Expert: Josh - 10/9/2003

Question
Please help with this question.  FInd the equation of the line that passes through
A(-1.8,2.1) and B(2.4,-0.5)
b) What is the length of AB?

Answer
Hi Katie,

My advise for you is not to focus on tackling just this question, but understand how to approach this type of problem in general.

Given two points, A=(a1,a2), B=(b1,b2), the cartesian equation for the straight line passing through these two points is given by,
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(x-x1)=m*(y-y1),
where gradient m=(b2-a2)/(b1-a1).
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Interpretation:
-On the LHS, the term (x-x1) measures the distance away from a known point x1 on the horizontal x-axis.
-Similarly, on the RHS, the term (y-y1) measures distance with respect to the point y1 on the vertical y-axis.
-You can pick any point known to be lying on the curve for (x1,y1). You are at liberty to choose (x1,y1) to be point A or point B, for instance.
-Example: suppose, we pick (x1,y1)=(a1,a2)=(-1.8,2.1), then we assign x1=-1.8, y1=2.1.

The straight line equation now becomes,
(x-x1)=m*(y-y1) ...watch out for the minus sign -(-)=+
(x-(-1.8))=m*(y-2.1)
(x+1.8)=m*(y-2.1) ...[#1]
which means, horizontal distance is measured from x=-1.8 on the x-axis, and vertical distance is measured separately, from y=2.1. OK?

-Finally, what is m? "m" is the slope of the straight line. Commonly, understood as "rise over run", this is precisely what m=(y2-y1)/(x2-x1) represents, given two points (x1,y1), (x2,y2) on the curve.
-The way we have defined the 2 points in this case is A=(a1,a2), B=(b1,b2). So, m=(b2-a2)/(b1-a1).
If you remember that the numerator is concerned with the vertical (y) coordinate and the denominator is concerned with the horizontal, (x) coordinate, you will not be confused.

Exercise: Calculate the slope m for yourself. Just plug in the numbers.
Solution: m=(-0.5-2.1)/(2.4-(-1.8))=-2.6/4.2~=0.6190 ...[#2]

If m is negative, (m<0), the slope is like this If m is positive, (m>0), the slope is like this /
If m is zero, (m=0), in the case of a straight line, it is flat like this _.

Exercise: Put everything together, using [#1],[#2] to obtain the answer for the equation of a straight line.

b) Distance formula (motivated by Pythagoras theorem)
distance between point A and B, which we denote as d(A,B) is given by,
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d(A,B)=sqrt[(b2-a2)^2+(b1-a1)^2] ...[#3]
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sqrt reads square root, ()^2 means square the term enclosed by parenthesis.

Question: How do you appreciate this result and not have to remember another piece of useless information that you already know?

Observe that if you square both sides of [#3], it is nothing more than Pythagorean theorem, which you would have learnt in your foundation study of trigonometry.

For a right angle triangle (RAT), the squared distance of the hypotenuse (longest slanted side) is equal to the sum of the respective squared distance of the other two sides.

In the simplest case, if (b1,b2) is measured with respect to A=(a1,a2), where A is the origin (0,0), then, the equation collapses to
d(0,B)=sqrt[(b2)^2+(b1)^2].
Upon squaring both sides, you must recognize
d^2=(b2)^2+(b1)^2
it is of the same form as
d^2=x^2+y^2, right?
which describes a circle centered at (0,0) with radius d.
A circle which is positioned at (a1,a2) (off-centered with respect to some reference grid) has non-zero a1, a2 values.

Can you reconcile the ideas of a circle and Euclidean distance? Answer: the circle is a locus, namely, a collection of points (b1,b2) which are equidistant (d) from some reference point (a1,a2). This is the whole point behind using a compass in geometry.

Exercise: I could have given you the answer, but I think you can work it out by yourself and indeed, all questions of the same nature (go over what we have discussed above). I want you to understand where the results come from and make sure that the results are intuitive to you, not just blindly follow some rules given in the textbook. When it makes sense to you, you will not find it difficult.

Cheers,
Josh