You are given the two points A (1, 4) and B (13, y ). The two points are 13 units apart. What are two possible locations of point B ?

Question

You are given the two points A (1, 4) and B (13, y ). The two points are 13 units apart. What are two possible

locations of point B ?

I'm really confused please help!

Answer 1

These types of problems always come down to drawing right triangles to use Pythagorean’s theorm

Draw a triangle with point A at (1,4) and point B at (13,y). Now drop a vertical line from point B and a horizontal line from point A. They will meet at a right angle at point C which will have coordinates of (13,4)

Now you have a right triangle ABC with AB=13(given) AC is 12(13-1)

The length of CB is (y-4) which by Pythagorean theorem is 5

So y-4=5 y=9 so point B is at (13,9)

Can draw the mirror image triangle where y-4= -5 y=-1 which would give coordinates of (13,-1)

Answer 2

Hi Bee!

Let's try to envision the problem as we work through it.

Imagine a graph with a point and a line. The point is (1, 4) - that's 1 step to the right from the y-axis, and 4 steps up from the x-axis. There's a line 12 more steps to the right hitting every point where x=13. We want to know where you would land on that line if you were to walk exactly 13 steps, starting from Point A moving in whatever directions.

Here's how we find that mathematically:

The direct distance between any two points can be found using the Pythagorean theorem. Let's draw a couple triangles with these three points to better illustrate:

Y

↑ B ·

| / |

| c / | b

| / |

| / a |

| A ·-------

| \ |

|-------\----|-------------→ X

| c \ | b

| \ |

| B ·

↓

a² + b² = c², where a is the left-right width, b is the up-down length, and c is the direct distance between points A and B. (For the Pythagorean theorem, it's important that b is perpendicular to a.)

Our goal is the two possible y-values that are 13 steps away from Point A AND on the x=13 line. (You could, of course, take 13 steps in any direction and draw a whole circle around Point A, but only 2 points would be on x=13.)

We are given that c = 13 for "13 units apart."

We can find a from the known x-coordinates, 1 and 13. And we can set up a relationship between our known and unknown y-coordinates.

a = 13 - 1 = 12

b = y - 4

Now we're left with our 2 unknowns, so let's plug it all into the Pythagorean theorem and solve for y.

12² + (y - 4)² = 13²

(y-4)² = 169 - 144 = 25

√(y-4)² = √25 = ±5 ; remember that +5 or -5 can both square to become 25.

y - 4 = ±5

Let's split this up into two equations:

y - 4 = 5 ; y - 4 = -5

+4 +4 +4 +4

y = 9 ; y = -1

Likely, you might be more used to the Distance Formula, whereas I focused more on the Pythagorean theorem, but that's how the P-theorem fits into the Distance Formula. You might expect more to see the solution started off like so before meeting up with what I've shown here:

a² + b² = c²

(x2 - x1)² + (y2 - y1)² = 13²

(13 - 1 )² +( y - 4 )² = 13²

So there you have it! Let me know if you have any questions, and please reach out any time.