This is a question from my school text book:-

 

The line y = 3x - 5 meets the x-axis at the point M. The line y = -2/3x + 2/3 meets the y axis at the point N. Find the equation of the line joining the points M and N. Write your answer in the form ax + by + c = 0.

 

For point M i got the coordinate (5/3, 0)

and for point N

(0, 2/3)

 

I next worked out the gradient of the line connecting M and N to be -2/5 and tried using y - y1 = m(x - x1) to get the equation for the line eventually getting to 6x - 15y - 3 = 0, however the book lists the answer as 6x + 15y - 10 = 0.

 

Is my answer the correct answer and the book's answer wrong, or did i mess up in my working out somewhere? I've been pondering on it for a while and cant see how the book got its answer.

 

Thanks for any help

Rob

Check your work again because I worked it out and got the books answer. But don't worry to much, a lot of the time the book can be wrong

The book's answer is correct. So are your steps up to everything that you've shown. (points are correct; slope is correct)

 

Perhaps you made a mistake substituting (x1,y1) in the final equation. Which point did you use - M or N ? I suggest you recheck this last bit of working, or show what you did.

y - 2/3 = -2/5 x 5/3

 

-5y + 2/3 = -5/3 - 2x

 

-15y + 2 = -5 -6x

 

6x -15y + 7 = 0

I think I see your problem.

y-y1=m(x-x1)

You have chosen the point (0,2/3)

therefore x1 = 0

y1=2/3

That will give the right answer

Whatever you just did really is confusing. Here's how I did it:

One of your points is (0, 2/3) so you know your y intercept is 2/3. The slope -2/5 is right, so I just plugged it into the equation y=mx+b and got:

y=-2x/5+2/3

Then multiply both sides by 15 and get:

15y=-6x+10

Then just solve for zero, 6x+15y-10=0

 

Edit:Well Mobius found your error and explained it before me, oh well. At least I showed you another method to do it.

Cheers guys, see where i went wrong now. Having just had a rather long break from maths over the summer months my brain is half asleep most of the time.