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The point $(4,1)$ is the midpoint of $(a,b)$ and $(-1,5)$.

Find the values of $a$ and $b$ considering this statement.

I know the midpoint formula is: $$ \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right) $$ But I do not know how to apply it.

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Wait..... How do i do fractions!!\ – Mark Zakhem Aug 14 '13 at 9:12
In mathjax, just type \frac{x}{y} with $s on the outside to do a fraction. – user150132 May 16 '14 at 1:59


$$(4,1)=\left(\frac{a-1}2\;,\;\frac{b+5}2\right)\iff \begin{cases}\frac{a-1}2=4\\{}\\\frac{b+5}2=1\end{cases}$$

and now solve the easy system above...

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Nice detailed answer. Helped me out because I've got a page of these to do. I understand the concept now. THANKS! – Mark Zakhem Aug 14 '13 at 9:20
Any time, @MarkZakhem . Now don't forget (1) to upvote all the answers you found helpful, and after some time has ellapsed, (2) accept the answer you find the most helpful of them all. – DonAntonio Aug 14 '13 at 9:23
Shouldn't there be single l in elapse? – Ramit Aug 14 '13 at 9:27
Shouldn't there be an "a" between "be" and "single"? Just kidding...yes, apparently that word has a single "l" but I don't worry too much about that: first, english is my third langauge, and second, I'm not writing a paper or something like that. – DonAntonio Aug 14 '13 at 9:30

Midpoint formula is $\frac{x_1 + x_2}{2}$ for x, so $8 = a - 1$

$a$ is thus $9$

Solve for $b$ in the same way.

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If $(4,1)$ is the midpoint, you can plug that in to the midsegment theorem: $$4=\frac{a-1}{2}$$ Similarly, you can do that to find b. $$1=\frac{b+5}{2}$$ Solve for these equations to get $(a,b)=(9,-3)$.

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The vector from the end point you are given to the midpoint you know is $(4,1)-(-1,5)=(5,-4)$

To go from one endpoint to the other you need to double the difference. I suggest drawing a diagram.

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Try drawing diagrams, and understand the formulas. Just memorizing formulas in grade 9 and 10 is a recipe for failing grade 11 and 12.

Notice that $(-1,5)$ is $5$ units to the left of $(4,1)$. Can you guess how many units to the right of $(4,1)$ that $(a,b)$ is?

Moving vertically, $(-1,5)$ is $4$ units above of $(4,1)$. Can you guess how many units below $(4,1)$ that $(a,b)$ is?

It is clear that the horizontal distance from $(-1,5)$ to $(4,1)$ is the same as the horizontal distance from $(4,1)$ to $(a,b)$. Another way of saying that is that $(4,1)$ lies half-way between the endpoints, so you could consider the midpoint $(4,1)$ as the average of the 2 endpoints.

Thus $$4=average(-1,a)=\frac{-1+a}{2}$$ and $$1=average(5,b)=\frac{5+b}{2}$$

I hope that helps

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