3. Various Forms of the Equation of a Line

Various Forms of the Equation of a Line:

Different forms of equations of a straight line

• Point-slope form equation of line. Consider a non-vertical line L whose slope is m, A(x,y) be an arbitrary point on the line and. ...
• Two-point form equation of line. ...
• Slope-intercept form equation of line. ...
• Intercept form. ...
• Normal form.

Horizontal and vertical lines:

If a horizontal line L is at a distance a from thex-axis then ordinate of every point lying on the line is either a or – a .

Therefore, equation of the line L is either y = a or y = – a. Choice of sign will depend upon the position of the line according as the line is above or below the y-axis.

Similarly,

the equation of a vertical line at a distance b from the y-axis is either x = b or x = – b

Example 6 Find the equations of the lines parallel to axes and passing through (– 2, 3).

Solution Position of the lines is shown in the Figure. The y-coordinate of every point on the line parallel to x-axis is 3, therefore, equation of the line parallel tox-axis and passing through

(– 2, 3) is y = 3. Similarly, equation of the line parallel to y-axis and passing through (– 2, 3)

is x = – 2.

Point-slope form

 Suppose that P0 (x0, y0) is a fixed point on a non-vertical line L, whose slope is m. Let P (x, y) be an arbitrary point on L  .

Then, by the definition, the slope of L is given by

is the equation of line.

Example : Find the equation of the line through (– 2, 3) with slope – 4.

Solution: Here m = – 4 and given point (x0 , y0) is (– 2, 3).

By slope-intercept form above formula , equation of the given line is

y – 3 = – 4 (x + 2) or

4x + y + 5 = 0, which is the required equation.

Two-point form

Let the line L passes through two given points P1 (x1, y1) and P2 (x2, y2).

Let P (x, y) be a general point on L . The three points P1, P2 and P are collinear, therefore, we have slope of P1P = slope of P1P2

 

 

is the equation of the line passing through the points (x1, y1) and (x2, y2).

Example: Write the equation of the line through the points (1, –1) and (3, 5).

Solution : Here x1 = 1, y1 = – 1, x2 = 3 and y2 = 5. Using two-point form  above formula

for the equation of the line, we have

 

• –3x + y + 4 = 0, which is the required equation.

 

Slope-intercept form Sometimes a line is known to us with its slope and an

intercept on one of the axes. We will now find equations of such lines.

Case I Suppose a line L with slope m cuts the y-axis at a distance c from the origin

. The distance c is called the y-intercept of the line L. Obviously, coordinates of the point where the line meet the y-axis are (0, c). Thus, L has slope m and passes through a fixed point (0, c).

Therefore, by point-slope form, the equation of L is

y - c = m( x - 0 ) or y = mx + c

 

 

Thus, the point (x, y) on the line with slope m and y-intercept c lies on the line if and only if

 y = mx +c ......................(3)

Note that the value of c will be positive or negative according as the intercept is made

on the positive or negative side of the y-axis, respectively.

Case II Suppose line L with slope m makes x-intercept d. Then equation of L is

y = m(x – d) …………………... (4)

Students may derive this equation themselves by the same method as in Case I.

 

Intercept – form

 Suppose a line L makes x-intercept a and y-intercept b on theaxes. Obviously L meets x-axis at the point (a, 0) and y-axis at the point (0, b) . By two-point form of the equation of the line, we have,

y – 0 =   b-00-a( x – a)

• ay = – bx + ab
• xa  + yb  = 1

is equation of the line making intercepts a and b on x-and y-axis, respectively

Normal form

 Suppose a non-vertical line is known to us with following data:

(i) Length of the perpendicular (normal) from origin to the line.

(ii) Angle which normal makes with the positive direction of x-axis.

Let L be the line, whose perpendicular distance from origin O be OA = p and the

angle between the positive x-axis and OA be ÐXOA = w.

Draw perpendicular AM on the x-axis in each case.

 

In each case, we have OM = p cos w and MA = p sin w, so that the coordinates of the

point A are (p cos w, p sin w).

Further, line L is perpendicular to OA. Therefore

The slope of the line L

 

the equation of the line L at point A(pcos w, psin w ) is

 

x cos w + y sin w = p.

Hence, the equation of the line having normal distance p from the origin and angle w