Two dimensional coordinate geomatry- Coordinate Geomatry - Math - Coordinate Geomatry

Subject: Matrix and Coordinate Geomatry

Subject Code: MATH 4201

Topics: Two dimensional coordinate geomatry

Lecturer Name : Mohammad Abu Jabed (MAJ)

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STRAIGHT LINES

We have read about lines, angles and rectilinear figures in geometry. Recall that a line is the join of two points in a plane continuing endlessly in both directions. We have also seen that graphs of linear equations, which came out to be straight lines

Interestingly, the reverse problem of the above is finding the equations of straight lines, under different conditions, in a plane. The analytical geometry, more commonly called coordinate geomatry, comes to our help in this regard. In this lesson. We shall find equations of a straight line in different forms and try to solve problems based on those.

OBJECTIVES

After studying this lesson, you will be able to :

·         derive equations of a line parallel to either of the coordinate axes;

·       derive equations in different forms (slope-intercept, point-slope, two point, intercept, parametric and perpendicular) of a line;

·         find the equation of a line in the above forms under given conditions;

·         state that the general equation of first degree represents a line;

·         express the general equation of a line into

(i) slope-intercept form (ii) intercept form and (iii) perpendicular form;

·       derive  the formula for the angle between two lines with given slopes;

·         find the angle between two lines with given slopes;

·         derive the conditions for parallelism and perpendicularity of two lines;

·         determine whether two given lines are parallel or perpendicular;

·         derive an expression for finding the distance of a given point from a given line;

·         calculate the distance of a given point from a given line;

·       derive the equation of a line passing through a given point and parallel/perpendicular to a given line;

·         write the equation of a line passing through a given point and:

(i) parallel or perpendicular to a given line (ii) with given x-intercept or y-intercept

(iii) passing through the point of intersection of two lines; and

·         prove various geometrical results using coordinate geometry.

EXPECTED BACKGROUND KNOWLEDGE

·         Congruence and similarity of traingles

STRAINGHT LINE PARALLEL TO AN AXIS

If you stand in a room with your arms stretched, we can have a line drawn on the floor parallel to one side. Another line perpendicular to this line can be drawn intersecting the first line between your legs.

In this situation the part of the line in front of you and going behind you is the y-axis and the one being parallel to your arms is the x-axis.

The direction part of the y-axis in front of you is positive and behind you is negative.

The direction of the part x-axis to your right is positive and to that to your left is negative. Now, let the side facing you be at b units away from you, then the equation of this edge will be

y = b (parallel to x-axis)

where b is equal in absolute value to the distance from the x-axis to the opposite side.

If b > 0, then the line lies in front of you, i.e., above the x-axis. If b < 0, then the line lies behind you, i.e., below the x-axis.

If b = 0, then the line passes through you and is the x-axis itself.

Again, let the side of the right of you is at c units apart from you, then the equation of this line will be x = c (parallel to y - axis)

where c is equal in absolute value, to the distance from the y-axis on your right.

If c > 0, then the line lies on the right of you, i.e., to the right of y-axis. If c < 0, then the line lies on the left of you, i.e., to the left of y-axis

If c = 0, then the line passes through you and is the y-axis.

Example: Find the equation of the lines passing through (2, 3) and is

(i) parallel to x-axis.   (ii) parallel to y-axis.

Solution :

(i)          The equation of any line parallel to x-axis is y = b

Since it passes through (2, 3), hence b = 3

\       The required equation of the line is y = 3

(ii)         The equation of any line parallel to y-axis is x = c

Since it passes through (2, 3), hence c = 2

\       The required equation of the line is x = 2.

Example: 
Find the equation of the line passing through (–2, –3) and
(i) parallel to x-axis    (ii) parallel to y-axis

Solution :

(i)          The equation of any line parallel to x-axis is y = b

Since it passes through (–2, –3), hence –3 = b

\       The required equation of the line is y = –3

(i)          The equation of any line parallel to y-axis is x = c

Since it passes through (–2, –3), hence –2 = c

\       The required cquation of the line is x = –2

CHECK YOUR PROGRESS 10.1

1.     If we fold and press the paper then what will the crease look like

2.      Find the slope of a line which makes an angle of  (a)  45° with the positive direction of x-axis. (b) 45° with the positive direction of y-axis. (c)45° with the negative direction of x-axis.

3.     Find the slope of a line joining the points (2, –3) and (3, 4).

4.     Determine x so that the slope of the line through the points (2,5) and (7, x) is 3.

5.     Find the equation of the line passing through (–3, –4) and

(a)         parallel to x-axis.          (b)       parallel to y-axis.

6.     Find the equation of a line passing through (5, –3) and perpendicular to    x-axis.

7.     Find the equation of the line passing through (–3, –7) and perpendicular to y-axis.

➤So far we have studied about the inclination, slope of a line and the lines parallel to the axes. Now the questions is, can we find a relationship between x and y, where (x, y) is any arbitrary point on the line?

