Basic Concepts Of Straight line Part 3

Posted on - 04-01-2017

IIT JEE

Straight Line

Any equation of first degree of the form Ax + By + C = 0, where A, B, C are constants always represents a straight line (at least one out of A and B is non zero).

Slope:

If q is the angle at which a straight line is inclined to the positive direction of x axis, slope of the line is m = tanq, 0 £ q < 180° (q > 90° ).

Intercept of a straight line on the axis:

If a line AB cuts the x-axis and y-axis at A and B respectively and O be the origin then OA and OB with proper sign are called the intercepts of the line AB on x and y axes respectively.

Notes:

l If we say that the length of intercept of a line on x-axis is 3, it means that intercept on x-axis is either 3 or –3.

· If intercept of a line on x-axis –3, then length of intercept of axis is | –3 | = 3.

Standard equations of Straight Lines:

Slope-Intercept Form :

y = mx + c,

where m = slope of the line = tanq

c = y intercept

Intercept Form :

x/a + y/b = 1

x intercept = a

y intercept = b

Normal Form :

x cosa + y sina = p, where a is the angle which the perpendicular to the line makes with the axis of x and p is the length of the perpendicular from the origin to the line. p is always positive. .

Slope Point Form:

Equation: y - y₁ = m(x - x₁), where

(a) One point on the straight line is (x₁, y₁)

(b) The direction of the straight line i.e., the slope of the line = m.

Two Points Form:

Equation: ,

where (x₁, y₁) and (x₂, y₂) are the two given points. Here .

Parametric Form:

To find the equation of a straight line which passes through a given point A(h, k) and makes a given angle q with the positive direction of the x-axis.
P(x, y) is any point on the line LAL¢.

Let AP = r. x – h = r cosq y – k = r sinq.

is the equation of the straight line LAL¢.

Any point on the line will be of the form (h+r cosq, k+ r sinq) and for any value of r

this will represent a point on the straight line. Here | r | gives the distance of the point P from the fixed point (h, k).

Note:

If point P is taken relatively upward to A then r is positive otherwise negative. If line is parallel to x-axis then for the point right to A, r is positive and for left to A, r is negative. .

Reduction of the General Equation to Different Standard Forms:

Slope-Intercept Form:

To reduce the equation Ax + By + C = 0 to the form y = mx + c

Given equation is Ax + By + C = 0
&⇒ y =

which is of the form y = mx + c

where m = ( B > 0)

Note:

Slope of the line Ax + By + C = 0 is –, i.e. –,

y intercept of the line = -C/B

Intercept Form:

To reduce the equation Ax + By + C = 0 to the form

This reduction is possible only when C > 0

Given equation is Ax + By = –C

&⇒ which is of the form . where a =

Note:

Intercept on the x-axis = , Intercept on the y-axis =

Thus intercept of a straight line on the x-axis can be obtained by putting y = 0 in the equation of the line and then finding the value of x. Similarly, intercept on the y-axis can be obtained by putting x = 0 and solving for y.

Normal Form:

To reduce the equation Ax + By + C = 0 to the form xcosa + ysina = p

Given equation is Ax + By + C = 0 Or, Ax + By = –C

Case I: When –C>0, then normal form is

where cosa = sina = , p =

Case II: When –C < 0, then write the equation as -Ax –By = C

where cosa = , sina = p =

Note:

In the normal form xcosa + ysina = p, p is always taken as positive. .

Illustration - 7:

Reduce the line 2x - 3y + 5 = 0, in slope-intercept, intercept and
normal forms.

Solution:

Slope-Intercept Form:

Intercept Form:

Normal Form :

, ,

Example 1

Find the equations of the lines which pass through the point (3, 4) and the sum of their respective intercepts on the axes is 14. .

Solution:

Let the equation of the line be . . . (i)

This passes through (3, 4), therefore . . . (ii)

It is given that a + b = 14
&⇒ b = 14 –a . Putting b = 14 - a in (ii),we get
&⇒ a² -13 a + 42 = 0

&⇒ (a – 7)(a –6) = 0
&⇒ a = 7, 6
&⇒ two such lines are there.

For a = 7, b = 14 –7 = 7 and for a = 6 , b = 14 – 6 = 8. .

Putting the values of a and b in (i), we get the equations of lines

and or x + y = 7 and 4x + 3y = 24.

Example 2

A line joining two points A(2, 0) and B(3, 1) is rotated about A in the anticlockwise direction through an angle of 15°. Find the equation of the line in the new position. If B goes to C, in the new position, what will be the coordinates of C.

