Midpoint Formula: Example, Coordinate Geometry, Statistics

Midpoint Formula is used to find the exact midpoint of a line segment. Sometimes you will have to find the number that is exactly half of two numbers. For that, you find an average of the two numbers. In a similar manner, we use the midpoint formula in coordinate geometry to find the midway number (i.e., point) of two coordinates.

The formula for midpoint is used to find the middle point between two points for which the coordinates are familiar to us. The midpoint lies between the two points on the line and is at the same distance from both points. This formula is also used to find the endpoint coordinates if we know the coordinates of the other endpoint and the midpoint.

Midpoint Formula in Coordinate Geometry

In coordinate geometry, we know about the Cartesian plane and the representation of points using the coordinate system. Thus, several times we need to find the position of the midpoint between two points. Likewise, the midpoint of a line segment may be required. This midpoint will work as the centre point of a straight line. Occasionally we will have to find the number that is half of two numbers. Likewise, we are using the midpoint formula in coordinate geometry to find the middle number of two coordinates. 

The midpoint formula is a mathematical equation that is used to locate the halfway point between two points.

Let us consider a line segment with its endpoints, \(\left( {{x_1},{y_1}} \right)\) and \(\left( {{x_2},{y_2}} \right).\) For every line segment, the midpoint is midway between its two endpoints.

The expression for the \(x-\)coordinate of the midpoint is  \(\frac{{\left( {{x_1} + {x_2}} \right)}}{2},\) which is the average of the \(x-\)coordinates.

Similarly, the expression for the \(y-\)coordinate is  \(\frac{{\left( {{y_1} + {y_2}} \right)}}{2},\) which is the average of the \(y-\)coordinates.

Thus, the coordinate for the midpoint is, \((x,y) = \left[ {\frac{{\left( {{x_1} + {x_2}} \right)}}{2},\frac{{\left( {{y_1} + {y_2}} \right)}}{2}} \right]\) 

The next two formulas are closely linked to the midpoint formula.

1. The centroid of the triangle formula
2. The section formula

Centroid of a Triangle Formula

The centroid of a triangle takes the average of the \(x\) coordinates and the \(y\) coordinates of all the three vertices. Thus, the centroid formula can be mathematically stated as \(G(x,y) = \left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\)

Note: The centroid of a triangle is the point of intersection of its medians, and it (centroid) divides each median in the ratio \(2:1.\)

Section Formula

Think a telephone company wants to position a transmit tower at \(P\) between \(A\) and \(B\) is in such a way that the distance of the tower from \(B\) is twice its distance from \(A.\) Here, \(P\) lies on \(AB,\) which will split \(AB\) in the ratio \(1:2.\) If we take \(A\) as the origin \(O,\) and \({\rm{1}}\,{\rm{km}}\) as one unit on both the axis, the coordinates of \(B\) will be \((36, 15).\) To know the tower’s position, we must understand the coordinates of \(P.\) How are we going to find these coordinates?

Let the coordinates of \(P\) be \((x, y).\)  Draw perpendiculars from \(P\) and \(B\) to the \(x-\)axis, meeting at \(D\) and \(E,\) each. Get \(PC\) perpendicular to \(BE.\) Then, by the \(AA\) similarity principle, \(∆POD\) and \(∆BPC\) are similar.

Therefore, \(\frac{{OD}}{{PC}} = \frac{{OP}}{{PB}} = \frac{1}{2},\) and \(\frac{{PD}}{{BC}} = \frac{{OP}}{{PB}} = \frac{1}{2}\)

So, \(\frac{x}{{36 – x}} = \frac{1}{2}\) and \(\frac{y}{{15 – y}} = \frac{1}{2}\)

These equations give \(x=12\) and \(y=5.\) You can verify that \(P(12, 5)\) satisfies the condition that \(OP:PB=1:2.\) Now allow us to understand that you may have evolved through this example to get the general formula. Take into account any two points \(A\left( {{x_1},{y_1}} \right)\) and \(A\left( {{x_2},{y_2}} \right)\) and is based on the assumption that \(P(x,y)\)  divides \(AB\) internally within the ratio  \(m:n,\) i.e.,

Consider any two points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right)\) and assume that \(P(x, y)\) divides \(AB\) internally in the ratio \(m:n,\) i.e.,

\(\frac{{PA}}{{PB}} = \frac{m}{n}\)

Draw \(AR, PS\) and \(BT\) perpendicular to the \(x-\)axis. Get \(AQ\) and \(PC\) in parallel with the \(x-\)axis. Later, in accordance with the \(AA\) similarity criterion, \(\Delta PAQ \sim \Delta BPC\)

Therefore, \(\frac{{PA}}{{BP}} = \frac{{AQ}}{{PC}} = \frac{{PQ}}{{BC}}………..\left( 1 \right)\)

Now,
\(AQ = RS = OS – OR = x – {x_1}\)
\(PC = ST = OT – OS = {x_2} – x\)
\(PQ = PS – QS = PS – AR = y – {y_1}\)
\(BC = BT – CT = BT – PS = {y_2} – y\)

