2 lines passing Q and R meets at the mid-point,

Question

Consider the straight line whose parametric equation is $$(x, y) = (1, 1)+ t(12,−1)$$ Show that the above line and a line passing Q and R meets at the mid-point.

$Q = (5, 5)$ and $R = (9,−4)$

How do I approach this problem? Any help and direction would be appreciated.

Answer 1

Here's line $1$: $r_1(s) = (1+12s,1-s)$ and

here's line $2$: $r_2(t) = (5+4t, 5-9t)$.

To find out where these two lines intersect, we just need to find values of $s$ and $t$, so that $1+12s = 5+4t$ and $1-s = 5-9t$. Rearranging the terms in these two equations, we have

$4t - 12s = -4$ and $-9t + s = -4$

Solving systems of linear equation using gaussian elimination, we get that $s = \frac{1}{2}$ and $t = \frac{1}{2}$.

Remember that when $s$ varies from $0$ to $1$, this parametrization of line $1$ traces a line segment between two points. The midpoint of this line segment happens when $s = \frac{1}{2}$. Same thing happens for line $2$.

Answer 2

The mid-point of $Q$ and $R$ is just $M = (Q+R)/2$. You can work this out yourself Now you want to see if there exists $t$ such that $(1,1)+t(12,-1) = M$: choose $t$ so that $1+12t$ equal the $x$-coordinate of $M$, and check whether $1-t$ equals the $y$-coordinate of $M$.