Solution:

Slope of the line AB is m = 1

&⇒ tanq = 1
&⇒ q = 45°

AB = Ö2
&⇒ AC = Ö2

Slope of AC = tan(45° + 15°) = tan(60°)

[because angle between AB and AC=15°]

A is (2, 0)
&⇒ h = 2, k = 0. .

Here, C is given by the formula

(h + r cosq, k + r sinq), (r = Ö2)

C = (2 + Ö2 cos 60°, 0 + Ö2 sin 60°), C =

&⇒ Equation of the line AC is y = Ö3(x – 2).

Example 3

If the straight line through the point P (3, 4) makes an angle p/6
with the x-axis and meets the line 12x + 5y + 10 = 0 at Q, find the
length of PQ.

Solution:

The equation of a line passing through P (3, 4) and making an angle p/6 with the x-axis is

, where r represents the distance of any point on this line from the given point P (3, 4). The co-ordinates of any point Q on this line are

If Q lies on 12x + 5y + 10 = 0, then

&⇒
&⇒ length PQ = .

Example 4

A canal is 4½ kms from a place and the shortest route from this place to the canal is exactly north-east. A village is 3 kms north and 4 kms east from the place. Does it lie on canal?.

Solution:

Let AB be the canal and O be the given place. Let L be the foot of perpendicular from O to AB. Given, OL = 9/2. And ∠AOL = 45°.

Now equation of line AB will be

xcos45° + ysin45° = 9/2

Or x + y = 9/Ö2 ……..(1).

Let S be the given village, then
Sº (4, 3)

Putting x=4 and y = 3 in equation (1), we get 4 + 3 = 9/Ö2, which is not true. Thus the co-ordinates of S .

doesn’t satisfy equation (1) and hence the given village does not lie on the canal.

Example 5

Find the equation of the straight line which makes an angles of 15° with the positive direction of x-axis and which cuts an intercept of length 4 on the negative direction of y-axis. .

Solution:

Here m = tan15° = tan(45° – 30°)

= = 2 –

and c = intercept on y-axis = -4

Therefore, equation of the required line will be

y = (2 – ) x – 4

or (2 –)x – y – 4 = 0.

Example 6

Find the equation of the straight line which passes through the point
(3, 2) and whose gradient is . Find the co-ordinates of the points on the lines that are 5 units away from the point (3, 2).

Solution:

Let P º (3, 2)and let the required line make an angle q with the positive direction of x-axis . .

Given tanq =

\ cos q =and sinq =

Now equation of the required line will be

or 3 (x – 3) = 4( y – 2) or 3x – 4y –1 = 0 . . . (1).

Now the co-ordinates of the points which are at a distance of 5 units from P are (3 ± 5 cosq, 2 ± 5 sinq) or (3 ± 4, 2 ± 3)

or (7, 5 ) and ( –1, –1) . .

Position of two points with respect to a given line:

Let the given line be ax + by + c = 0 and P(x₁, y₁), Q(x₂, y₂) be two points. If the quantities ax₁ + by₁ + c and ax₂ + by₂ + c have the same signs, then both the points P and Q lie on the same side of the line ax + by + c = 0.

If the quantities ax₁ + by₁ + c and ax₂ + by₂ + c have opposite signs, then they lie on the opposite sides of the line.

Example 7

Find the range of q in the interval (0, p) such that the points (3, 5) and (sinq, cosq) lie on the same side of the line x + y - 1 = 0.

Solution:

Here 3 + 5 - 1 = 7 > 0

Hence sinq + cosq - 1 > 0
&⇒ sin(p/4 + q) > 1/Ö2

&⇒ p/4 < p/4 + q < 3p/4

&⇒ 0 < q < p/2.

Angle between two straight lines:

If q is the acute angle between two lines, then
tanq = where m₁ and m₂ are the slopes of the two lines and are finite.

Notes:

If the two lines are perpendicular to each other then m₁m₂ = -1.
Any line perpendicular to ax + by + c = 0 is of the form bx – ay + k = 0.
If the two lines are parallel or coincident, then m₁ = m₂.
Any line parallel to ax + by + c = 0 is of the form ax + by + k = 0.
If any of the two lines is perpendicular to x-axis, then the slope of that line is infinite. Let m₁ = ¥. Then = or q = |90° - a|
where tana = m₂ i.e. angle q is the complimentary to the angle which the oblique line makes with the x-axis.

