Understanding Engineering Mathematics

y

0 AD x

B

C

ab

l 2 l 1

Figure 7.9Gradients of perpendicular lines multiply to−1.

On the other hand the gradient ofl 2 is

−

BD

CD

=−

1

tanβ

=−

1

tanθ

=−

1

m

Solution to review question 7.1.5

A.Any line parallel toy= 1 − 3 xhas the same gradient,m=−3and

so can be written as

y=− 3 x+c

We now have to choose the interceptcso that the line passes through

(−1, 2), which we do by substituting these values in the equation:

2 =− 3 (− 1 )+c

So

c=−1 and the equation required is

y=− 3 x− 1

B. If the linel through (−1, 2), (0, 4) has gradientmthen any line

perpendicular to it has gradient− 1 /m. The gradient oflis

m=

4 − 2

0 −(− 1 )

= 2

so any line perpendicular to it has gradient−^12 and has an equation of

the form

y=−^12 x+c

If this line passes through (−1, 2) then substituting this point in the

equation gives

2 =−^12 (− 1 )+c