Straight lines
. Line
through two points
The line through two distinct points (x1, y1) and (x2,
y2) is given by
(1)
𝑦2− 𝑦1
y = y1 + [
]·(x - x1),
𝑥2− 𝑥1
where x1 and x2 are assumed to be different. In case
they are equal, the equation is simplified to
x = x1
and does not require a second point.
Equation (1) can also be written as
𝑦2− 𝑦1
y - y1 = [
]·(x - x1),
𝑥2− 𝑥1
or even as
(x2 - x1)·(y - y1) = (y2 - y1)·(x - x1),
where one does not have to worry whether x1 = x2 or
not. However, the simplest for me to remember is this
(y - y1)/(y2 - y1) = (x - x1)/(x2 - x1)
which is not as universal is the one before.
Intercept-intercept
Assume a straight line intersects x-axis at (a, 0) and yaxis at (0, b). Then it is defined by the equation
x/a + y/b = 1,
which also can be written as
xb + ya = ab.
The latter form is somewhat more general as it allows
either a or b to be 0. a and b are defined as xintercept and y-intercept of the linear function. These
are signed distances from the points of intersection of
the line with the axes.
Point-slope
The equation of a straight line through point (a, b) with
a given slope of m is
y = m(x - a) + b, or y - b = m(x - a).
As a particular case, we have
Slope-intercept equation
The equation of a line with a given slope m and the yintercept b is
y = mx + b.
This is obtained from the point-slope equation by
setting a = 0. It must be understood that
the point-slope equation can be written for any point
on the line, meaning that the equation in this
form is not unique. The slope-intercept equation is
unique because if the uniqueness for the line of
the two parameters: slope and y-intercept.
.
POINT-SLOPE FORM
Suppose that we want to find the equation of a straight line
that passes through a known point
and has a known slope. Let (x,y) represent the coordinates
of any point on the line, and let (x,,y,)
represent the coordinates of the known point. The slope is
represented by m.
Recalling the formula defining slope in terms of the
coordinates of two points, we have
EXAMPLE: Write the equation of a line parallel to 3x - y
- 2 = 0 and passing through the point (5,2).
SOLUTION: The coefficients of x and y in the desired
equation are the same as those in the given equation.
Therefore, the equation is
3x-y+D=0.
EXAMPLE: Find the equation of a line passing through
the point (2,3) and having a slope of 3.
SOLUTION:
The point-slope form may be used to find the equation of
a line through two known points.
The values of x 1 , x 2 , y 1 , and Y2 are first used to find the
slope of the line; then either known
point is used with the slope in the point-slope form.
EXAMPLE: Find the equation of the line through the
points ( - 3,4) and (4, - 2).
SOLUTION:
Letting (x,y) represent any point on the line and using ( 3,4) as
Using (4, - 2) as the known point will also give 7y + 6x =
10 as the linear equation.
SLOPE-INTERCEPT FORM
Any line that is not parallel to the Y axis intersects the Y
axis at some point.
The x coordinate of the point of intersection is 0, because
the Y axis is vertical
and passes through the origin. Let the y coordinate of the
point of intersection be
represented by b. Then the point of intersection is (0,b), as
shown in figure. The y coordinate, b, is called the
y intercept.
The slope of the line in figure is
The value of y in this expression is y - b, where y
represents the y coordinate of any point on the line. The
value of
x is x - 0 = x, so
Slope-intercept form.
This is the standard slope-intercept form of a straight line.
EXAMPLE: Find the equation of a line that intersects the
Y- axis at the point (0,3) and has a slope of 5/3.
SOLUTION:
PRACTICE PROBLEMS:
Write equations for lines having points and slopes as follows:
General equation
A straight line is defined by a linear equation whose
general form is
Ax + By + C = 0,
where A, B are not both 0.
Different forms of General form:
(i) If B ≠0 , then y =
it is slope
intercept form where m = -A/B and y-intercept = -C/B
If B=0 , which is a vertical line whose slope is
undefined and x-intercept is -C/A.
(ii) If C ≠0, then x/a+y/b = 1 which is intercept form
where a = x-intercept = -C/A , b = y-intercept= -C/B
If C =0 , then it becomes a line passing through
origin, has zero intercepts on the axes.