Note 

• The relationship between x and y which is satisfied by the co-ordinates of arbitrary point on the line is called the equation of a straight line. The equation of the line can be found in various forms under the given conditions, such as

(a)    When we are given the slope of the line and its intercept on y-axis.

(b)    When we are given the slope of the line and it passes through a given point. 

(c)    When the line passes through two given points.

(d)    When we are given the intercepts on the axes by the line.

(e)    When we are given the length of perpendicular from origin on the line and the angle which the perpendicualr makes with the positive direction of x-axis.

(f)     When the line passes through a given point making an angle a with the positive direction of x-axis. (Parametric form).

We will discuss all the above cases one by one and try to find the equation of line in its standard forms.

(A)   SLOPE-INTECEPT FORM

Let AB be a straight line making an angle q with x-axis and cutting off an intercept OD =

c from OY.

As the line makes intercept OD = c on y-axis, it is called y-intercept. Let AB intersect OX' at T.

Take any point P(x, y) on AB. Draw PM ^ OX.

The OM = x, MP = y.

Draw DN ^ MP.

•  From the right-angled triangle DNP, we have

\   y = x tan q + c

tan q = m (slope)

\   y = mx + c

Since, this equation is true for every point on AB, and clearly for no other point in the plane, hence it represents the equation of the line AB.

Note : (1)  When c = 0 and m ¹ 0 Þ the line passes through the origin and its equation is   y = mx

(2) When c = 0 and m = 0 Þ the line coincides with x – axis and its equation is of the

(3) When c ¹ 0 and m = 0 Þ the line is parallel to x-axis and its equation is of the form

           y = c

Example 10.3

Find the equation of a line with slope 4 and y-intercept 0.

Solution : 

Putting m = 4 and c = 0 in the slope intercept form of the equation, we get y = 4 x

This is the desired equation of the line

Example 10.44   

Determine the slope and the y-intercept of the line whose equation is

8x + 3y = 5

Solution : 

The given equation of the line is 8x + 3y = 5

Example 10.5  

Find the equation of the line cutting off an intercept of length 2 from the negative direction of the axis of y and making an angle of 1200 with the positive direction x-axis

Solution : From the slope intercept form of the line

                   \ y = x tan 120° + (–2)

Here m = tan 120°, and c = –2, because the intercept is cut on the negative side of y -axis.

(b) POINT-SLOPE FORM

Here we will find the equation of a line passing through a given point A(x , y ) and having the slope m.

Let P(x, y) be any point other than A on given the

     

Note : Since, the slope m is undefined for lines parallel to y-axis, the point-slope form of the equation will not give the equation of a line though A (x , y ) parallel to y-axis. However, this presents no difficulty, since for any such line the abscissa of any point on the line is x . Therefore, the equation of such a line is x = x .

Example 10.6  

Determine the equation of the line passing through the point (2,– 1) and having (2/3)

 

slope

Example 10.7  

Find the equation of the line passing through (3, – 7) and (– 2,– 5).

 Solution :  

The equation of a line passing through two points (x1, y1) and (x2, y2) is given by
 

 










(d) INTERCEPT FORM
We want to find the equation of a line which
cuts off given intercepts on both the
co-ordinate axes.              Let PQ be a line meeting x-axis in A and    y-axis in B. Let OA = a,     OB = b.
Then the co-ordinates of A and B are
(a,0) and (0, b,) respectively.


The equation of the line joiningAand B is

This is the required equation of the line having interceptsa and b on the axes.


Example 10.8 

 
Find the equation of a line which cuts off intercepts 5 and –3 on x and y axes respectively.

 Solution : 

The intercepts are 5 and –3 on x and y axes respectively. i.e., a = 5, b = – 3
The required equation of the line is

Example 10.9  

Find the equation of a line which passes through the point (3, 4) and makes
intercepts on the axes equl in magnitude but opposite in sign.

 Solution : 

Let the x-intercept and y-intercept be a and –a respectively

∴ The equation of the line is

Example 10.10  

Determine the equation of the line through the point (– 1,1) and parallel to x - axis. 

Solution :  Since the line is parallel to x-axis its slope ia zero. Therefore from the point slope
form of the equation, we get
y – 1 = 0 [ x – (– 1)]
y – 1 = 0
which is the required equation of the given line

Example 10.11 
Find the intercepts made by the line
3x –2y + 12 = 0 on the coordinate axes

Solution :  
Equation of the given line is
3x – 2y = – 12.
Dividing by – 12, we get 

Comparing it with the standard equation of the line in intercept form, we find a = –4 and 
b =6. Hence the intercepts on the x-axis and y-axis repectively are –4. and 6.

