Math Is Fun Forum

Yes. That's right.

Ok, cool.

So the problem reduces to finding the equation of the perpendicular bisector of two points.

I'll do it with a numerical example.

Say the two points are A(1,3) and B(2,9).

We use this simple fact:

The product of the gradients of perpendicular lines is -1.

So, the gradient ot the line through AB is (9-3)/(2-1) = 6

So the gradient of a line perpendicular to AB is -(1/6) because 6 * -(1/6) = -1

Cool, so you know the gradient of the perpendicular bisector - call it m.

Now you need the intercept of the bisector - call it c. Well, we know the bisector contains the midpoint of AB which in this case is ((1 + 2) / 2,(3 + 9)/2) = (3/2,6) so to find the intercept we use y = mx + c with say x = 1 and y = 3 and m = -(1/6) which gives c = 19/6 so the equation of the perpendicular bisector is y = -(1/6)x + 19/6.

I'm sure you'll be fine now. Let me know.

In the Cartesian system, a straight line has an equation of the form y = mx + c where m and c are contants.

So in other words, if you are given a point (x1,y1) and you can show that
y1 = m * x1 + c then the point (x1,y1) must be on the line. Further, if x1 and y1 cannot be related by the expression y1 = m * x1 + c then the point (x1,y1) is not on the line. So the equation y1 = m * x1 + c defines a propertiy which all points on the infinite line share and all points off of the infinite line do not share. This is why the expression is a useful one for defining a straight line in Cartesian space.

So , say we were asked to find the Cartesian equation of the line which passes through the point A(2,3) in the direction of the vector i+j.

This is pretty staight forward:

(2,3) is on the line as is (3,4). We know (3,4) is on the line as we can move to point (2,3) and travel in the direction i+j (one unit in the x direction and one unit in the y direction) to arrive at point (3,4).

Now the m in our Cartesian equation is the gradient and this will be (4 - 3) / (3 - 2) = 1 so the equation we seek is y = x + c. To find c just substitute in a known point on the line (say (2,3)) to give 3 = 2 + c therefore c = 1 so the equation we seek is y = x + 1. In other words, y is one more than x. So, (10,14) is not on this line as 14 is not one more than 10. But (-4,-3) is on the line as -3 is one more than -4.

Now, what if we were asked to find the vector equation of the line which passes through the point A(2,3) in the direction of the vector i+j.

Well, the direction of the line is i+j, let's call this vector b. So b = i + j. Now, just to be conventional let's find the unit vector in this direction. Well, that's easy enough. It's:

Now, the position vector of A is a = 2i + 3j. So we need to define a property which is true for all points on the desired line and not for all points off of the desired line.

So we could write r = a + tb' where t is any scalar number.

So we are saying that for all points on the line, r (remember, r is the position vector of any point on the line and is analogous to the (x,y) of any point on the line in Cartesian space) = a + tb'.

So, if a point is on the line, it's position can be written as a + tb' but if it is off of the line then it's position cannot. So, how about (-4,-3) is that on the line?

Well, its position vector is -4i-3j. and we can write that as:

So we have shown that the point (-4,-3) is on the line defined by the vector equation r = a + tb' as it's poition r can be expressed as the vector a plus minus-six-root-two lots of b'.

A point such as (10,14) with position vector 10i+14j cannot be so expressed and so is not on the line.

Incidentally, one need not use the unit vector (here b') in the direction of the line, but could use any vector in that direction. Obviously this would simply mean that each point on the line would have a different value for t. This is analogous to y = 2x + 5 and 3y = 6x + 15 being two expressions for that same line. they are the same equation.

So equation of a tangent to the curve at any point x is:

So say we have two points which have a common tangent, (a,b) and (c,d) then the x coordinates will have to simultaneously satisfy:

So we "just" have to solve these for a and c where a is not equal to b.

Ok, then I "cheat" and use mathematica:

eqns = {4a^3 - 4a - 1 == 4c^3 - 4c - 1, -3a^4 + 2a^2 == -3c^4 - 2c^2}
Solve[eqns, {a, c}]

Whicih gives six solutions including:

and

The others are similarly unpleasant.

This solution gives a is approx. 0.853 and c is approx 0.247 and if you look at the graph of the function at these points:

Plot of x^4-2x^2-x

This looks reasonable.

Plot of (3X^2-X+1)/(X+3)

Here you can see that as x tends to -3 from -infinity the value of the function is heading to -infinity.

You might have entered the function incorrectly?

Mitch.

Hi,

There's a mistake in you're second line of algebra. The proof to the point you have taken it should read:

Now. A fraction will be negative when the numerator and denominator are of different signs. So we need to look for where

and

So we see that the denominator and numerator will be of different signs when:

This is you're answer.

An easy way to see when the two functions are of different sign is to draw out a number line and above it draw a line where the numerator is less than zero, and below it draw a line where the denominator is less then zero. Then, for whichever ranges the lines are not both absent or both present, these are the areas of the number line which are your answer.

5)

Hence:

3)

Hence:

Call point on x-axis P. It will have coordinates (x,0). We need distance from P to (6,2) to equal distance from P to (10,4).

Formula for distance between to points (a,b),(c,d) is:

So we require:

So square bot sides and expand giving:

Cancel square of x from both sides and rearrange giving:

therefore

Therefore P is the point:

Hi.

Think of percentages as hundreths. So "2 percent of" means "two hundreths of" or "2/100 of".

So, one hundreth of 4.91 is 0.0491.

Now however many times this goes into 0.74 is how many hundreths 0.74 is of 4.91

So, 0.74 is 0.74/0.0491 = 15.071 hundreths of 4.91 i.e. 0.74 is 15.071% of 4.91 to three decimal places.

So in general divide your total amount by 100 to get one hundreth of it and then divide this into the number you want as a percentage.

So if you want M as a percentage of N you calculate M*100/N = 100M/N

we need:

Okay. Cross multiply the RHS giving:

Now subtract RHS from both sides giving:

Cross multiply to give:

Tidying gives:

Factoring top and bottom gives:

Calcelling gives:

Now a quotient of two functions will be less than zero when the two functions are of opposite sign. We can easily see that -(x-3) <= 0 when x >= 3 and also that 3(x-5) <= 0 when x <= 5 therefore the quotient of the two functions will be less than zero when 3 => x >= 5.

So the answer is:

Not equal to 5 as this is an asymptote as is -5 from the pre-cancelled equation.

Mitch.

Hi,

No, I never got a reply to this or the subsequent one. Apart from a triple angle substitution and lots of algebra, I have no other ideas. The context of the problem is in a caculus text. I have arrived at the first formula and the answer given is in the second form. Given that it's a calculus question I would be suprised if one had to do lots af algebra at the end just to get into the form given - still, it's possible. Moreover. I was looking for a quicker method which I had missed.

Thanks,
Mitch.