# Equation of a Straight Line in different forms

### Slope-Intercept form

The equation of a straight line having slope m and making an intercept c on
y-axis is

y = m x +c

### Point-slope form

The equation of a straight line passing through the fixed point (x_{1},
y_{1}) and having slope m is

y -y_{1} = m (x -x_{1})

**Corollary.** The equation of a line passing through origin and having
slope m is y = m x.

### Two-point form

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

y - y_{1} = [(y_{2} - y_{1})/(x_{2} - x_{1})](x
-x_{1})

### Intercept form

The equation of the line cutting off intercepts a and b on the axes is

x/a + y/b = 1

### Normal (or perpendicular) form

The equation of a straight line in terms of the length of perpendicular
from the origin upon it and the angle which this perpendicular makes with the
positive direction of x-axis is given by

x cos +
y sin = p

### General form

Every straight line can be represented by an equation of the first degree
in x and y, and conversely every first degree equation in x, y represents a
straight line.

The equation A x +B y +C = 0 (where at least one of A and B is non-zero) is
called the general form.

## Illustrative Examples

### Example

Find the equation of a straight line whose inclination is 5/6 and which cuts off an intercept of 4 units on negative direction of y-axis.

### Solution

Let *m* be the slope of the line, then

*m* = tan 5/6 = tan 150° = tan (180° -30°) =
-tan 30° = 1/3

Also c = y-intercept = -4

So the equation of the line is y =(-1/3) x +(-4)
(As y = m x +c)

i.e. x + 3y +
43 = 0

### Example

Show that the points (a, 0), (0, b) and (3 a, -2 b) are collinear. Also find the equation of the line containing them.

### Solution

The equation of the line through (a, 0) and (0, b) is

y -0 =[(b-0)/(0-a)](x -a)
y - y_{1} = [(y_{2} - y_{1})/(x_{2} - x_{1})](x
-x_{1})

i.e. -ay = bx -ab
i.e. bx +ay -ab = 0

The third point (3 a, -2 b) lies on it if b. 3 a + a. (-2 b) - ab = 0 i.e. if
0 = 0, which is true.

Hence the given points are collinear and the equation of the line containing them is

bx +ay -ab = 0.

### Example

In what ratio is the line joining the points (2, 3) and (4, -5) divided by the line joining the points (6, 8) and (-3,-2)?

### Solution

The equation of the straight line joining the points (6, 8) and (-3, -2) is

y -8 = [(-2 -8)/(-3 -6)](x -6)
(Two-point form)

=> y -8 = (10/9) (x -6) => 9 y -72 = 10 x -60

=> 10 x -9 y +12 = 0 ...(i)

Let the line joining the points (6, 8) and (-3, -2) i.e. the line (i) divide
the line segment joining the points (2, 3) and (4, -5) at the point P in the
ratio k : 1, then the co-ordinates of point P are

((k.4 +1.2)/(k+1), (k (-5) +1 .3)/(k+1)) *i.e.* ((4k +20)/(k+1), (-5k +3)/(k+1))

Since P lies on (i), we get

10[(4k +20)/(k+1)] -9[(-5k +3)/(k+1)] + 12 = 0

=> 40 k +20 +45 k -27 +12 k +12 = 0

=> 97 k +5 = 0 => k = -5/97

Hence the required ratio is -5/97 i.e. 5 : 97 externally.

### Example

Prove that the locus of the point which is equidistant from the points (-3, 7) and (2, -5) is a straight line.

### Solution

Let A (-3, 7) and B (2, -5) be the given points and P (x,
y) be any point on the locus, then | AP | = | BP | (given)

=> (x +3)² +(y -7)² = (x -2)² +(y +5)²

=> x² +6 x +9 + y² -14 y +49 = x² -4 x +4 + y² +10 y +25

=> 10 x -24 y +29 = 0, which is a first degree equation in x and y, and so it
represents a straight line.

Hence the locus is a straight line. In fact, this Is the perpendicular
bisector of the line segment joining the two given points.

## Exercise 1

- Find the equation of a straight line parallel to x-axis at a distance

(i) 3 units above it

(ii) 3 units below it. - Find the equation of a straight line parallel to y-axis at a distance

(i) 2 units to the right

(ii) 2 units to the left of it. - Find the equation of a horizontal line passing through (5, -2).
- Find the equation of a vertical line passing through (-7, 3).
- Find the equation of a line which is equidistant from the lines y +5 = 0 and y -2 = 0.
- Find the equation of a line whose

(i) gradient = -1, y-intercept = 3

(ii) slope = - 2/7, y-intercept = -3

(iii) inclination = 3/4, y-intercept = -5 - Find the equations of the straight lines cutting off an intercept of 3
units from the negative direction of y-axis and equally inclined to the axes.