Substituting these values in \((1),\) we get
\(\frac{m}{n} = \frac{{x – {x_1}}}{{{x_2} – {x_1}}} = \frac{{y – {y_1}}}{{{y_2} – {y_1}}}\)
Taking, \(\frac{m}{n} = \frac{{x – {x_1}}}{{{x_2} – {x_1}}},\) we get \(x = \frac{{m{x_2} + n{x_1}}}{{m + n}}\)
Equally taking, \(\frac{m}{n} = \frac{{y – {y_1}}}{{{y_2} – {y_1}}},\) we get \(y = \frac{{m{y_2} + n{y_1}}}{{m + n}}\)

Therefore, the coordinates of the point \(P(x, y)\) that separates the line segment joining the points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right),\) internally, in the ratio \(m:n,\) are
\(\left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\) 

This is referred to as the section formula.

Specific case: The mid-point of a line segment separates the line segment in the ratio \(1:1.\)
Hence, the coordinates of the mid-point \(P\) of the merge of the points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right),\) is \(\left( {\frac{{1 \times {x_2} + 1 \times {x_1}}}{{1 + 1}},\frac{{1 \times {y_2} + 1 \times {y_1}}}{{1 + 1}}} \right) = \left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)\)

Midpoint Formula Example

The midpoint formula is referring to finding the coordinates of the midpoint of a line or line segment. Given the coordinates of two points on a line, the midpoint formula calculates the midpoint coordinates quickly.

Example: Find the coordinates of the mid-point of the line segment joining the points \(P(4, -6)\) and \(Q(-2, 4).\)

Mid-point \( = \left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right) = \left( {\frac{{4 – 2}}{2},\frac{{ – 6 + 4}}{2}} \right) = (1, – 1)\)

Midpoint Formula in Statistics

The midpoint formula in statistics is the classmark formula. Classmark is the midway in the class interval. Classmarks can be calculated using,
\({\rm{Mid}}\,{\rm{point}} = {\rm{Class}}\,{\rm{mark}} = \frac{{{\rm{ Lower}}\,{\rm{limit}}\, +\, {\rm{Upper}}\,{\rm{limit }}}}{2}\)

Example: Find the class mark of the class interval \(10-20.\)

\({\rm{Mid}}\,{\rm{point}} = {\rm{Class}}\,{\rm{mark}} = \frac{{10 + 20}}{2} = \frac{{30}}{2} = 15\)

Solved Examples – Midpoint Formula

Q.1. Find the coordinates of the centroid of a triangle whose vertices are \((0,6,),(8,12)\) and \((8,0.)\)
Ans: We know that the coordinates of the centroid of a triangle whose angular points are \(\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)\) are \(\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\)
So, the coordinates of the centroid of a triangle whose vertices are \((0,6),(8,12)\) and \((8,0)\) are \(\left( {\frac{{0 + 8 + 8}}{3},\frac{{6 + 12 + 0}}{3}} \right) = \left( {\frac{{16}}{3},6} \right)\)
Hence, the centroid of a triangle for the given vertices is \(\left( {\frac{{16}}{3},6} \right).\)

Q.2. Find the ratio where the \(y-\)axis splits the line segment joining the points \((5,-6)\) and \((-1,-4.)\) Likewise, find the point of intersection.
Ans: Let the ratio be \(k:1.\) Then, by the section formula, the coordinates of the point which divides the line segment in the ratio \(k:1\) are \(\left( {\frac{{ – k + 5}}{{k + 1}},\frac{{ – 4k – 6}}{{k + 1}}} \right).\)
This point is situated on the \(y-\)axis, and we understand that on the \(y-\)axis the abscissa is \(0.\)
Therefore, \(\frac{{ – k + 5}}{{k + 1}} = 0\)
So, \(k=5\)
That is, the ratio is \(5:1.\) Substituting the value of \(k=5,\) we obtain the point of intersection as \(\left( {0,\frac{{ – 13}}{3}} \right)\)

Q.3. The coordinates of one endpoint of a diameter of a circle are \((4,-1)\) and the coordinates of the centre of the circle are \((1,-3.)\) Determine the coordinates of the opposite end of the diameter.
Ans: Let \(AB\) be a diameter of the circle considering its centre at \(C(1,-3)\) in such a way that the coordinates of the one end \(A\) are \((4,-1)\).
Allow the coordinates of \(B\) be \((x,y)\).