NORMAL FORM
Methods for determining the equation of a line usually depend
upon some knowledge
of a point or points on the line. Let's now consider a method
that does not require advance
knowledge concerning any of the line's points. All that is
known about the line is its perpendicular distance
from the origin and the angle between the perpendicular and
the X axis, where the angle is
measured counterclockwise from the positive side of the X
axis.
line AB is a distance p away from the origin, and line OM
forms an angle Ѳ
(the Greek letter theta) with the X- axis. We select any point
P(x,y) on line AB and develop the
Normal form.
Equation of line AB in terms of the x and y of P. Since P
represents any point on the line,
the x and y of the equation will represent every point on the
line and therefore will represent the line itself.
PR is constructed perpendicular to OB at point R. NR is
drawn parallel to AB and PN is parallel to OB.
PS is perpendicular to NR and to AB. A right angle is formed
by angles NRO and PRN.
Triangles ONR and OMB are similar right triangles.
Therefore, angles NRO and MBO are equal
and are designated as Ѳ’. Since Ѳ + Ѳ’ = 90' in
triangle OMB and angle NRO is equal to Ѳ’,
then angle PRN equals Ѳ. Finally, the x distance of point P is
equal to OR, and the y distance of P is equal to PR.
To relate the distance p to x and y, we reason as follows:
This final equation is the normal form. The word "normal" in
this usage refers to the perpendicular relationship
between OM and AB. "Normal" frequently means
"perpendicular" in mathematical and scientific usage.
The distance p is always considered to be positive, and Ѳ is
any angle between 0' and 360'.
EXAMPLE: Find the equation of the line that is 5 units away
from the origin, if the perpendicular
from the line to the origin forms an angle of 30' from the
positive side of the X- axis.
SOLUTION.
DISTANCE FROM A POINT TO A LINE
We must often express the distance from a point to a line in
terms of the coefficients in
the equation of the line. To do this, we compare the two forms
of the equation of a straight line, as follows:
General equation: Ax + By + C = 0
Normal form:
The general equation and the normal form represent the same
straight line.
Therefore, A (the coefficient of x in the general form) is
proportional to cos Ѳ (the coefficient of x in the normal form).
By similar reasoning, B is proportional to sin 0, and C is
proportional to -p. Recalling
that quantities proportional to each other form ratios
involving a constant of proportionality,
let k be this constant. Thus, we have
Squaring both sides of these two expressions and then
adding, we have
Y
OR
WE can prove perp. distance from appoint p(x1,y1) on
A line Ax+By+C=0 is
Y
R (0, -C/B)
Ax+By+C=0
P( x1,y1)
( -C/A ,0) Q
X
The coefficients in the normal form, expressed in terms of A,
B, and C, are as follows:
The sign of
is chosen so as to make p (a
distance) always positive.
The conversion formulas developed in the foregoing
discussion are used in finding
the distance from a point to a line. Let p represent the
distance of line LK from the origin.
To find d, the distance from point P, to line LK, we
construct a line
Distance from a point to a line.
through P, parallel to LK. The distance of this line from the
origin is OS, and the difference between OS and p is d.
We obtain an expression for d, based on the coordinates of P1,
as follows:
and
Returning to the expressions for sin Ѳ , cos Ѳ, and p in terms
of A, B, and C (the coefficients in the general equation), we
have
In the formula for d, the denominator in each of the
expressions is the same. Therefore, we may combine terms as
follows:
We use the absolute value, since d is a distance, and thus avoid
any confusion arising from the ± radical.
Note that the absolute value,| | of a number is defined as
follows:
and
That is, for the positive number 2,
|2|=2
For the negative number -2,
The absolute value of
is
EXAMPLE: Find the distance from the point (2,1) to the
line 4x+2y+7=0.
SOLUTION:
PRACTICE PROBLEMS:
In each of the following problems point to the line:
find the distance from the
1. (5,2), 3x - y + 6 = 0
2. (-2,5), 3x + 4y - 9 = 0
ANSWERS:
PARALLEL AND PERPENDICULAR
LINES ( GENERAL FORM)
The general equation of a straight line is often written with
capital letters for coefficients, as follows:
Ax+By+C=0
These literal coefficients, as they are called, represent the
numerical coefficients encountered in a typical linear
equation.
Suppose we are given two equations that are duplicates
except for the constant term, as follows:
Ax +By+C=0
Ax +By+D=0
By placing these two equations in slope-intercept form, we
can show that their slopes are equal, as follows:
y=
Thus, the slope of each line is -A/B.
Since lines having equal slopes are parallel, we reach the
following conclusion:
In any two linear equations, if the coefficients of the
x and y terms are
identical in value and sign, then the lines
represented by these equations are parallel.
