Find the value of $(a,b)$

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 5 Answers Hints: $$(4,1)=\left(\frac{a-1}2\;,\;\frac{b+5}2\right)\iff \begin{cases}\frac{a-1}2=4\\{}\\\frac{b+5}2=1\end{cases}$$ and now solve the easy system above... - 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$8 = a - 1a$is thus$9$Solve for$b$in the same way. - 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$(a,b)=(9,-3)$. - 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. - 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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