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.
 A: 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.
A: 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...
A: 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 $(a,b)=(9,-3)$.
A: 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.
A: 
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
