Straight lines

.Line through two points

The line through two distinct points (x1, y1) and (x2, y2) is given by

(1) y = y1 + [y2− y1

x2− x1

]·(x - 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

y - y1 = [y2− y1

x2− x1

]·(x - 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 y-axis 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 x-intercept 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 y-intercept 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 x1 , x2, y1, 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(x2,y2) 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 L2, 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, L2, is parallel to the Y axis and the other, L1, has a positive slope, then

If L2 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

cosθ = y− y1

sinθ = r , where r is the distant of the point (x,y) on

the line 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 cos and Ѳ 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 cos /6 , 2 + п 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 2100 from the positive side of the X axis.

14. Find the equation of the line through (2,3) and perpendicular to 3x-2y=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 , cos =4/5 , equation Ѳpassing 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) (-3,5) (2,3) (x,y)

C B A 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 co-ordinates

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) B(X,2X-4)

(AB)2 + (BC)2 = (AC)2 ⇨ x=4 or 2 ∴ B(4,4) then D (2,0), if B(2,0) then D(4,4)]

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 C

11x+7y=9 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 slope=7/4

A(-3,1) 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 4 A(4cosα, 4sinα)

M O 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:

( −aba+b , −aba+b ) D line (iii) C ( aba−b , −aba−b )

Slope=1

Line(iv)

Slope=-1 line(ii)

(−aba−b ,

aba−b ) A line(i) B (

aba+b ,

aba+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+13−m |

⇨ m =1/2 or -2 , equations of BD ,AC, CD AD & BC are x-2y+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 8x-15y=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 = m1−

815

1+m18

15

⇨ m1 = 23/7 ,

(1,2) A B

m1 8x-15y=0

450 m2 =8/15

D C

Equations of AD & AB ( Perp. to each other) are 23x-7y-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 radius-vector 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-y-6=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 - 2c3 , 2 - c3) , (3

2 - c6 , 3+ 2c3 ) , ( 2+ c3 , 4+ 5c

3 ) resp. ⇨ AB = AC.

Answer 2 (i) If lines are concurrent then p1 q1 −1p2 q2 −1p3 q3 −1

=

0 , points are collinear.

(ii) ( x+3y+4)+ k(3x+y+4) = 0 ……..(1) , take slope of eqn.(1) = - 1+3k

3+k = ± 1 ( tan450 or tan1350)

then put the value of k = 1, -1 in eqn. (1) and eqns are 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

B 3x-4y+1=0 C

BY using angle b/w two lines formula ,we

get=slope of AC= m=- 52/89 ( −4−3

4

1−4.34

= 34−m

1+3m4

)

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 M

α Ѳ Ѳ β x – 2y-3=0

P N Q

A(1,-1) by solving LM & PA and

Tanα = tanβ ⇨ 12−3

2

1+12.32

= 32−m

1+32.m

⇨ 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 H 4x+5y=20

B 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 + sin2Ѳ/cos )/Ѳ√sin ²Ѳ+cos²Ѳ | = (mcosѲ+sinѲ)2

cosѲ similarly find p2 , p3 , then show p1 . p3 = (p2 )2.

Answer 10 same formula of perp. distant from a point to

the line i.e., P = A x1+B y1+C

√A2+B2

length of perp. from (1,2) to line 25x-5y-147=0 ( by solving given eqns.) is 132/√650 .