View Single Post
# 6 19-12-2010 , 12:49 AM
ctbram's Avatar
Moderator
Join Date: Jan 2004
Location: Michigan, USA
Posts: 2,998
There are generally two useful forms to describe a line, the one everyone remembers (slope-intercept), which is generally less useful and the more useful (point-slope) form. The reason (point-slope) form is more useful is because we are almost always given enough information in a question to derive a point and the slope directly.

Here are the two forms:

(slope-intercept): y = mx + b ;where m=slope=dy/dx and b=the y-intercept (eq.1)

(point-slope): y-y1 = m(x-x1); where m=slope=dy/dx and (x1,y1) is a point on the line (eq.2)

So for your first problem we are given p1(10,15), p2(150,300)

therefore the slope m=dy/dx=(y2-y1)/(x2-x1)=(300-15)/(150-10)=285/140 which is ~2.04. (eq.3))

So now we have all we need to determine the equation for the line using (point-slope) form (eq.2) and using p1(10,15) we get:

L1: y-15 = (285/140)(x-10)
y=(285/140)x-(2850/140)+15
y=(285/140)x-(2850/140)+(2100/140)
y=(285/140)x-(750/140) <==== here is your first answer the eq of L1

Therefore the y-intercept of this line using (eq.1) is b=-(750/140) ~ -5.4

To check lets use p2(150,300) and (point-slope) (eq.2) to verify that

y=(285/140)x-(750/140) substituting p2(150,300) we get

300=(285/140)(150) - (750/140) ................ which is true


"If I have seen further it is by standing on the shoulders of giants." Sir Isaac Newton, 1675

Last edited by ctbram; 20-12-2010 at 01:36 PM.