Understanding Engineering Mathematics

The gradient ism=AP/BP=tanθ.
Two special cases require comment. For a horizontal line (i.e. parallel to thex-axis) the
equation isy=c(a constant), since the gradient is zero. For a vertical line (i.e. parallel
to they-axis) the equation is given byx=c.
Sometimes, if we want to tidy up fractional coefficients for example, we write the
equation of a line as

ax+by+c= 0

Solution to review question 7.1.4

A.The equation of the straight line through the points(x 1 ,y 1 ),(x 2 ,y 2 )is

y−y 1 =

(

y 2 −y 1

x 2 −x 1

)

(x−x 1 )=m(x−x 1 )

wheremis the gradient. We can choose either of the given points for

(x 1 ,y 1 ).

(i) With (x 1 ,y 1 )=( 0 , 0 ), (x 2 ,y 2 )=( 1 , 1 )the gradient is m= 1

(210

➤

) so the required equation is

y− 0 = 1 (x− 0 )

or y=x

(ii) This question illustrates the limitations of general formulae. Using

the general result would give

y− 2 =∞(x− 1 )

which is not defined (6

➤

). In such circumstances we go back to

first principles, and look again at the points, (1, 2), (1, 3). Here

thex-coordinate always remains the same, so the line must be

parallel to they-axis. It is in fact the linex=1.

(iii) For the points (−2, 4), (1−3) the gradient is−^73 and the equation

becomes

y− 1 =−^73 (x− 3 )

or, tidying this up

7 x+ 3 y− 24 = 0

(iv) For the points (−1,−1), (−2,−3) the gradient ism=2andthe

line is

y−(− 1 )= 2 (x−(− 1 ))

or

2 x−y+ 1 = 0