Midpoint of a Line Segment

Sometimes you need to find the point located at the center of a line segment joining two points, i.e. the midpoint of a line segment. For instance, you might need to know if two line segments bisect each other given the midpoints and the endpoints of both the line segments. The concept doesn't come up often, but the formula is quite simple and obvious, so you should be able to remember it easily for later use.

The midpoint of the points \((x_1 , y_1)\) and \((x_2 , y_2)\) is

\[ \left(\frac {x_1 + x_2} {2} , \frac {y_1 + y_2} {2}\right).\ _\square\]

One important point to remember: It does not matter in which order you add the \(x\) values and \(y\) values. This is true because of the commutative property of addition, which states that \(a + b = b + a\). All that matters is that you average the \(x\) values of the points and the \(y\) values of the points and use the new \(x\) and \(y\) values as the coordinates of the midpoint.

Find the midpoint between \((–1, 2)\) and \((3, –6)\).

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To solve this problem, we need to use the midpoint formula. Plug in the numbers into the midpoint formula to get

\[\left(\frac {-1 + 3} {2} , \frac {2 - 6} {2}\right)=\left(\frac {2} {2} , \frac {-4} {2}\right)=(1,-2).\ _\square\]

Find the midpoint between \((6.4, 3)\) and \((–10.7, 4)\).

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To solve this problem, we need to use the midpoint formula. Plug in the numbers into the midpoint formula to get

\[\left(\frac {6.4 - 10.7} {2} , \frac {3 + 4} {2}\right)=\left(\frac {-4.3} {2} , \frac {7} {2}\right)=(-2.15,3.5).\ _\square\].

Find the value of \(p\) such that \((–2, 2.5)\) is the midpoint between \((p, 2)\) and \((–1, 3)\).

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To solve this problem, we need to use the midpoint formula. Plug in the values into the midpoint formula to get \(\left(\frac {p - 1} {2} , \frac {2 + 3} {2}\right)\). To get the value of \(p,\) we need to set \(\frac {p - 1} {2} = -2,\) which gives \(p -1 = -4 \Rightarrow p = -3.\ _\square\)

The line segment joining \(A(-3,1)\) and \(B(5,-4)\) is a diameter of a circle whose centre is \(C\). Find the coordinates of the point \(C\).

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The centre of a circle is the midpoint of the diameter of the circle. So, \(C\) divides the line segment \(AB\) in the ratio \(1:1\). Let the coordinates of \(C\) be \((x,y)\). Using section formula, we have:
\[ \begin{align} x & = \frac {(-3) + 5}{2} \\ & = 1 \end{align}\]

\[\begin{align} y & = \frac {1 + (-4)}{2} \\
& = - \frac {3}{2} \end{align}\]

Therefore, the coordinates of \(C \) are \(\left(1, -\frac {3}{2} \right) \). \(_\square\)

The mid point of the line joining \((3a,4)\) and \((-2,2b)\) is \((2,2a+2)\). Find the values of \(a\) and \(b\).

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The midpoint of a line segment with coordinates \((x_1,y_1)\) and \((x_2,y_2)\) is \( \left( \dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right).\) Thus, \[ (2,2a+2)=\left(\dfrac{3a-2}{2},\dfrac{4+2b}{2}\right) \implies a=2, b=4. \ _\square\]