[**Hint.**The lines which are equally inclined to the axes have slope = ± 1] - Find the equation of the line passing through the point (2, -5) and making an intercept of -3 on the y-axis.
- If the straight line y =
*m*x +c passes through the points (2, 4) and (-3, 6) find the values of m and c. - Find the equation of a straight whose y-intercept is -5 and which is

(i) parallel to the line joining the points (3, 7) and (-2, 0)

(ii) perpendicular to the line joining the points (-1, 6) and (-2, -3). - Find the equation of a straight line passing through origin and making an angle of 120° with the positive direction of x-axis.
- Find the equations of the bisectors of the angles between the co-ordinate axes.
- Find the equation of the straight line passing through the point (5, 7) and inclined at 45° to x-axis. If it passes through the point P whose ordinate is -7, what is the abscissa of P?
- Find the equation of the line through the point (-5, 1) and parallel to the line joining the points (7, -1) and (0, 3).
- Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining (1, 4), (2, 3).

## Answers

**1.**(i) y -3 = 0 (ii) y +3 = 0

**2.**(i) x -2 = 0 (ii) x +2 = 0

**3.**y +2 = 0

**4.**x +7 = 0

**5.**2y +3 = 0

**6.**(i) x +y - 3 = 0 (ii) 2 x +7 y +21 = 0 (iii) x +y +5 = 0

**7.**x +y +3 = 0, x -y -3 = 0

**8.**x +y +3 = 0

**9.**m = -2/5, c = 24/5

**10.**(i) 7 x -5 y -25 = 0 (ii) x +9 y +45 = 0

**11.**3 x +y = 0

**12.**x - y = 0, x +y = 0

**13.**x - y +2 = 0; -9

**14.**4 x +7 y +13 = 0

**15.**x -y +3 = 0

## Exercise 2

- State the geometrical meaning of the constants involved in

(i) x/a + y/b = 1

(ii) x cos + y sin = p

(iii) (x -x_{1)/}cos = (y -y_{1)/}sin = r - Find the equation of the line which cuts off intercepts 3 and 4 from the axes.
- Prove that the straight line whose intercepts on the axes are 2 and -3 respectively passes through the point (4, 3).
- Find the equation of a straight line which passes through the point (1, -3) and makes an intercept on y-axis twice as long as on x-axis.
- Find the equation of the straight line which passes through the point (3, -4) and makes

(i) equal intercepts on the axes

(ii) intercepts equal in magnitude but opposite in sign on the axes. - Find the equation of the straight line which passes through the point (3, -2) and cuts off positive intercepts on the x-axis and y-axis which are in the ratio 4 : 3.
- A straight line passes through the point (2, 3) and the portion of the line intercepted between the axes is bisected at this point, find its equation.
- A straight line is such that the portion of it intercepted between the
axes is bisected at the point (x
_{1 , }y_{1}). Find its equation. - A straight line passes through the point (1, 1) and the portion of the line intercepted between the axes is divided at this point in the ratio 3 : 4, find its equation.
- Find the equation of a line which passes through the point (-5, 2) and whose segment between the axes is divided by this point in the ratio 2 : 3.
- If the straight line x/a + y/b = 1passes through the points (12, -15) and (8, -9), find the values of a and b.
- Find the equation of a line which passes through the point (22, -6) and whose intercept on the x-axis exceeds the intercept on y-axis by 5.
- Find the equation of a line which passes through (-3, 10) and sum of its intercepts on the axes is 8.
- A straight line passes through the points (a, 0) and (0, b), the length of the line segment contained between the axes is 13 and the product of the intercepts on the axes in 60. Calculate the values of a and b and find the equation of the straight line.
- The area of a triangle formed by a line and the co-ordinate axes is 6 sq. units and the length of the segment intercepted between the axes is 5 units. Find the equation (s) of line.

## Answers

**1.**(i) a and b are the intercepts made by the line on x-axis and y-axis respectively.

(ii) p is the length of the perpendicular drawn from origin on the line and the angle which this perpendicular makes with the x-axis.

(iii) (x

_{1}, y

_{1}) is the point through which the line passes, is the inclination of the line and r is the directed distance of any point (x, y) on the line from the point (x

_{1}, y

_{1}).

**2.**4 x +3 y = 12

**4.**2 x +y +1 = 0

**5.**(i) x +y +1 = 0 (ii) x -y = 7

**6.**3 x +4 y = 1

**7.**3 x +2 y -12 = 0

**8.**x/x

_{1}+ y/y

_{1}= 2

**9.**4 x +3 y = 7

**10.**3 x -5 y +25 = 0

**11.**a = 2, b = 3

**12.**6 x +11 y -66 = 0, x +2 y -10 = 0

**13.**2 x - y +16 = 0, 5 x +3 y -15 = 0

**14.**a = 12, b = 5, 5 x +12 y = 60; a = 5, b = 12, 12 x +5 y = 60;

a = -12, b = -5, 5 x +12 y +60 = 0; a = -5, b = -12, 12 x +5 y +60 = 0

**15.**3 x +4 y = 12, 4 x +3 y = 12; 3 x +4 y +12 = 0, 4 x +3 y +12 = 0