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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.

alkabary
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almost a beginner
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2 Answers2

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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 $M$.

TonyK
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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 $2$.

MathNewbie
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  • Thanks, that helped a lot, now I just have to get used to Gaussian elimination. – almost a beginner Jun 08 '15 at 08:25
  • No worries. I'm glad you got something out of it. :) – MathNewbie Jun 08 '15 at 08:29
  • I tried the Gaussian elimination but didn't get the right answer, I was wonder if you could help out, here is what I did: 4t-12s = -4 : -9t+s = -4 : (r2*12)+r1 -- > -104t = -4 : -9t-s = -4 : r1/-104 --> t = 0.04 : -9t-s = -4 : -9(0.04)-s = -4 : -0.36+4 = s : s = -3.64 which is clearly wrong lol.. Where am I going wrong with this. Thanks in advance. – almost a beginner Jun 08 '15 at 09:15
  • Sure. Let's look at equation $1: 4t-12s = -4$ and equation $2: -9t + s = -4$. We can divide both sides of equation $1$ by $4$, so that you get that equation $1$ becomes $t - 3s = -1$. We can multiply equation $1$ by $9$ and add it to equation $2$ to get $-26s = -13$. Divide both sides by $-26$ and we get $s = \frac{1}{2}$. Looking at our original equation $1$, we have after substituting $s = \frac{1}{2}$ that $4t-12(\frac{1}{2}) = -4$. This simplifies to $4t - 6 = -4$. Simplifying it, we have $4t = 2$. Divide both sides by $4$, we get $t = \frac{1}{2}$. – MathNewbie Jun 08 '15 at 09:36
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    @almostabeginner, I see your error in what you wrote. In the first step, you wrote $(r2*12)+r1 -- > -104t = -4$. It should read $-104t = -52$. Double check this step. It should come out really nice. – MathNewbie Jun 08 '15 at 09:43
  • Ohh, I can't believe I missed it. Thanks a lot :) – almost a beginner Jun 08 '15 at 11:10