Pythagoras theorem

But

Now,

For case of three of dimension

– – A midpoint of line segment, is the point bisect a line segment.

OR

– Is the point which divide a certain line into two equal parts.

Proof,

-Consider point A(x1,y1), B(x2,y2) and R(x,y)

-Consider the figure lie low.

From, similarities of ΔADR and ΔRCB.

but,

From equation

But

AD=X-X1

RC=X2-X

RC=X2-X

For X

For

From equation

But

Then,

EXAMPLE.

1. Find the distance between A and B

2. Show that the point and are vertices of isosceles triangle.

3. Show that the point and are vertices of right angled triangles.

4. Show that the point and (0,1) are vertices of square.

Solution.

Given

** **

Since

Then the point A (1,2), B (3,5), C (4,4) are vertices of isosceles triangle.

Hence shown.

TRI – SECTION

-Is the process of dividing a certain line into three section or equal parts

** **

Where the coordinate P intersection depends on the two conditions.

i) When P is close to A

ii) When P is close to B

For P close to B.

RATIO THEOREM

– -Is the theory based on a division of a lines segment either internally or externally.

1. INTERNALLY DIVISION

Is a division of a lines segment internally under the given condition of ratio.

-Let line AB being divided at R in ratio M: N.

Where

Ratio M: N.

-Consider the figure below.

From similarities of ΔADR and ΔRCB.

II. EXTERNALLY DIVISION

-Is the theory based on a division of a lines segment externally under the given ratio.

Let,

A (X1,Y1), B (X2,Y2) and R (X, Y) under ration R (X,Y).

-Consider the graph below.

** **

EXAMPLES

1. Find the coordinate of point and divided internally or externally in the ration

2. Find the point of in-section of a line a joining, point and if P is closed of A.

3. Find the coordinate of in – section of a line AB at point P. If B is closer to A given that Aand B

SOLUTIONS

1. For internally division

GRADIENT

– Is the ration expressed as vertical change over horizontal change.

OR

– Is the ratio between change in Y over change in X.

Mathematically

Mathematically

gradient denoted as

i.e

However the gradient can explained by using three different methods.

i) . GRADIENT FROM ANGLE OF INCLINATION

ii). GRADIENT FROM THE CURVE (calculus method)

iii). GRADIENT BETWEEN TWO POINTS.

– Consider the figure below.

GRADIENT FROM ANGLE OF INCLINATION

-Let θ be angle of inclination

** **

GRADIENT FROM THE CURVE

This is explained by using calculus notation idea where;

of a curve at a given point.

However gradient can be obtained directly from the equation of a straight line as coefficient of x from the equation in form of

y =mx + c

This is explained by using calculus notation idea where;

of a curve at a given point.

However gradient can be obtained directly from the equation of a straight line as coefficient of x from the equation in form of

y =mx + c

BEHAVIOUR OF GRADIENT BETWEEN TWO POINT

-Lets two points be which can be used to form the line AB.

(i) If , and the line increase from left to right imply positive slope.

** **

(ii) If the line decrease from right to left imply slope.

** **

(iii) If the line is horizontally with zero gradient

(iv) If the line is vertical with infinity gradient

COLLINEAR POINTS

– – Are point which lie on the same straight line

Where,

A, B, and C are collinear.

– – The condition of collinear points have the same slope/ gradient

Note:

If A (); B and C are collinear then the area of = 0.

Example 1

1. 1.Determine the value of K such that following points are collinear:-

a) and

b) and

2. Show that the points , and are collinear.

3. The straight line Cut the curve at P and Q. Calculate the length PQ.

4. If A and B are products of OX and OY respectively. Show that xy=16. If the area of is 8 units square.

Solution:

A B, C

For collinear point

** **

2. Give

; B C

Alternatively

** **

Since the area of ΔABC is 0 unit hence the points are collinear.