Since the line passes through (5,2), the values x = 5
and y = 2 must satisfy the equation. Substituting these,
we have
3(5) - (2) + D = 0
D= -13
Thus, the required equation is
3x – y - 13 =0
A situation similar to that prevailing with parallel lines
involves perpendicular lines. For example, consider
the equations
Ax +By+C=0
Bx - Ay+D=0
Transposing these equations into the slope-intercept
form, we have
y=
Since the slopes of these two lines are negative
reciprocals, the lines are perpendicular.
The conclusion derived from the foregoing discussion
is as follows:
If a line is to be perpendicular to a given line, the
coefficients
of x and y in the required equation are found by
interchanging
the coefficients of x and y in the given equation
and changing the sign of one of them.
EXAMPLE: Write the equation of a line
perpendicular to the line x + 3y + 3 = 0 and having a y
intercept of 5.
SOLUTION: The required equation is
3x-y+D=0
Notice the interchange of coefficients and the change
of sign. At the point where the line crosses the Y axis,
the value of x is 0 and the value of y is 5. Therefore,
the equation is
3(0) - (5) + D = 0 D=5
The required equation is 3x - y+5=0
PRACTICE PROBLEMS:
1. Find the equation of the line whose perpendicular
forms an angle of 135' from the positive side of the X
axis and whose perpendicular distance is V-2-units
from the origin.
Find the equations of the following lines:
2. Through (1,1) and parallel to 5x - 3y = 9.
3. Through (- 3,2) and perpendicular to x + y = 5.
ANSWERS:
SUMMARY (STRAIGHT LINES)
The following are the major topics covered in this chapter:
1. Distance between two points:
where (x,,y,) and (x2,y2) are given points on a line.
2. Division of a line segment:
where k is the desired proportion of the distance between points
(x1,y1) and(x 2 ,y 2 ) and (x,y) is the desired point.
3. Midpoint of a line segment:
4. Inclination: The angle of inclination is the angle the line crossing
the X- axis makes
with the positively directed portion of the X axis, such that 0 ° < α <
180'.
5. Slope:
The slope of a horizontal line is zero.
The slope of a vertical line is meaningless.
6. Slopes of parallel lines: Slopes are equal or
where m1 and m2 are the slopes of the lines L, and L 2, respectively.
7. Slopes of perpendicular lines: Slopes are negative reciprocals or
8. Acute angle between two lines:
( arctan means tan-1 )
However, if one line, L 2 , is parallel to the Y axis and the other, L1, has a
positive slope, then
If L 2 is parallel to the Y axis and L1, has a negative slope, then
9. Obtuse angle between two lines:
where + is the acute angle between the two lines.
10. Point-slope form of a straight line:
11. Slope-intercept form of a straight line:
y = mx + b
where b is the y- intercept.
12. Normal form of a straight line:
where p is the line's perpendicular distance from the origin and Ѳ is the
angle
between the perpendicular and the X- axis.
13. Parallel lines:
In any two linear equations, if the coefficients of the x and y terms are
identical in value and sign,
then the lines represented by these equations are parallel; that is,
Ax+By+C=0 and Ax+By+D=0
are parallel lines.
Perp. distant b/w two ii lines.
14. Perpendicular lines:
If a line is to be perpendicular to a given line, the coefficients of x and y
in the
required equations are found by interchanging the coefficients of x and y
in the given
equation and changing the sign of one of them; that is,
Ax+By+C=0 and Bx-Ay+D=0
are perpendicular lines.
15. Distance from a point to a line:
where A, B, and C are the coefficients of the general equation of a line Ax
+ By + C = 0 and
(x,,y,) are the coordinates of the point.
SYMMETRIC FORM AND PARAMETRIC EQUATIONS OF A
LINE
The equation of the straight line passing through (x1,y1) and making
an angle Ѳ wih the positive direction of x-axis is
𝒙− 𝒙𝟏
𝒚− 𝒚
𝟏
=
= r , where r is the distant of the point (x,y) on the line
𝒄𝒐𝒔𝜽
𝒔𝒊𝒏𝜽
from the point (x1,y1).
REMARK: If P(x,y) be a point at a distant of r units from a given
point Q( x1,y1), then x = x1 + r cosѲ and y = y1 + rsinѲ
EXAMPLE: If the straight line drawn through the point P(√3 ,2) and
making an angle п/6 with the x-axis meets the line
√3 x – 4y + 8=0 at Q, find the length of PQ.
SOLUTION: Let the line through P making an angle п/6 with the x-axis
meets the line √3 x – 4y + 8=0 at Q.
