Coordinate geometry with area of triangle

[americast]

Let A(1, 2), B (3,4), C( x, y) be points such that (x- 1) (x-3) +(y-2) (y-4)=0. Area of triangle ABC=1.
maximum number of positions of C in the xy plane is

(a) 2

(b) 4

(c) 8

(d) None of these

I have tried using the staircase formula which gives me something like x-y=2. Therefore I see only two possibilities: (1,3) & (2,4) which are in accordance with the equation given. So, acc to me, ans is a. But according to the book,the answer is b.

Any help would be appreciated. Thanx in advance...

 

[Mentallic]

Let A(1, 2), B (3,4), C( x, y) be points such that (x- 1) (x-3) +(y-2) (y-4)=0. Area of triangle ABC=1.
maximum number of positions of C in the xy plane is

(a) 2

(b) 4

(c) 8

(d) None of these

I have tried using the staircase formula which gives me something like x-y=2. Therefore I see only two possibilities: (1,3) & (2,4) which are in accordance with the equation given. So, acc to me, ans is a. But according to the book,the answer is b.

Any help would be appreciated. Thanx in advance...

Your choices don't satisfy the condition.

(x,y) = (1,3)

(x-1)(x-3)+(y-2)(y-4) = 0+(-1)(-3) = 3 $\neq$ 0

Anyway, the condition given is a circle and it can be shown that A and B both lie on the circle and AB is in fact the diameter. Hence ABC would create a right angled triangle. By inspection, ABC has max area when it's isosceles (and can be shown to be greater than 1 at this point) and of course a min value of 0 when C=A or C=B. Hence the area must be 1 somewhere in between. Due to the symmetry of the problem, it can do this in each quarter of the circle, so the answer must be a multiple of 4.
If you can prove that it happens to have an area of 1 only once in each quarter, then you've found your answer b), else it would be c) or d).

 

[americast]

Great! It worked... B is the correct answer...!

And yes, I was terribly wrong in my previous answer...

Thanx again...

 

[Mentallic]

What was your argument to conclude that it was b) and not c) or d)?

 

[thelema418]

Another method is to use the formula $A = 1/2 \cdot b \dot h$. You can calculate the base using the distance formula for two points. You know the area, so you need to calculate the "height".

Next you would find the equation of the line created by the points A and B. Then you would write equations for lines that are parallel to A and B, but higher and lower by distance $h$. You can do all of this on graph paper without exact calculation of the points.

 

[Mentallic]

Another method is to use the formula $A = 1/2 \cdot b \dot h$. You can calculate the base using the distance formula for two points. You know the area, so you need to calculate the "height".

The triangle created is a right triangle, so $A = 1/2\cdot ab$ where a and b are the non-hypotenuse legs is sufficient and easier.

I don't believe this helps get us closer to the answer though.

 

[thelema418]

The benefit is that the method relies mainly on distance measured in the plane, and not many other geometric propositions.

There is a lot of sophistication in calculations that involve geometric propositions. You found A and B are on the circle. What would you do if they weren't? How do you know that they are the diameter of the circle? The use of the inscribed angle property is contingent upon this finding this information. You report a max area and minimum areas and then essentially use a squeezing principle. That is based on a lot of other assumptions. Yes, it is a method that works -- and it involves a lot of big concepts.

My distance method works whether or not A and B creates a diameter, a chord, a tangent, or no intersection with the function for a circle. You can also change the function to a hyperbola, parabola, periodic wave, etc.

Unfortunately, I can't upload a Geogebra file here -- just pictures. If you drag the point C across either parallel line, the triangle keeps an area of 1. You can just count the intersection points of the two parallels with the circle. Or you can manually calculate them if you wish.

You can do this on grid paper easily with a straightedge and compass without reasoning about minima and maxima and the continuity of the function, etc.

 

[americast]

What was your argument to conclude that it was b) and not c) or d)?

The radius is 2sqrt(2). The height required for the triangle to be of area 1 unit is 1/sqrt(2). The max height Is 2sqrt(2) (radius).1/sqrt(2) being smaller can be on either side. Since the base it the diameter, it is a right triangle. For both sides it would be 2*2=4.

Thanx...