3. Given

4. Given

Since

Area of ΔOAB = 8 square units

Then,

5. If and are the collinear of midpoint of the line forming the points and show that, x-y+1=0.

Solution

M.P = (x, y)

A and B

Required

** **

EQUATION OF THE LINE

– Mostly depends on different format those under one of the following

1. EQUATION OF A POINT AND SLOPE.

Let point be and slope be M.

-Let the two points being denoted as A and B

-Consider the figure below

For the line only if

=

Multiply by both side

=

3. EQUATION OF A SLOPE AND Y – INTERCEPT FORM

– -Consider the slope and – intercept,

** **

4. EQUATION OF 2 INTERCEPTS FORM

-Let consider X – intercept C, and Y intercept B

5. FAMILY EQUATION

– Is the equation formed from intersection of two lines passing through a certain points.

E.g. equation for intersection of L1 and L2 passing through points (a,b) can be obtained by using the formula.

Where K is constant

*

Important steps of determining family equation.

Important steps of determining family equation.

1. Solve for K by regarding equations of two lines and the point passing through.

2. Form family equation by using the value of K without regarding the point passing through.

Example.

1. Find the equation passing through the point (2,3) from the intersection to provided

Solution.

Point

But,

(

From

The equation is

ANGLE BETWEEN TWO LINES

-Let consider and where Q is angle made between and and is the angle of and is the angle of .

-Consider the figure below

Our intentions to find the value of which always should be acute angle.

where

** **

PARALLEL AND PERPENDICULAR LINES

(a) PARALLEL LINES

-Are the lines which never meet when they are produced

Means that is parallel to symbolically //

– -However condition for two or more lines to be parallel state that they posses the same gradient.

b) PERPENDICULAR LINES

– -Are the lines which intersect at right angle when they are produced.

Means that is perpendicular to

– -Symbolically is denoted as L1⊥L2

– -However the condition for two or more lines to be perpendicular states that “The product of their slopes should be equal to negative one”.

– -Let consider the figure below

NOTE:

1. The equation of the line parallel to the line = 0 passing through a certain point is of the form of . Where – is constant.

2. The equation of the line perpendicular to the line pass through a certain point is of the form of when – is constant.

3. The calculation of K above done by substitution certain point passing through.

2. The equation of the line perpendicular to the line pass through a certain point is of the form of when – is constant.

3. The calculation of K above done by substitution certain point passing through.

THE EQUATION OF PERPENDICULAR BI SECTOR

– Let two point be A and B.

Where,

Line L is perpendicular bisector between point A and B.

Now our intention is to find the equation of L.

IMPORTANT STEPS

1. Determine the midpoint between point A and B.

2. Since L and are ⊥ to each other then find slope of L.

for

for

3. Get equation of L as equation of perpendicular bisector of by using and mid point of A and B.

THE COORDINATE OF THE FOOT OF PERPENDICULAR FROM THE POINT

THE POINT TO THE LINE

– -Our intention is to find the coordinate of the foot (x,y) which act as the point if intersection of and .

– Let consider the figure below.

IMPORTANT STEPS

1. Get slope of formatted line i.e. and then use if to get slope of L2. Since

2. From equation of by using and point provided from.

3. Get coordinate of the food by solving the equation and simultaneously as the way Y please.

EXAMPLE

1. Find the acute angle 6. between lines

and

2. Find the acute angle between the lines represented by

3. find the equation of the line in which such that X – axis bisect the angle between the with line

4. find the equation of perpendicular bisector between A and B

5. Find the coordinate of the foot perpendicular from of the line

6. Find the equation of the line parallel to the line 3x – 2y + 7 = 0 and passing through the point

7. find the equation if the line perpendicular to the line and passing through the point

8. Find the equation of perpendicular bisector of AB. where A and B are the point and respectively.

Solution

Given

Consider

From

Also

Recall

=

θ2tan-1 1

Therefore;

2)

Solution

Given

Factorize completely

From

Recall

Given

Consider the figure below

From

The slope is negative then at x –axis y=0

4) Solution

Given A B

From

Also

Midpoint =

M.p =

Then

The equation is

Solution

Given

From

But

For the equation

The coordinate of the foot is

The perpendicular line from point A to the straight line intersect the line at point B. if the perpendicular is extended to C in such a way that AB = . Determine line coordinate of C.

