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?
On
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".
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.