By above formula Q = (√3 + r cosп/6 , 2 + r sinп/6) lies on
equation ⇨ r =6.
ADDITIONAL PRACTICE PROBLEMS
1. Find the distance between P1 (- 3, - 2) and P2(-7,I).
2. Find the distance between P1 (- 3/4, - 2) and P2(1, - 1/2). 3.
Find the coordinates of a point 115 of the way from P1(- 2,0) to P2(3, 5).
4. Find the midpoint of the line between P1(-8/3,4/5) and P2(- 4/3,6/5).
5. Find the slope of the line joining P1(4,6) and P2(-4,6).
6. Find the slope of the line parallel to the line joining P1(7,4) and
P2(4,7).
7. Find the slope of the line perpendicular to the line joining P1(8,1) and
P2(2,4).
8. Find the obtuse angle between the two lines which have m, = 7 and
m2 = - 3 for slopes.
9. Find the obtuse angle between the Y axis and a line with a slope of m
= -1/4.
10. Find the equation of the line through the points ( - 6,5) and (6,5).
11. Find the equation of the line whose y intercept is (0,0) and whose
slope is 4.
12. Find the slope and y intercept of the line whose equation is
4y+8x=7.
13. Find the equation of the line that is 3/2 units away from the origin,
if the perpendicular
from the line to the origin forms an angle of 210 0 from the positive side
of the X axis.
14. Find the equation of the line through (2,3) and perpendicular to 3x2y=7.
15. Find the equation of the line through (2,3) and parallel to 3x-2y=7.
16. Find the distance from the point (3, - 5) to the line 2x+y+4=0.
17. Find the distance from the point (3, - 4) to the line 4x +3y= 10
ANSWERS
.
ASSIGNMENT
Question 1 In what ratio is the line joining the points A(4,4)
and B(7,7) divide by P(-1,-1)?
[Hint: use section formula after assuming ratio k:1 , k= -5/8]
Question 2 Determine the ratio in which the line 3x+y – 9 =
0 divides the segment joining the points (1,3) and (2,7).
[ Hint: use section formula , k=3/4]
**Question 3 The area of a triangle is 5. Two of its vertices
are (2,1) and (3,-2). Third vertex is (x,y) where y = x+3. Find
The co-ordinates of the third vertex.
[Let the vertices are A(x,y), B(2,1), C(3,-2) , Area of triangle
ABC = ½ |3x+y – 7|=5 , in case (i) x=7/2,y=13/2
In case (ii) x=-3/2 , y=3/2]
Question 4 Find the equation of the straight lines which pass
through the origin and trisect
the intercept of line 3x+4y=12 b/w the axes.
[ Hint: Let the line AB be trisected at P and Q, then AP : PB =
1:2 , A(4,0) , B(0,3) BY using section formula we get P(8/3 , 1)
AQ : QB = 2 : 1 ⇨ Q=(4/3 ,2) then equation of line OP and
OQ passing through (0,0) is 3x – 8y =0 and 3x – 2y =0 resp.]
Question 5 If the straight line drawn through the point P(2
,3) and making an angleп/4 with the x-axis meets the line
x + y + 1=0 at Q, find the length of PQ. [ answer is r =
3√𝟐 ]
** Question 6 Find the distant of the point (2,5) from the line
3x+4y+4=0 measured parallel to a line having slope ¾.
[Hint: tanѲ= ¾ ⇨ sinѲ =3/5 , cosѲ=4/5 , equation passing
through A(2,5) by symmetric form is
X= 2+(4/5)r , y= 5+(3/5)r they lie on a given line ⇨ r = 5
Question 7 The line segment joining A(2,3), B(-3,5) is
extended through each end by a length equal to its original
length.
Find the co-ordinates of the new ends.
[ Hint: answer is x= 7, y= 1 and α = -8 , 𝜷 = 7
(𝜶 , 𝜷)
C
(-3,5)
(2,3)
B
A
(x,y)
D ]
Question 8 The line segment joining A (6,3) to B (-1,-4) is
doubled in length by having added to each end. Find the coordinates
Of the new ends.
[ Hint : given CB = ½ BA , AD = ½ BA ∴ B divides CA
internally in 1:2 and A divides BD internally in 2:1
C( -9/2 ,-15/2) , D( 19/2 , 13/2)]
Question 9 Two opposite vertices of a square are (3,4) and
(1,-1). Find the co-ordinates of other vertices.