Solution

Given

Let

From

Then

The coordinate of is since point B

Recall

Since

Then

For x

Compare off

∴

For y

The coordinate of C is

THE SHORTEST/PERPENDICULAR DISTANCE FROM THE POINT TO THE LINE

Introduction:

From the figure above it be loved that PC is the only one posses shortest distance rather than the other as perpendicular to the line that is way, our intention is to get shortest / perpendicular distance from the point P to the line.

Our intention is to find shortest distance from point to the line .

From

Since

From the trigonometric identity

Then

But

Also

Since the point R is on the line

Hence;

Substitute the value of into 1 above

Then

Apply square root both sides

=

= +

Hence

Hence

At the point

Note:

The distance from the origin to the line ax +by + c = 0 is given by;

The distance from the origin to the line ax +by + c = 0 is given by;

Example:

1. i) Find the perpendicular distance from point for the line

2. ii) Find the value of K if perpendicular distance from point for the line is units.

3. iii) Find the shortest distance from the origin to the line

4. If the shortest distance from the point to the line is 3 units. Find the value fm.

Solution

Given: point

Line

Recall:

The perpendicular distance Is 4 unit

Solution:

Given point

Required k

Then

THE EQUATION OF ANGLE BISECTOR BETWEEN TWO LINES

* Consider the figure below.

Where, PM and PN are perpendicular distance from point P. which are always equal.

Since

Then

=

NOTE;

i) for the equation take +ve

i ii) for the equation take –ve

THE CONCURRENT LINES

These are the lines which intersect at the same point.

Example:

– where and are concurrent line.

– However the point of intersection if concurrent line normally calculated under the following steps.

1. Select two equation of straight line which relate to each other from the those equation provided.

2. The get point of inter section of selected equation as usual. Points of intersection into the third equation in such a way that if the result of L.H.S is equal R.H.S imply that these line are currents lines.

Example;

i.Show that the lines

, and are current lines

, and are current lines

ii. Determine the value of M for which the lines, – 3 = 0 and are current.

iii. Find the equation of bisect of angle formed by the lines represented by pair of the following.

a) and

b) and

Solution:

1)Given

By solving since simultaneous equation

)

=

For the first equation take the it be cones

Then for the equation take cones from

=

=

The equations of base equation of the angle are

THE AREA OF TRIANGLE WITH THREE VERTICES

By geometrical method.

Consider the figure below.

Our intention is to find the area of

Now,

Area of = area of trapezium ABED area of trapezium ACED

But area of trapezium

Also consider, Area of trapezium ABED

Area of trapezium DCEF

Area of trapezium

Then

But simplification the formula becomes

If ABC has A (x1, y1), B (x2, y2) and C (x3, y3) for immediately calculation of area the following technique should be applied by regarding three vertices of as A (x1, y1), B (x2, y2) and C (x3, y3)

Area =

TERMINOLOGIES OF TRIANGLE

is the line which divide sides of triangle at two equal points

–

If ABC has A (X1, Y1), B (X2, Y2) and C (X3, Y3) for immediately calculation of area the following technique should be applied by regarding three vertices of as A (x1,y1), B (x2, y2) and C (x3, y3)

Area =

TERMINOLOGIES OF TRIANGLE

1.MEDIAN

1.MEDIAN

Median is the line which divide sides of triangle at two equal points

Where:

AQ, BR and CP are the median of

Also G is the centered of triangle

AQ, BR and CP are the median of

Also G is the centered of triangle

CENTROID FORMULA

Consider the figure below with verities A (x1, y1); B(x2, y2) and C (x3, y3)

Let line BM divide by point G in ration M: N = 2 : 1

Internally

Consider the figure below with verities A (x1, y1); B(x2, y2) and C (x3, y3)

Let line BM divide by point G in ration M: N = 2 : 1

Internally

The centroid formula is

=

2.