Since \(C\) is the mid-point of \(AB.\) Therefore, the coordinates of \(C\) are \(\left( {\frac{{x + 4}}{2},\frac{{y – 1}}{2}} \right).\)
Then the coordinates of \(C\) are provided to be \((1,-3).\)
Therefore, \(\frac{{x + 4}}{2} = 1\) and \(\frac{{y – 1}}{2} =\, – 3\)
\( \Rightarrow x + 4 = 2\) and \(y – 1 = \,- 6\)
\( \Rightarrow x =\, – 2\) and \(y = \, – 5\)
Hence, the coordinates of \(B\) are \((-2,-5).\)

Q.4. If \(A( – 2, – 1),B(a,0),C(4,b)\) and \(D(1,2)\) are the vertices of the parallelogram, determine the values of \(a\) and \(b.\)
Ans: We understand that the diagonals of a parallelogram bisect each other. Hence, the coordinates of the mid-point of \(AC\) are the same as the coordinates of the mid-point of \(BD,\) i.e.,
\(\left( {\frac{{ – 2 + 4}}{2},\frac{{ – 1 + b}}{2}} \right) = \left( {\frac{{a + 1}}{2},\frac{{0 + 2}}{2}} \right)\)
\( \Rightarrow \left( {1,\frac{{ – 1 + b}}{2}} \right) = \left( {\frac{{a + 1}}{2},1} \right)\)
\( \Rightarrow \frac{{a + 1}}{2} = 1\) and \(\frac{{ – 1 + b}}{2} = 1\)
\( \Rightarrow a + 1 = 2\) and \( – 1 + b = 2\)
\( \Rightarrow a = 1\) and \(b = 3\)
Hence, \(a=1\) and \(b=3.\)

Q.5. Find the coordinates of points that trisect (three equal parts) the line segment joining \((1,-2)\) and \((-3,4.)\)
Ans: Let \(A(1,-2)\) and \(B(-3,4)\) be the given points. Allow the points of trisection as \(P\) and \(Q.\) Then, \(AP=PQ=QB=α\)

Therefore, \(PB=PQ+QB=2α\) and \(AQ=AP+PQ=2α\)
\(⟹AP:PB=α:2α=1:2\) and \(AQ:QB=2α:α=2:1\)
Therefore, \(P\) divides \(AB\) internally in proportion \(1:2\) while \(Q\) divides internally in proportion \(2:1.\) Hence, the coordinates of \(P\) and \(Q\) are
\(P\left( {\frac{{1 \times ( – 3) + 2 \times 1}}{{1 + 2}}\frac{{1 \times 4 + 2 \times ( – 2)}}{{1 + 2}}} \right) = P\left( {\frac{{ – 1}}{3},0} \right)\)
\(Q\left( {\frac{{2 \times ( – 3) + 1 \times 1}}{{2 + 1}}\frac{{2 \times 4 + 1 \times ( – 2)}}{{2 + 1}}} \right) = Q\left( {\frac{{ – 5}}{3},2} \right)\)
Thus, the two points of trisection are \(\left( {\frac{{ – 1}}{3},0} \right)\) and \(\left( {\frac{{ – 5}}{3},2} \right)\)

Summary

In this article, we learnt about midpoint formula example, midpoint formula in coordinate geometry, midpoint formula in statistics, the centroid of the triangle formula, the section formula, solved examples on midpoint formula and FAQs midpoint formula.

The learning outcome of this article is that the midpoint formula is a mathematical equation that is used to locate the halfway point between two data points.

FAQs

Q.1. How do you find the midpoint formula?
Ans: Let us think about a line segment with its endpoints, \(\left( {{x_1},{y_1}} \right)\) and \(\left( {{x_2},{y_2}} \right).\) For every line segment, the midpoint is midway between its two endpoints.
The phrase for the \(x-\)coordinate of the midpoint is \(\frac{{\left( {{x_1} + {x_2}} \right)}}{2},\) which makes up the average of the \(x-\)coordinates.
Likewise, the phrase for the \(y-\)coordinate is \(\frac{{\left( {{y_1} + {y_2}} \right)}}{2},\) which makes up the average of the \(y-\)coordinates.
Thus, the formula for midpoint is, \((x,y) = \left[ {\frac{{\left( {{x_1} + {x_2}} \right)}}{2},\frac{{\left( {{y_1} + {y_2}} \right)}}{2}} \right]\)

Q.2. What is the midpoint and distance formula?
Ans: \({\rm{Midpoint}}\,{\rm{formula}} = (x,y) = \left[ {\frac{{\left( {{x_1} + {x_2}} \right)}}{2},\frac{{\left( {{y_1} + {y_2}} \right)}}{2}} \right]\) and
\({\rm{Distance}}\,{\rm{formula}} = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2}} \)

Q.3. What is the midpoint in statistics?
Ans: Midpoint formula in statistics is the classmark formula, classmark is the midway in the class interval. Classmarks can be calculated using,
\({\rm{Midpoint}} = {\rm{Classmark}} = \frac{{{\rm{ Lower}}\,{\rm{limit + Upper}}\,{\rm{limit }}}}{2}\)

Q.4. What is the midpoint between two numbers?
Ans: The midpoint between two numbers is the number precisely in the middle of the two numbers. Computing the midpoint is the same thing as determining the average of two numbers. Therefore, you can estimate the midpoint between any two numbers by adding them to each other and dividing them by two.

Q.5. What is the midpoint formula used for?
Ans: The midpoint formula is a mathematical formula used to locate the midway point between two points.

We hope you find this detailed article on the midpoint formula helpful. If you have any queries or doubts regarding this topic, ask them in the comment section. We will assist you at the earliest. Happy learning!