[ Let A(3,4) , C(1,-1) then slope of AC = 5/2 , M mid
point of AC & BD =(2, 3/2)
Let m is the slope of a line making an angle of 450 with
AC i.e., m is slope of lines AD or CD.
USE formula of angle b/w two lines ⇨ m= -7/3, 3/7
then equations of AD , CD are
7x+3y – 33=0 , 3x – 7y – 10 =0 and D(9/2 ,1/2) ,M is
mid point of BD ⇨ B(-1/2 , 5/2)]
Question 10 (i) Find the co-ordinates of the orthocenter of
the ∆ whose angular points are (1,2), (2,3), (4,3).
(ii) Find the co-ordinates of the circumcenter
of the ∆ whose angular points are (1,2), (3,-4), (5,-6).
[ answer (i) (1,6) (ii) (11,2)]
Question 11 Find the equation of the line through the
intersection of the lines x -3y+1=0 and 2x+5y -9=0 and whose
distant from the origin is √𝟓 .
[ Hint: (x -3y+1) +k(2x+5y -9)=0 -------(1) , then find
distant from (0,0) on the line (1) is √𝟓 ⇨ k=7/8 ,put in (1)
Answer is 2x+y – 5=0]
Question 12 The points (1,3) and (5,1) are the opposite
vertices of a rectangle.The other two vertices lie on the line y
= 2x+c. Find c and the remaining vertices.
[Hint: Use mid point formula , it lies on BD ∴ c = -4 , use
M(3,2) (by mid point
D(𝜶, 𝜷)
C(5,1)
y=2x+c
M (3,2)
A(1,3)
(AB)2 + (BC)2 = (AC)2
B(2,0) then D(4,4)]
B(X,2X-4)
⇨ x=4 or 2 ∴ B(4,4) then D (2,0), if
Question 13 The consecutive sides of a parallelogram are
4x+5y=0 and 7x+2y=0. If the equation of one diagonal be
11x+7y=9, find the equation of other diagonal .
[ Hint:
D
11x+7y=9
C
P
7x+2y=0
O
4x+5y=0
B
B(5/3,-4/3) , D(-2/3,7/3) by solving equations of OB & BD and
OD & BD resp.
Then find point P (1/2,1/2) & equation of OC i.e, OP is
y=x.]
Question 14 One side of a rectangle lies along the line
4x+7y+5=0. Two of vertices are (-3,1) & (1,1).
Find the equation of other three sides.
[ Hint:
D
slope=-4/7
C(1,1)
Slope=7/4
A(-3,1)
slope=7/4
4x+7y+5=0
B
Equation of BC is 7x – 4y -3=0 , equation of AD & CD are
7x – 4y +25=0 & 4x+7y=11=0 resp.]
Question 15 The extremities of the base of an isosceles ∆
are the points (2a,0) & (0,a). The equation of the
one of the sides is x=2a. Find the equation of
the other two sides and the area of the ∆.
[ Hint:
y
C
B(0,a)
x+2y-2a=0
o
A(2a,0)
X
by solving CA2 = CB2 ⇨ Y=(5a)/2 i.e, C is (2a,5a/2),
equation of BC is 3x – 4y+4a=0 & area of ∆ACB is 5a2/2
sq.units.]
Question 16 One side of a square is inclined to x-axis at an
angle α and one of its extremities is at origin.
If the sides of the square is 4, find the equations of
the diagonals of the square.
[ Hint: Take ∆OLA , find A as OL/4=cosα & AL/4=sinα
and in ∆OMC ,find point C, then find equation of OB & AC
( by using mid pointof OB & AC) equations of OB & AC are
x(cosα+sinα) – y(cosα – sinα)=0 ,
x(cosα - sinα) + y(cosα + sinα)=4 resp
Y
B(h,k)
(-4sinα, 4cosα)
C
4
M
4
O
A(4cosα, 4sinα)
L
Angle COM=900-α , angle AOL=α ]
Question 17 Prove that the diagonals of the //gm. Formed
by the four lines.
x/a + y/b = 1 ……(i) , x/b + y/a = 1 …….(ii) , x/a +
y/b = -1 ……(iii) , x/b + y/a = -1 ……..(iv) are perp. to each
other.