ALTITUDE OF TRIANGLE

Are the perpendicular drawn Y vertexes to the opposite side of triangles.

ALTITUDE OF TRIANGLE

Are the perpendicular drawn Y vertexes to the opposite side of triangles.

3. ORTHO CENTRE

Is the point of intersection of altitude of triangle.

Is the point of intersection of altitude of triangle.

4. CIRCUM CENTRE

Is the point of intersection of perpendicular bisector of the side of triangle.

Is the point of intersection of perpendicular bisector of the side of triangle.

EXAMPLE.

1. find the area of triangle with vertices A (0,2), B (3,5) and C (-1, 9)

2. Find the centered of the where A (1,1); B (3,2) and C (5,4)

3. From triangle ABC, A (2,1) B (6, -9) and C (4,11) find the equation of altitude through A.

2. Find the centered of the where A (1,1); B (3,2) and C (5,4)

3. From triangle ABC, A (2,1) B (6, -9) and C (4,11) find the equation of altitude through A.

Area =

Area of = 12 Square units

Area of = 12 Square units

Solution 2:

=

B(X2, Y2)=

C(X3, Y3)=

Recall

Solution :

=

B(X2, Y2)=

C(X3, Y3)=

Recall

Solution :

Given

A , B C

Required the equation of latitude consider the figure below

Since

MAQ.MBC = –1

MAQ =

But

MBC =

MBC =

MBC=

MBC =‾6

MBC=

MBC=

Y = + 1

Y = – + 1

Y= +

Y= +

− Locus is a free point which moves on x – y plane under a given condition.

However the locus can be described by using various properties as vertical line (X = ), H horizontal line (Y =B) straight line (Y = mx + C) on X – axis, (Y = C), on Y – axis as well as equation of the circle x2 + y2 + 2gx + 2fy + C = 0 or x2 + y2 = r2

However the locus can be described by using various properties as vertical line (X = ), H horizontal line (Y =B) straight line (Y = mx + C) on X – axis, (Y = C), on Y – axis as well as equation of the circle x2 + y2 + 2gx + 2fy + C = 0 or x2 + y2 = r2

Note:

The point of locus is Locus equal distance means equal distance.

EXAMPLE.

1. Find the locus of P (x, y) which is equal distance from A (2, 3) and B (4, 7).

2. A point P moves so that it perpendicular distance from a line 3x + 5y + 4 = 0. Is proper final to square of its distance from point Q (1, 2) if point (2,1) is

one possible position of P, prove that the equation of locus P is given by x2 + y2 – 8x – 14y -3 = 0

one possible position of P, prove that the equation of locus P is given by x2 + y2 – 8x – 14y -3 = 0

3. Show that A (7, -2) and B (2, 6) are all equal distance from the line 3x -4y – 4 = 0.

4. A point Q moves such that its distance from point (5, 3) is equal to twice. Its distance from the line X = 2. Find the equation of locus.

5. Find the locus of the point which moves so that the sum of the square of its distance from point (-2, 0) and (2, 0) is 26 units.

Solution 1:

1. Given

A

B

Then consider

=

=

=

= =

Square both sides

+=+ ²

Expand

x² -4x + 4 +y² -6y +9 = x² – 8x + 16+ y² – 14y + 49

4x + 8y – 52 =0

The locus is straight line

2.

2. Given that 3x + 5y + 4 =0

2. Given that 3x + 5y + 4 =0

P

Q

Point (2,-1) one possible position of P.

⊥ dPL

⊥dPl = IC²

⊥dPl= ————— (i)

Also

=

= +²

⊥dPl = K

= K

But =

= K

= K

= 10K

=

⊥dDL=K

= K

=

=

6x + 10y + 8=

6x + 10y +8 = x² – 2x + 1 + y² – 4y + 4

² + y² – 8y- 14y -3=0

² + y² – 8y – 14y -3 =0

² + y² – 8y – 14y -3=0

Solution 4:

Q

Solution 4:

Q

A

L: =2

2QA = QL

Solution 5:

Solution 5:

P

Points A& B (2, 0)