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I have two points, A (4,3) and B (k,h). B is an unknown point that is located 10 units along the slope of A (4,3) which is y = y = -4/3 x + 25/3.

The question asks me to move 10 units (distance = 10) from point (4,3) located on the line represented by y = -4/3 x + 25/3. Any ideas on how to tackle this one?

Imagine Dragons
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2 Answers2

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The points at a distance of $10$ from $(4, 3)$ are given by $(x - 4)^2 + (y - 3)^2 =100$. The points on your line are $y = \frac{25 - 4x}{3}$. Two equations in two unknowns. Solve them, and you will get two solutions (a circle and a line passing through its center will intersect in two points). Choose the solution that is "upslope".

Paul Sinclair
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This problem is easy using the concept of a vector. Note, there are two solutions to this answer.

A vector [x,y] represents a change in x and change in y. We can represent movement along a line of slope -4/3 with the vector [3,-4] or [-3,4] which point in opposite directions (two of infinite possibilities). These vectors (like slope) can be thought of as making two triangles with legs 3 and 4, so their length is the hypotenuse of such a triangle of length 5. Therefore, we need just apply each vector twice to (4,3). In the first case, we get (4,3)+2[3,-4] = (10,-5) and for the opposite direction, we get (4,3) + 2[-3,4]= (-2,11).

Screw systems of equations. Effective, but take way too long.

Winston
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  • Oh my gosh, that's really fast! But what if I don't know anything about vectors. In this case, they act like slopes? @Winston – Imagine Dragons Sep 19 '15 at 20:49
  • Vectors and slopes while different concepts have similar meanings in two dimensions. A vector can be thought of as a pointed line segment with an arrow end with a length, where (sort of opposite to slope) its described by the amount it changes in x and the amount it changes in y from start to arrow. One key distinction between a vector and slope is this sense of direction. – Winston Sep 19 '15 at 20:53
  • How did you get this: "We can represent movement along a line of slope -4/3 with the vector [3,-4] or [-3,4]." And how do you know you have to apply these vectors twice? @Winston – Imagine Dragons Sep 19 '15 at 20:57
  • Yes. A simplified way to look at it is, -4/3 represents rise in 4 (change in y) and run of -3 (change in x) i.e. the vector [-3,4]. The other vector just has the signs switched. The length of the vector is 5, so if I apply it twice I've gone 10. It'd be super useful to draw it out. – Winston Sep 19 '15 at 20:59
  • But how do you know that this vector needs to be multiplied twice then added to the coordinate? Why does each vector represent half the distance? @Winston – Imagine Dragons Sep 19 '15 at 21:01
  • Sorry, see my edit above. – Winston Sep 19 '15 at 21:02
  • How can you calculate a length of a vector? @Winston – Imagine Dragons Sep 19 '15 at 21:04
  • Draw an arrow or segment on a piece of graph paper by going 3 left and up 4 from a point, and connect the start and finish. How far have you gone? Think right triangles. – Winston Sep 19 '15 at 21:05
  • Pythagorean Theorem, then because 3 ^ 2 + 4 ^ 2 = 5 ^ 2, got it. Can this method be applied to any problem like this?

    I'm just confused about how you created those vectors [3,-4] or [-3,4] for slope -4/3.

    So if I were to do the same thing for something like slope -25/8, the vectors would be [-8,25] and [-25,8]. And the length of each vector would be -25 ^2 + 8 ^ 2 = c ^ 2 ... Therefore c ^ 2= 689 hence distance per vector is 26.25? @Winston

     – Imagine Dragons Sep 19 '15 at 21:10
  • Yes good! You switched the second vector. It should read [8,-25]. And the lengths are correct. Here's the thing, consider a line say y=5x. The slope of this line is 5, take the point (0,0) on the line and the points (1,5) and (1,-5). How have we gotten there in terms of change in x and change in y? Unfortunately, we are oversimplifying vectors, there are really an infinite number of vectors that can describe the slope of a line. We've just chosen convenient ones. – Winston Sep 19 '15 at 21:17
  • You are really helpful @Winston these solutions were amazing. I never thought about it this way, then again, I never learned about vectors yet but this motivates me now :). Is there a place where you learned all of this? or was it just your school? Thanks again, you are awesome! – Imagine Dragons Sep 19 '15 at 21:21
  • @ImagineDragons thank you! I am a teacher, and I had the fortune of seeing some neat curricula in my time. Look at the Exeter math texts. – Winston Sep 19 '15 at 21:33