[HINT:
(
−𝒂𝒃 −𝒂𝒃
,
)D
𝒂+𝒃 𝒂+𝒃
line (iii)
𝒂𝒃
−𝒂𝒃
C ( 𝒂−𝒃 , 𝒂−𝒃 )
Slope=1
Line(iv)
Slope=-1
−𝒂𝒃
(
𝒂−𝒃
,
𝒂𝒃
𝒂−𝒃
)A
line(ii)
B(
line(i)
𝒂𝒃
𝒂+𝒃
,
𝒂𝒃
𝒂+𝒃
)
Find all co-ordinates & for perpendicularity of diagonals show
product of slopes of AC & BD = -1× 𝟏 = −𝟏]
Question 18 On the portion of the line x+3y – 3 =0 which is
intercepted b/w the co-ordinates axes, a square is
constructed on the side of the line away from the origin.
Find the co-ordinates of the intersection of its diagonals.
Also find the equations of its sides.
[HINT: P is the mid point of BD
Y
C
D(4,3)
(0,1)B
45
P(2,2)
X+3y-3=0
A(3,0)
angle ABD=450 , use formula tan 450 =|
𝟑𝒎+𝟏
𝟑−𝒎
|
⇨ m =1/2 or -2 , equations of BD ,AC, CD AD & BC are x2y+2=0, 2x+y-6=0, x+3y-13=0, 3x-y-9=0 & 3x-y+1=0 resp.]
Question 19 The hypotenuse of a right angled ∆ has its ends
at the points (1,3) and (-4,1).
Find the equation of the legs (perpendicular sides) of the
triangle.
[ Hint: The legs (perpendicular sides) of the triangle , we
can consider as \\ to x-axis & y-axis resp. line \\ to x-axis is
y=k
It passes through (-4,1) ⇨ k=1 ∴ y=1 similarily x=1 as x=k
passes through (1,3)]
Question 20 If one diagonal of a square is along the line 8x15y=0 and one of its vertex is at (1,2), then find the equation
of sides of the square passing through this vertex.
[ Hint: use formula tan 450 =
𝟖
𝟏𝟓
𝟖
𝟏+𝒎𝟏
𝟏𝟓
𝒎𝟏 −
⇨ m1 = 23/7 ,
(1,2) A
m1
B
8x-15y=0
450
D
m2 =8/15
C
Equations of AD & AB ( Perp. to each other) are 23x7y-9=0 , 7x+23y-53=0]
Question 21 Find the reflection of (4,-13) about the line
5x+y+6=0.
[Hint: use mid point formula & concept of product of
slopes of perp. lines , then answer is (-1,-14)]
General equation
A straight line is defined by a linear equation whose
general form is
Ax + By + C = 0,
where A, B are not both 0.
The coefficients A and B in the general equation are the
components of vector
n = (A, B) normal to the line. The pair r = (x, y) can be
looked at in two ways: as a point or as a
radius-vector joining the origin to that point. The latter
interpretation shows that a straight line
is the locus of points r with the property
r·n = const.
That is a straight line is a locus of points whose radiusvector has a fixed scalar product
with a given vector n, normal to the line. To see why
the line is normal to n, take two distinct
but otherwise arbitrary points r1 and r2 on the line, so
that
r1·n = r2·n.
But then we conclude that
(r1 - r2)·n = 0.
In other words the vector r1 - r2 that joins the two
points and thus lies on the line is perpendicular to n.
Normalized equation
The norm ||n|| of a vector n = (A, B) is defined
via ||n||2 = A2 + B2 and has the property that, for any
non-trivial vector n, n/||n|| is a unit vector,
i.e., || n/||n|| || = 1.
Note that the line defined by a general equation would
not change if the equation
were to be multiplied by a non-zero coefficient. This
property can be used to keep
the coefficient A non-negative. It can also be used to
normalize the equation by dividing it by ||n||.
As a result, in a normalized equation
Ax + By + C = 0,
A2 + B2 = 1.
(In the applet, the coefficients of the normalized
equation are rounded to up to 6 digits, for which reason
the above identity may only hold approximately.)
The normalized equation is conveniently used in
determining the distance from a point to a line
Parametric equation
A line through point r0 = (a, b) parallel to vector u = (u,
v) is given by
(x, y) = (a, b) + t·(u, v),
where t is any real number. In the vector form, we have
r = r0 + t·u,
where r = (x, y).
Implicit equation
A line through point r0 = (a, b) perpendicular to
vector n = (m, n) is given by
m(x - a) + n(y - b) = 0,
or if we take r = (x, y), a generic point on the line, we
see that
n·(r - r0) = 0,
where dots indicates the scalar product of two vectors.
A function whose graph is a straight line is linear and
continuous.
A continuous linear function must have the
form f(x) = ax. Discontinuous linear
functions look dreadful.