Example 8

Find the equation to the straight line which is perpendicular bisector of the line segment AB, where A, B are (a, b) and (a¢, b¢) respectively.

Solution:

Equation of AB is y - b =

i.e. y(a¢ - a) - x(b¢ - b) = a¢b - ab¢.

Equation to the line perpendicular to AB is of the form

(b¢ - b)y + (a¢ - a)x + k = 0 ...(1).

Since the midpoint of AB lies on (1)

Hence the required equation of the straight line is

2(b¢ - b)y + 2(a¢ - a)x = (b¢² - b² + a¢² - a²).

Equation of Straight Lines passing through a given point and equally inclined to a given line:

Let the straight lines passing through the point (x₁, y₁) and make equal angles with the given straight line y = m₁x + c. .

If m is the slope of the required line and a is the angle which this line makes with the given line, then tana = ±.

1. The above expression for tana, gives two values of m, say m_A and m_B. .

The required equations of the lines through the point (x₁, y₁) and making equal angles a with the given line are y – y₁ = m_A(x - x₁), y - y₁ = m_B(x - x₁).

Example 9

Find the equation to the sides of an isosceles right-angled triangle, the equation of whose hypotenuse is 3x + 4y = 4 and the opposite vertex is the point (2, 2).

Solution:

The problem can be restated as:

Find the equations to the straight lines passing through the given point (2, 2) and making equal angles of 45° with the given straight line 3x + 4y - 4 = 0. Slope of the line 3x + 4y - 4 = 0 is m₁ = -3/4.

&⇒ tan 45° =±

m_A = and m_B = -7

Hence the required equations of the two lines are

y - 2 = m_A (x - 2) and y - 2 = m_B(x - 2)

&⇒ 7y - x - 12 = 0 and 7x + y = 16.

Example 10

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 equation to the line AC so that AB = AC. .

Solution:

Clearly, AB and AC both pass through A(2, -7) and are equally inclined to 3x – 4y +1 = 0. So, they form an isosceles triangle and their equations are given by .

(y + 7) = ,

where m = slope of BC = 3/4 and

tana = =

&⇒ (y +7) =

&⇒ (y +7) = and y +7 =

&⇒ y+ 7 = -4(x –2) and y + 7 = -

&⇒ 4x + y = 1 and 52 x + 89y + 519 = 0

Clearly, 4x + y = 1 is the equation of AB. So, the equation of AC is .

52x + 89 y + 512 = 0

Alternatively:

Slopes of the lines AB and BC are –4 and 3/4. Let the slope of the line AC be m. .

Then = ± .

Hence equation of the line AC is

y + 7 = (x - 2)
&⇒ 52x + 89y + 519 = 0.

The distance between two parallel lines:

ax + by + c₁ = 0 and ax + by + c₂ = 0 is .

Length of the Perpendicular from a Point on a Line:

The length of the perpendicular from P(x₁, y₁) on ax + by + c = 0 is .

The length of the perpendicular from origin on ax + by + c = 0 is .

Example 11

Find all points on x + y = 4 that lie at a unit distance from the line
4x + 3y – 10 = 0. .

Solution:

Coordinates of an arbitrary point on the given line are P(t, 4-t). Let
P(t, 4-t) be the required point. Then, distance of P from the line
4x+3y– 10 = 0 is unity i.e. = 1

&⇒ | t + 2| =5
&⇒ t + 2 = ± 5
&⇒ t = -7 or t = 3. .

Hence, the required points are (-7, 11)and (3, 1). .

Example 12

Three lines x + 2y + 3 = 0, x + 2y - 7 = 0 and 2x - y - 4 = 0 form 3 sides of two squares. Find the equation of remaining sides of these squares.

Solution:

Distance between the two parallel lines is . The equations of sides A and C are of the form
2x – y + k = 0. Since distance between sides A and B = distance between sides

B and C
&⇒
&⇒ k = 6, -14.

Hence the fourth sides of the two squares are

(i) 2x - y + 6 = 0, (ii) 2x - y - 14 = 0.

Notes:

(i) In order to find the foot of the perpendicular from a given point to a given line, the most convenient method is to write the equation of the given line in such a way so that the coefficient of y becomes negative and then write the equation of the perpendicular in the parametric form. Then put r = the algebraic length (without the modulus) of the perpendicular = p. In this way, we get the foot of the perpendicular directly.

(ii) To find the mirror image of a given point in a given line, we use the same process, taking r = 2p.