Expert: Josh - 4/14/2006

Question
I need the help with the following questions

1. The graph of the equation 2x-y+4=0 crossed the x axis at x=

2. An equation of th eilne passing through (4,-5) having slope 3/7 is

Thank you very much fo your help

Answer
Hi Tim,

Q1. The given equation has the form a*x+b*y=c, where a=2,b=-1 and c=-4. So, it represents a straight line. It crosses the x-axis where y=0. Putting this into the equation, solving for x will give you the x-intercept.

2x-y=-4, substitute y=0.
Immediately, we get 2x=-4. This implies the line crosses the x-axis at x=-2.

Q2. Remember that the equation for a straight line may be written in the gradient form, as

y=mx+b, where m=slope, b=y-intercept (i.e., where the line crosses the y-axis).

Depending on the info you've been given, sometimes, it's easier to consider (y-y1)=m(x-x1).

Use this when you're given a point and the slope.
Here, let (x1,y1)=(4,-5) and m=3/7.

(y-(-5))=(3/7)*(x-4)
7(y+5)=3(x-4)
7y+35=3x-12

You can rearrange this as
7y=3x-47.

Check that this indeed is the equation for a line passing through (4,-5). How? If the point satisfies the equation, it must lie on the line.

Does 7y=3x-47 when x=4,y=-5?
Well, the left hand side LHS=7*-5=-35.
The right hand side RHS=3*4-47=-35.
LHS=RHS, so we've got the equation right:)