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For Collin Li and n.f's question here, I had the same problem as the original question (maybe for a different reason?)

Exercise 1D - Question 3A

Find the equation in the form ax+by+c=0 for the straight line passing through points (-3,-4) and (-1,-10)

The answer in the book (and the one that I got) is: 3x+y+13=0

I would have gone: (y2-y1)/(x2-x1) to find the gradient, then just substituted it into the equation y=mx+c, along with the any one of the co-ordinates to get the c - then I would have just written out the equation.

So: -4 = (-14/-4)(-3) + c

Correct?

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Great use of Merspi to chase up a follow-up question - new questions should be asked as new questions, great work! I've added the relevant link (for others to peruse) for you. – Collin Li♦ Dec 22 at 10:30

2 Answers

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By subbing into the formula y-y1=m(x-x1) after finding your gradient (y2-y1)/(x2-x1) you will get the result of 3x+y+13=0

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I see where your problem is james.

Your gradient was calculated incorrectly. You should have:

  • m = [-10 - (-4)]/[-1 - (-3)] = -6 / 2 = -3

You had m = (-14/-4), because you added y2 and y1 together instead (and same for x too).

If you proceed with your method with m = -3, you will have no problems getting this right. Please don't hesistate to comment on this if you are still having troubles getting there.

BTW, n.f's answer is also adequate, but if you prefer your way (y = mx + c) instead of the bigger (y-y1=m(x-x1)) equation, then you can stick with your way!

Personally, I do it your way james, I find the big equation clumsy and hard to remember sometimes.

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WOW! that's exactly what I think! appreciate the help! – Impractical Dec 22 at 13:41

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