Is there a better way to do this?

[allegansveritatem]

Here is problem:


Here is my solution:


The way I did it seems to rely a little to much on mother with than on algebra---not that algebra has no connection with mother wit, mind you.

 

[MarkFL]

Another approach would be to observe that the angle of inclination of the line \(y=x\) is:

[MATH]\theta=\arctan(1)=\frac{\pi}{4}[/MATH]
Hence:

[MATH]\cos(\theta)=\frac{x}{4}\implies x=4\cdot\frac{1}{\sqrt{2}}=2\sqrt{2}\implies y=2\sqrt{2}[/MATH]
These are equivalent to the coordinates you found.

 

[Romsek]

\(\displaystyle C = \left(4\cos\left(\frac \pi 4\right),~4\sin\left(\frac \pi 4\right)\right)=

\left(2\sqrt{2},~2\sqrt{2}\right)\)

 

[MarkFL]

You could also use the fact that the diagonal of a square is \(\sqrt{2}\) times the length of the sides of the square, and write:

[MATH]x=y=\frac{4}{\sqrt{2}}=2\sqrt{2}[/MATH]

 

[allegansveritatem]

Another approach would be to observe that the angle of inclination of the line \(y=x\) is:

[MATH]\theta=\arctan(1)=\frac{\pi}{4}[/MATH]
Hence:

[MATH]\cos(\theta)=\frac{x}{4}\implies x=4\cdot\frac{1}{\sqrt{2}}=2\sqrt{2}\implies y=2\sqrt{2}[/MATH]
These are equivalent to the coordinates you found.

I don't know trigonometry...I was aware that if I knew more about triangles I could have done this differently. But is there a way to solve this for someone who has studied algebra for several years but is not quite up to trigonometry yet.

 

[allegansveritatem]

\(\displaystyle C = \left(4\cos\left(\frac \pi 4\right),~4\sin\left(\frac \pi 4\right)\right)=

\left(2\sqrt{2},~2\sqrt{2}\right)\)

no trig.

 

[Otis]

Is there a better way to do this? ...

That's a subjective question. Its answer depends upon what a person knows and their definition of "better".

?

 

[Otis]

You could also use the fact that the diagonal of a square is \(\sqrt{2}\) times the length of the sides of the square, and write:

[MATH]x=y=\frac{4}{\sqrt{2}}=2\sqrt{2}[/MATH]

That's what I did, Mark, after realizing the exercise is the same as asking, "Where does a circle of radius 4 units intersect the line y=x?". So, I forgot about circles -- visually placing a 4-unit square in Quadrant I with its lower-left corner at the origin. Using the fact above allows calculating the upper-right coordinates via mental calculation.

I'm not saying that approach is better; it's just different.

?

 

[pka]

no trig.

I agree no trig. Let \(\displaystyle C: (c,c)\) be the centre of the red circle.
Then the length of \(\displaystyle \|\overline{OC}\|=4\) which implies that \(\displaystyle c=2\sqrt{2}\).
So your work is correct.