Page 156 - DMTH201_Basic Mathematics-1

Unit 5: Equations of Straight Lines

Notes
Example: Show that the path of a moving point such that its distances from two lines
3x 2y = 5 and 3x + 2y = 5 are equal is a straight line.
Solution:
Given lines are
3x – 2y = 5 …(1)

and 3x + 2y = 5 … (2)
Let (h, k) is any point, whose distances from the lines (1) and (2) are equal. Therefore

3h 2k 5 3h 2k 5
or 3h 2k 5 3h 2k 5 ,
9 4 9 4
which gives 3h 2k 5 = 3h + 2k 5 or (3h 2k 5) = 3h + 2k 5.

5
Solving these two relations we get k = 0 or h . Thus, the point (h, k) satisfies the equations y
3
5
= 0 or x , which represent straight lines. Hence, path of the point equidistant from the lines
3
(1) and (2) is a straight line.

5.6 Summary

Slope (m) of a non-vertical line passing through the points (x , y ) and (x , y ) is given by
1 1 2 2
y y y y
m 2 1 1 2 , x x .
x x x x 1 2
2 1 1 2
Slope of horizontal line is zero and slope of vertical line is undefined.
An acute angle (say ) between lines L and L with slopes m and m is given by
1 2 1 2
m m
tan 2 1 ,1 m m 0.
1 m m 1 2
1 2
Two lines are parallel if and only if their slopes are equal.
Two lines are perpendicular if and only if product of their slopes is 1.

Three points A, B and C are collinear, if and only if slope of AB = slope of BC.
Equation of the horizontal line having distance a from the x-axis is either y = a or y = a.
Equation of the vertical line having distance b from the y-axis is either x = b or x = b.
The point (x, y) lies on the line with slope m and through the fixed point (x , y ), if and only
0 0
if its coordinates satisfy the equation y y = m (x x ).
0 0
Equation of the line passing through the points (x , y ) and (x , y ) is given by
1 1 2 2
y y
y y 1 2 1 (x x 1 ).
x 2 x 1
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 .

LOVELY PROFESSIONAL UNIVERSITY 149