Jtdcftul

So, I've spent hours on this question and it's frustrating me way too much so I created an account for StackExchange just to understand how you solve this problem.

Let me start by sharing the question: "Sally has hidden her brother's birthday present somewhere in the backyard. When writing instructions for finding the present, she used a coordinate system with each unit on the grid representing 1 m. The positive y-axis of this grid points north. The instructions read "Start at the origin, walk halfway to (8,6), turn 90 degrees left, and then walk twice as far." Where is the present?" By the way, you have to answer the question algebraically and the answer in the textbook is (-2,11)

What I Know

Let P represent the coordinate of the present

Let M represent the midpoint between (0,0) and (8,6)

• M is at (4,3) which is solved using the midpoint formula.


• The distance from P to M is 10 m.


• The equation for the line from the origin to the midpoint is:
  y= $frac{3x}{4}$
  


• The equation for line PM is:
  y= $frac{-4x}{3}$ + $frac{25}{3}$


• Using the distance formula for two coordinates I know that the distance from M to P can be represented in this equation: $100=(4-x)^2 + (3-y)^2$ The x and y in the equation represents the x,y coordinates of the point P. I don't know how to simplify this equation further.

My troubles

I come to this point where I don't know what to do anymore. Please explain how you solved this problem.

Thank you so much

Equivalent to 4 steps east, 3 steps north, then 8 steps north, 6 steps west, resulting in (4-6, 3+8) or (-2, 11). After plotting, it's much easier to see the triangles you have to describe algebraically.

Go to (4,3), and then walk in the direction of slope -4/3, i.e., left 3 and up 4, twice as far, i.e., left 6 and up 8. So, 4-6=-2, and 3+8=11. There’s a 3,4,5 right triangle and a 6,8,10 right triangle involved.

Solving quadratic equations is serious overkill.

If you know complex numbers, the problem becomes almost trivial to solve. Complex numbers lend themselves very naturally to mapping to vectors and points on the 2-d plane.

So you start at the origin (point $0+0i$) and walk halfway to $8+6i$, which means you end up at $4+3i$. At this point you turn ninety degrees to your left (which is counter-clockwise), and the vector for the new direction can be found by multiplying the previous one by $i$. And since you're walking twice as far, this is equivalent to again multiplying the result by two. So you're now taking the path $2i(4+3i) = -6 + 8i$. The final position is simply the sum of $4+3i$ and $-6+8i$, i.e. $4+3i -6+8i = -2 + 11i$. This is the point $(-2,11)$ as required.

It is quickly checked that an orthogonal vector to $binom{4}{3}$ in the right direction is $binom{-3}{4}$ and this vector has also length $5$.

So, the algebraic solution is
$$binom{4}{3} + 2binom{-3}{4} = binom{-2}{11}$$

Hint:
You can make a second equation by considering the equation of the line because you know that the point has to be on that line.