Straight Line: Any first degree equation 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 θ is the angle at which a straight line is inclined to the positive direction of x axis, slope of the line is m = tanθ, 0 ≤ θ < 180° (θ ≠ 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:

∎ 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 = tanθ

c = y intercept

∎ **Intercept Form : **

$ \displaystyle \frac{x}{a} + \frac{y}{b} = 1 $

x intercept = a

y intercept = b

∎** Normal Form : **

x cosα + y sinα = p, where α 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 − y_{1} = m(x − x_{1}), is the equation of line passing through the point (x_{1}, y_{1}) and having the slope ‘ m ‘

∎ **Two Points Form: **

Equation: $ \displaystyle y – y_1 = \frac{y_2 – y_1}{x_2 – x_1}(x – x_1)$

is the equation of line passing through two points (x_{1}, y_{1}) and (x_{2}, y_{2})

∎ ** Parametric Form: **

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

Let AP = r

x − h = r cosθ , y − k = r sinθ

$ \displaystyle \frac{x – h}{cos\theta} = \frac{y – k}{sin\theta} = r $

is the equation of the straight line LAL’

Any point on the line will be of the form (h + rcosθ, k + rsinθ), where |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.

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

Solution: Slope-Intercept Form:

3y = 2x + 5

$ \displaystyle y = \frac{2x}{3} + \frac{5}{3}$

$ \displaystyle tan\theta = m = \frac{2}{3} \; , \; c = \frac{5}{3} $

**Intercept Form:**

$ \displaystyle \frac{x}{-5/2} + \frac{y}{5/3} = 1 $

$ \displaystyle a = \frac{-5}{2} \; , \; b = \frac{5}{3} $

**Normal Form:**

$ \displaystyle \frac{-2x}{\sqrt{13}} + \frac{3y}{\sqrt{13}} = \frac{5}{\sqrt{13}} $

$ \displaystyle cos\alpha = \frac{-2}{\sqrt{13}} \; , \; sin\alpha = \frac{3}{\sqrt{13}} \; , p = \frac{5}{\sqrt{13}}$

Problem : 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

$ \displaystyle \frac{x}{a} + \frac{y}{b} = 1 $ . . . . . . (i)

This passes through (3, 4), therefore

$ \displaystyle \frac{3}{a} + \frac{4}{b} = 1 $ . . . . . . (ii)

It is given that a + b = 14 => b = 14 –a.

Putting b = 14 − a in (ii),

we get ,

$ \displaystyle \frac{3}{a} + \frac{4}{14-a} = 1 $

⇒ a^{2} −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

$ \displaystyle \frac{x}{7} + \frac{y}{7} = 1 $ and $ \displaystyle \frac{x}{6} + \frac{y}{8} = 1 $

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

Problem : A rod of steel is fixed at A(4, 0) and a toy is placed on it at B(0, 4). Now rod is rotated about A through an angle of 15° in clockwise direction, then find the new position of a toy.

Solution: Let new position of a toy be C.

Slope of AB = (4−0)/(0−4) = −1 => θ = 135°

Rod is rotated through 15° in clockwise direction.

θ_{new} = 135° − 15° = 120°

AB = 4√2

⇒ h = 4, k = 0

Hence C ≡ (h + rcos θ, k + rsin θ)

= (4 + 4√2 cos 120°, 0 + 4√2 sin 120°) = (4 − 2√2 , 2√6 ).

Problem : 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 π/6 with the x-axis is

(x−3)/cosπ/6 = (y−4)/sinπ/6 = r

, 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 (r cosπ/6 + 3 , r sinπ/6 + 4)

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

⇒ r = −132/(12√3 + 5)

⇒ length PQ = 132/(12√3 + 5)

Problem: A canal is 4½ km from a place and the shortest route from this place to the canal is exactly north-east. A village is 3 km north and 4 km 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

x cos45° + y sin45° = 9/2

Or , x + y = 9/√2 ……..(1)

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

Putting x=4 and y = 3 in equation (1),

we get 4 + 3 = 9/√2, 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.

### Position of two points with respect to a given line :

Let the given line be ax + by + c = 0 and P(x_{1 }, y_{1}) , Q(x_{2 }, y_{2}) be two points. If the quantities ax_{1} + by_{1} + c and ax_{2} + by_{2} + 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 ax_{1} + by_{1}+ c and ax_{2} + by_{2} + c have opposite signs, then they lie on the opposite sides of the line.

Problem : Find the range of θ in the interval (0, π) such that the points (3, 5) and (sinθ, cosθ) lie on the same side of the line x + y – 1 = 0.

Solution: Here 3 + 5 – 1 = 7 > 0

Hence sinθ + cosθ – 1 > 0

⇒ sin(π/4 + θ) > 1/√2

⇒ π/4 < π/4 + θ < 3π/4

⇒ 0 < θ < π/2

### Exercise :

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

(ii) 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.

(iii) If the image of the point (2, 1) with respect to the line mirror be (5, 2), find the equation of the mirror.

(iv) Find the equation of the straight line which passes through the point (3, 2) and whose slope is 3/4. Find the co-ordinates of the points on the line that are 5 units away from the point (3, 2).

(v) A line through the origin intersects x = 1, y = 2 and x + y = 4, in A, B and C respectively, such that OA. OB.OC = 8√2 . Find the equation of the line.