# Two points: Distance, midpoint, equations of line passing through them and perpendicular line [closed]

Consider two points (-2,3) and (-1,6).

1. Calculate the distance between these two points.
2. Find the midpoint.
3. Obtain the equation of the line passing through the given points.
4. Find the equation of the perpendicular line passing through (1,1).
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## closed as off-topic by TZakrevskiy, Daniel Rust, Lord_Farin, Did, Davide GiraudoSep 23 '13 at 12:03

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• "Homework questions must seek to understand the concepts being taught, not just demand a solution. For help writing a good homework question, see: How to ask a homework question?." – TZakrevskiy, Daniel Rust, Lord_Farin, Did, Davide Giraudo
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please post whatever you have tried to solve the problem rather than just order other uses to solve it for you. –  Praphulla Koushik Sep 23 '13 at 11:23
Hello, welcome to Math.SE. Please note that homework should not be used as a standalone tag; see tag-wiki and meta. Also, please try to make the title of your question more informative. E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. –  Lord_Farin Sep 23 '13 at 11:54

The situation can be depicted as follows:

To answer these questions, we:

1. Use Pythagoras's Theorem on the depicted right angle triangle to find the distance between $(-1,6)$ and $(-2,3)$.

2. To find the midpoint, we take the component-wise average of $(-1,6)$ and $(-2,3)$. I.e., the first component is the average of $-1$ and $-2$, and the second component is the average of $3$ and $6$.

3. To find the equation of the line connecting $(-1,6)$ and $(-2,3)$, we know it has form $y=mx+c$, so we can solve the system of equations \begin{align*} 6 &= m \times (-1)+c & \text{since } (-1,6) \text{ is on the line} \\ 3 &= m \times (-2)+c & \text{since } (-2,3) \text{ is on the line} \\ \end{align*}

Alternatively, we know that $m=\text{rise}/\text{run}$ which we can also find from the diagram, then we can substitute in either point to find the $c$-value.

4. If $m$ is the gradient of a line in 2D Euclidean space, then $-\frac{1}{m}$ is the gradient of any perpendicular line. So the equation of the perpendicular line going through $(1,1)$ is $$y=-\frac{1}{m}x+d$$ where we can calculate $d$ using the point $(1,1)$; we have: $$1=-\frac{1}{m}+d.$$

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The equation of a straight line is "y = a + b * x". Since you know two points on the line, write the conditions; this will give you two equations for the two unknowns "a" and "b". Can you continue with this ?

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Actually, this is only true for lines nor parallel to the $y$-axis. –  Carsten Schultz Sep 23 '13 at 11:32