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### Coordinates of a Point

```
Date: 6/12/96 at 9:28:16
From: Wong Cheong Siong
Subject: Coordinates of C

Hi Dr. Maths,

ABC is a right-angled triangle labeled counter-clockwise with its
point C lying on the line y=3x. A is (2,1) and B is (5,5). Find the
two possible coordinates of C.
```

```
Date: 6/12/96 at 13:16:35
From: Doctor Anthony
Subject: Re: Coordinates of C

You have not provided all the information that is required.  There are
in fact four possible points where C could be, depending on where the
right-angle is.  If it is at C there are two possibilities, if at A
there is a third possibility and if at B there is a fourth
possibility.  I will assume you meant point C, since that gives two
possibilities.

The easy way to do this is to write down the equation of the circle on
AB as diameter, and then find the two points where the line y = 3x
cuts this circle.  These will be the two possible points C.

The distance AB is sqrt((5-2)^2 + (5-1)^2) = sqrt(3^2 + 4^2) =
sqrt(25) = 5

So radius of the circle is 2.5, and centre of circle is given by

{(5+2)/2, (5+1)/2)} = {3.5, 3}

The equation of the circle is therefore

(x-3.5)^2 + (y-3)^2 = 2.5^2

Now we put y = 3x in this and solve the resulting quadratic to get the
coordinates of C

(x-3.5)^2 + (3x-3)^2 = 2.5^2

x^2 - 7x + 12.25 + 9x^2 - 18x + 9 = 6.25

10x^2 - 25x + 15 = 0

2x^2 - 5x + 3 = 0

(2x-3)(x-1) = 0

The two possible answers are x = 3/2 and x = 1
Corresponding y values are   y = 9/2 and y = 3

Thus C has two possible values:

(1, 3)  or  (3/2, 9/2)

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 6/12/96 at 23:42:51
From: Wong Cheong Siong
Subject: Re: Maths Question.

Thank you Dr. Maths. Your help is greatly appreciated. However, this
method of solving has not been taught yet, so would you explain to me
more clearly how to apply this method of solving? I do not understand
where you will draw the circle and whether it will be a full or
semi-circle and where the circle will be touching which points. Your
answer to this question is correct but I wish you could explain more
clearly.

Best regards,
Wong Cheong Siong
```

```
Date: 6/13/96 at 10:53:6
From: Doctor Anthony
Subject: Re: Maths Question.

The circle is drawn on AB as diameter, and this circle will cut the
line  y = 3x in two positions, giving the possible positions of
point C.  As you know, the angle subtended by the diameter of a circle
at any point on the circumference is a right-angle, so the condition
that the angle at C is a right-angle is guaranteed.

If you don't like the circle method for solving this problem, here is
an alternative method.

Let the point C have coordinates (k, 3k) where k is to be found.  We
require the line CA and CB to be perpendicular so we write down this
condition in terms of k.

Slope of CA = (3k-1)/(k-2)     Slope of CB = (3k-5)/(k-5)

Condition for perpendicular lines is that product of slopes equals -1.

So we require {(3k-1)(3k-5)}/{(k-2)(k-5)} = -1

9k^2 - 18k + 5 = -{k^2 - 7k + 10}

10k^2 - 25k + 15 = 0

2k^2 - 5k + 3 = 0

(2k-3)(k-1) = 0

So k has the values 3/2 and 1

This gives two positions for C, namely  (3/2, 9/2) and (1, 3)

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Coordinate Plane Geometry
High School Equations, Graphs, Translations
High School Geometry

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