Page 149 - DMTH201_Basic Mathematics-1

Basic Mathematics – I

Notes Solution:
Given lines can be written as

a c
y = 1 x 1 …(1)
b b
1 1
a c
and y = 2 x 2 …(2)
b 2 b 2

a a
Slopes of the lines (1) and (2) are m 1 and m 2 , respectively. Now
1 2
b b
1 2
(i) Lines are parallel, if m = m , which gives
1 2
a 1 a 2 or a 1 a 2 .
b 1 b 2 b 1 b 2

(ii) Lines are perpendicular, if m .m = 1, which gives
1 2
a a
1 2 1 or a b b b 0
b 1 b 2 1 2 1 2

Example: Find the equation of a line perpendicular to the line x 2 y +3 = 0 and passing
through the point (1, 2).
Solution:
Given line x 2 y + 3 = 0 can be written as

1 3
y = x ...(1)
2 2
1
Slope of the line (1) is m . Therefore, slope of the line perpendicular to line (1) is
1
2
1
m = 2
2
m
1
Equation of the line with slope 2 and passing through the point (1, 2) is
y ( 2) = 2(x 1) or y = 2x,
which is the required equation.

5.5 Distance of a Point From a Line

The distance of a point from a line is the length of the perpendicular drawn from the point to the
line. Let L : Ax + By + C = 0 be a line, whose distance from the point P (x , y ) is d. Draw a
1 1
perpendicular PM from the point P to the line L (Figure 5.18). If the lines meets the x-and y-axes
C C
at the points Q and R, respectively. Then, coordinates of the points are Q ,0 and R 0, .
A B
Thus, the area of the triangle PQR is given by

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