To be more specific, I am going to discuss real valued
functions of one real variable,
i.e. f: R R, where R is, as usual, the set of all real
numbers. Such a function is called
linear provided the following condition holds:
(*)
For every two real x1 and x2, f(x1 + x2) = f(x1) + f(x2)
Assuming that the function f is also continuous I plan to
show that f(x) = ax for some real a.
Please note that if indeed f(x) = ax then a = f(1) which
provides a starting point for the proof.
But first let me note that (*) contains an unknown
which, as we are going to establish,
is equal to f(x) = ax. In other words, (*) serves as an
example of a functional equation –
an equation whose unknown is a function.
Proof
The proof proceeds in several steps.
1. x is 0.
f(0) = f(0 + 0) = f(0) + f(0) = 2f(0).
Therefore f(0) = 2f(0) and finally f(0) = 0.
2. x is negative.
Let x be negative, e.g., let x + y = 0, where y is
positive; so that -x = y. Then
0 = f(0) = f(x + y) = f(x) + f(y).
Therefore f(-x) = f(y) = -f(x).
3. x is an integer.
We have f(2) = f(1 + 1) = f(1) + f(1) = 2f(1).
By induction, assume f(k - 1) = (k - 1)f(1). Then
f(k) = f(1 + (k-1)) = f(1) + (k-1)f(1) = kf(1).
Let's denote a = f(1). We have shown that for all
integers n, f(n) = an.
4. x is rational
First of all, for any integer n≠0, we have 1 =
n/n. Then, as before, a = f(1) = f(n/n) =
nf(1/n).Hence, f(1/n) = a/n = a(1/n). For p =
m/n we similarly have
f(p) = f(m/n) = mf(1/n) = m·a/n = a(m/n) = ap.
5. x is irrational
Any irrational number r can be approximated by a
sequence of rational numbers pi.
The closerpi is to r, the closer api is to ar.
However, since api = f(pi) and
assuming f continuous we must necessarily get f(r)
= ar.
Continuity of the function is quite essential as it's
possible to show [Ref. 1, 2] that the graph
of any discontinuous solution to (*) is dense in the
plane R2. For the sake of reference,
the graph of a function f: R R is defined as a set of
pairs (x, y), i.e. elements of R2 such that y = f(x).
Formally,graph(f) = {(x, y)∈R2: y = f(x)}.
Remark
Generally speaking, a function that satisfies (*) is
called additive. The function that
satisfies f(x) = axfor some a is said to be homogeneous.
A function is said to be linear
if it's both additive and homogeneous. We have just
shown that a continuous additive
function is necessarily linear.
The graph of a linear function is a straight line whose
(linear) equation may be obtained
in different forms depending on the manner in which
the line is defined.
HOT SKILLS QUESTIONS **
Question 1 Show that the lines 4x+y-9=0 , x-2y+3=0, 5x-y6=0 make equal intercepts on any line of gradient 2.
Question 2 (i) If the lines p1x+q1y=1, p2x+q2y=1 and
p3x+q3y=1 be concurrent ,
show that the points (p1,q1), (p2,q2), (p3,q3) are
collinear.
(ii) Find the eqns. Of the lines passing through
the point of intersection of the lines x+3y+4=0 and 3x+y+4=0
and equally inclined to the axes.
Question 3 The line 2x-y=5 turns about the point on it, where
the ordinate is equal to the abscissa through an angle of 450 in
the anticlockwise direction. Find the equation of the line in
the new position.
Question 4 A line 4x+y=1 through the point A(2,-7) meets
the line BC whose equation is 3x-4y+1=0 at the point B.
Find the eqn. of the line AC so that AB=AC.
Question 5 Straight lines 3x+4y=5 and 4x-3y=15 intersect at
A. Points B & C are chosen on these lines such that AB=AC.
Find the possible eqns. Of BC passing through
the point (1,2).
Question 6 A ray of light is sent along the line x-2y-3=0.
Upon reaching the line 3x-2y-5=0, the ray is reflected from it.
Find the eqn. of the line containing the
reflected ray.
Question 7 The eqns. Of two sides of a ∆ are 3x-2y+6=0 and
4x+5y=20 and orthocenter is (1,1). Find eqn. of third side.
Question 8 The eqns. Of the perp. bisectors of the sides AB
& AC of ∆ ABC are x-y+5=0 and x+2y =0 resp. If the point
A is (1, -2) , find the eqn. of the line BC.