Example 10.12 

The segment of a line, intercepted between the coordinate axes is
bisected at the point (x1, y1). Find the equation of the line

Solution : 

Let P(x1, y1) be the middle point or the
segmentCD of the lineAB intercepted between the
axes. Draw PM ⊥ OX
∴ OM = x1 and MP = y1
∴ OC = 2x1 and OD = 2y1
 
Now, from the intercept form of the line









(e) PERPENDICULAR FORM (NORMAL FORM)
We now derive the equation of a line when p be the length of perpendicular from the origin on the line and a, the angle which this perpendicular makes with the positive direction of
x-a x is are given.

(i) LetA B be the given line cutting off interceptsa and b on x–axis and y –axis respectively.
Let OP be perpendicular from origin O on A B and ∠ POB = a (See Fig. 10.6 (i))

Note : 1. p is the length of perpendicular from the origin on the line and in always
taken to be positive.
2. a is the angle between positive direction of x-axis and the line perpendicularfrom the origin to the given line.

Example 10.13 

Determine the equation of the line with a = 135° and perpendicular distance p = √2 from the origin.

Solution :

From the standard equation of the line in normal form have
x cos 135° + y sin 135° = 2

 








Example 10.14 

Find the equation of the line whose perpendicular distance from the origin is
6 units and the perpendicular from the origin to line makes an angle of 30° with the positive
direction of x-axis.
 
Solution : 

Here, a= 30°, p = 6
∴ The equation of the line is
x cos 30° + y sin 30° = 6

 






(f) PARAMETRIC FORM
We now want to find the equation of a line
through a given point Q (x1, y1) which makes
an angle a with the positive direction of
x – axis in the form 
 




 where r is the distance of any point P (x, y) on the line from Q (x1, y1).
Let AB be a line passing through a given pointQ (x1, y1) making an angle a with the positive direction of x-axis.

Let P (x, y) be any point on the line such that QP = r (say)
Draw QL, PM perpendiculars to x-axis and draw QN ⊥ PM.
From right angled ∆ PNQ

Note : 
1. The co-ordinates of any point on this line can be written as
(x1+ r cos a , y1+ r sin a ). Clearly coordinates of the point depend on the value ofr. This variable r is called parameter. 
2. The equation x = x1+ r cos a , y = y1+ r sin a are called the parametric equations of the line. 
3. The value of 'r' is positive for all points lying on one side of the given point and
negative for all points lying on the other side of the given point.

Example 10.15 

Find the equation of a line passing through A (–1, –2) and making an angle
of 30° with the positive direction ofx-axis, in the parametric form. Also find the coordinates of a point P on it at a distance of 2 units from the point A.
Solution : Here x1= –1, y1= – 2 and a = 30°

∴ The equation of the line is

 Example 10.16 

Find the distance of the point (1, 2) from the line 2x – 3y + 9 = 0
measured along a line making an angle of 45° with  x – axis.
 
Solution : The equation of any line through A (1, 2) making an angle of 45° with x-axis is

 CHECK YOUR PROGRESS 10.2
1. 

• (a) Find the equation of a line with slope 2 and y – intercept is equal to –2. 
• (b) Determine the slope and the intercepts made by the line on the axes whose equation  is 4x + 3y = 6.
  

2. Find the equation of the line cutting off an interecept 1/3 on negative direction of axis of
y and inclined at 120° to the positive direction of x-axis.
 
3. Find the slope and y-intercept of the line whose equation is 3x – 6y = 12.
 
4. Determine the equation of the line passing through the point (–7, 4) and having the slope -3/7
 
5. Determine the equation of the line passing through the point (1, 2) which makes equal
angles with the two axes.
 
6. Find the equation of the line passing through (2, 3) and parallel to the line joining the
points (2, –2) and (6, 4).
 
7. 

• (a) Determine the equation of the line through (3, –4) and (–4, 3). 
• (b) Find the equation of the diagonals of the rectangle ABCD whose vertices are
  A (3, 2), B (11, 8), C (8, 12) and D (0, 6).
  

8. Find the equation of the medians of a triangle whose vertices are (2, 0), (0, 2) and (4, 6).
 
9. Find the equation of the line which cuts off intercepts of length 3 units and 2 units on
x-axis and y-axis respectively.
 
10. Find the equation of a line such that the segment between the coordinate axes has its mid
point at the point (1, 3)
 
11. Find the equation of a line which passes through the point (3, –2) and cuts off positive
intercepts on x and y axes in the ratio of 4 : 3.
 
12. Determine the equation of the line whose perpendicular from the origin is of length 2 units and makes an angle of 45° with the positive direction of x-axis.
 
13. If p is the length of the perpendicular segment from the origin, on the line whose intercept on the axes are a and b, then show that

14. Find the equation of a line passing through A (2, 1) and making an angle 45° with the
positive direction of x-axis in parametric form. Also find the coordinates of a pointPon
it at a distance of 1 unit from the point A.

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