Question 9 Prove that the length of perps. From points (m2,
2m), (mn, m+n) and (n2, 2n) to the line xcosѲ+ysinѲ + P=0
Where P = sin2Ѳ/cosѲ form G.P.
Question 10 Find the distant of the point (1,2) from the
straight line with slope 5 and passing through the point
Of intersection of x+2y=5 and x-3y = 7
ANSWERS WITH HINTS
Answer 1
B
A
C
Eqn. of line with gradient 2 is y = 2x +c , solve it with given
eqns.
We will have points B, A, C are (1 𝟐𝒄
𝟑
𝒄
) , ( 2+ 𝟑 , 4+
Answer 2 (i)
𝟓𝒄
𝟑
𝟐𝒄
𝟑
𝒄
𝟑
𝒄
, 2 - 𝟑) , (𝟐 - 𝟔 , 3+
) resp. ⇨ AB = AC.
If lines are concurrent then
𝒑𝟏
𝒑𝟐
𝒑𝟑
𝒒𝟏
𝒒𝟐
𝒒𝟑
−𝟏
−𝟏
−𝟏
= 0 , points are collinear.
(ii) ( x+3y+4)+ k(3x+y+4) = 0 ……..(1) , take
𝟏+𝟑𝒌
slope of eqn.(1) = - 𝟑+𝒌 = ± 1 ( tan450 or tan1350)
eqns are
then put the value of k = 1, -1 in eqn. (1) and
x-y=0 and 4x+4y+8=0 .
Answer 3 Let P (h,h) on the line AB is 2x-y =5 ∴ P(5,5) ,
CD is another line passing through P and making
An angle 450 with AB ∴ slope of CD = tan (Ѳ+450) = -3 ( slope
of AB is 2) by using angle b/w two lines
So eqn. of CD will be 3x+y-20=0.
Answer 4
A(2,-7)
4X+Y-1=0
α
α
3x-4y+1=0 C
B
BY using angle b/w two lines formula ,we
get=slope of AC= m=- 52/89 (
3
4
3
1−4.
4
−4−
3
−𝑚
4
3𝑚
1+
4
=
)
Eqn. of AC is 52x+89y+519=0.
Answer 5 same as Q. 4 Let m be the slope of BC , it will be
1/7, -7 (angle b/w AB & BC =angle b/w BC & AC)
Eqns of BC are 7x+y-9=0 , x-7y+13=0.
Answer 6
L
A 3x-2y-5=0
α
Ѳ Ѳ
M
β
x – 2y-3=0
P
A(1,-1) by solving LM & PA and
N
Q
Tanα = tanβ ⇨
𝟏 𝟑
−
𝟐 𝟐
𝟏𝟑
𝟏+ .
𝟐𝟐
=
𝟑
−𝒎
𝟐
𝟑
𝟏+ .𝒎
𝟐
⇨ m= 29/2 ∴
eqn. of AQ is 29x-2y-31=0.
Answer 7 Let eqns. of AB & AC are given and H(1,1) be
orthocenter
A
N
M
3x-2y+6=0
B
H
4x+5y=20
L
C
Eqns. Of BM & CN are 5x-4y-1=0 , 2x+3y-5=0 by using point
slope form , and B(-13, -33/2) , C(35/2, -10) ,
Then eqn. of BC will be 26x-122y-1675=0.
Answer 8 same as Q.7 by using point slope form & mid
point formula , we will get
Eqns. Of AB & AC are x+y=-1 , 2x-y=4 and B(-7, 6) & C(11/5,
2/5) , then find eqn. of BC by two point form .
Answer 9 USE perp. distant from a point A( m2, 2m) on
the line xcosѲ+ysinѲ + sin2Ѳ/cosѲ =0 is
p1 = | (m2 cosѲ + 2msinѲ +
sin Ѳ/cosѲ)/√𝒔𝒊𝒏²Ѳ + 𝒄𝒐𝒔²Ѳ | =
2
(𝒎𝒄𝒐𝒔Ѳ+𝒔𝒊𝒏Ѳ)𝟐
similarly find p2 , p3 , then show
𝒄𝒐𝒔Ѳ
p1 . p3 = (p2 )2.
Answer 10 same formula of perp. distant from a point to
𝑨𝒙𝟏 +𝑩𝒚𝟏 +𝑪
the line i.e., P =
𝟐
𝟐
√𝑨 +𝑩
length of perp. from (1,2) to line 25x-5y-147=0 ( by
solving given eqns.) is 132/√𝟔𝟓𝟎 .