# Section Formula

The **section formula** tells us the coordinates of the point which divides a given line segment into two parts such that their lengths are in the ratio \(m:n\).

The midpoint of a line segment is the point that divides a line segment in two equal halves. The section formula builds on it and is a more powerful tool; it locates the point dividing the line segment in any desired ratio.

The section formula is helpful in coordinate geometry; for instance, it can be used to find out the centroid, incenter and excenters of a triangle. It has applications in physics too; it helps find the center of mass of systems, equilibrium points, and more.

## Internal Divisions with Section Formula

If point \(P (x,y)\) lies on line segment \(\overline{AB}\) \((\)between points \(A\) and \(B)\) and satisfies \(AP:PB=m:n,\) then we say that \(P\) divides \(\overline{AB}\) internally in the ratio \(m:n.\) The point of division has the coordinates

\[P=\left( \dfrac{mx_2 + nx_1}{m+n}, \dfrac{my_2 + ny_1}{m+n} \right). \]

The formula can be derived by constructing two similar right triangles, as shown below. Their hypotenuses are along the line segment and are in the ratio \(m:n\).

The red and the green triangles are similar since the corresponding angles of the triangles are equal. This implies that the ratio of their corresponding sides are equal. Note that point \(P\) is \(\frac{m}{m+n} \times AB\) away from \(A\). That is,

\[\begin{align} x & = x_1 + \frac{m}{m + n} (x_2 - x_1) \\ & = \frac{(m + n) x_1 + m x_2 - m x _1}{m + n} \\ & = \frac{ m{ x }_{ 2 }+n{ x }_{ 1 }}{m + n}. \qquad (1) \end{align}\]

Similarly, solving for \(y\) gives

\[ y= \frac{m{ y }_{ 2 }+n{ y }_{ 1 }}{m + n}. \qquad (2)\]

Therefore, from \((1)\) and \((2)\)

\[P (x,y) = \left( \dfrac { m{ x }_{ 2 }+n{ x }_{ 1 } }{ m+n }, \dfrac { m{ y }_{ 2 }+n{ y }_{ 1 } }{ m+n } \right).\ _\square\]

As a special case of internal division, if \(P\) is the midpoint of \(\overline{AB}\), then it divides \(\overline{AB}\) internally in the ratio \(1:1\). Hence applying the formula for internal division and substituting \(m = n = 1\), we get

\[P=\left( \dfrac{x_1+y_1}{2}, \dfrac{x_2+y_2}{2} \right). \]

Given \(A=(-3,1)\) and \(B=(3,-6)\), what are the coordinates of point \(P=(x,y)\) which internally divides line segment \(\overline{AB}\) in the ratio \(1:2?\)

The point \(P\) is \(\frac{1}{1+2} \times AB\) away from point \(A\).

When measured parallel to the \(x\)-axis, we get

\[\begin{align} x & = -3 + \frac{1}{3} \times \big(3 - (-3)\big) \\

&= -1. \end{align}\]When measured parallel to the \(y\)-axis, we get

\[\begin{align} y & = 1 + \frac{1}{3} \times (-6 -1) \\ & = - \frac{4}{3}. \end{align}\]

Thus, the coordinates of \(P\) are \(\big( -1, -\frac{4}{3} \big) \) \(_\square\)

Given \(A=(-3,6)\), what are the coordinates of \(B=(x_2,y_2)\) if point \(P=(-2,4)\) divides line segment \(\overline{AB}\) internally in the ratio \(1:3?\)

In this example, we are to find one of the endpoints of the line segment. Drawing similar triangles will help us solve this problem too.

The sides of the triangle are in the ratio \(1 : 3\). The base of the pink triangle has length \(-2 - (-3) = 1\). The base of the green triangle is three times as long, that is, \(x - (-2) = 3 \times 1\). Solving this yields \(x = 1\).

The height of the pink triangle is \(4 - 6 = -2\). The height of the green triangle is three times as long, that is, \(y - 4 = 3 \times (-2)\). Solving this equation yields \(y = -2\).

Thus, the coordinates of \(B\) are \((1,-2).\) \(_\square\)

In triangle \(ABC,\) the midpoints of sides \( BC, \) \( CA \) and \( AB \) are \( (1, 0), \) \( (3, 5) \) and \( (-2, 4), \) respectively. Find the coordinates of the three vertices \(A,\) \(B\) and \(C.\)

Given \(A= (-2,-1)\) and \(B=(4,11)\), point \(P= (x,y)\) internally divides line segment \(\overline{AB}\) in the ratio \(m:n\). If \(P \) is the intersection point of \(\overline{AB}\) and the \(y\)-axis, what is the value of \(m : n?\)

Since point \(P\) is on the \(y\)-axis, its \(x\) coordinate is zero. We can write the coordinates of \(P\) as \((0, y)\).

The horizontal distance between \(P \) and \(A\) is \(0 - (-2) = 2\).

The horizontal distance between \(B\) and \(P\) is \(4 - 0 = 4\).The ratio of the bases of the right triangles is \(2 : 4\), or \(1 : 2\). Since the triangles are similar, the ratio of their hypotenuses is also \(1 : 2\).

Therefore, point \(P\) divides line segment \(AB\) in the ratio \(1 : 2\). \(_\square\)

In what ratio does the point \(P=(-3,7)\) divide the line segment joining \(A=(-5,11)\) and \(B=(4,-7)?\)

We can draw 2 similar right triangles: the red triangle with hypotenuse \(AP\) and the blue triangle with hypotenuse \(PB.\)

Point \(P\) divides line segment \(AB\) in the ratio \(AP : PB\), which is equivalent to \(a:b\) since the triangles are similar. Let us find the lengths of \(a\) and \(b:\)

\[a = (-3) - (-5) = 2, \quad b = 4 - (-3) = 7.\]

Thus, point \(P\) divides line segment \(AB\) in the ratio \(a : b = 2 : 7\).

Alternatively, the ratio \(AP : PB\) is also equal to \(c : d,\) i.e.

\[c = 7 - 11 = -4, \quad d = (-7) - 7 = -14 \implies c:d=2:7.\]

We get the ratio \(2 : 7\) again, which is consistent with our previous calculations. \(_\square\)

## External Divisions with Section Formula

If \(P = (x,y)\) lies on the extention of line segment \(\overline{AB}\) \((\)not lying between points \(A\) and \(B)\) and satisfies \(AP:PB=m:n,\) then we say that \(P\) divides \(\overline{AB}\) externally in the ratio \(m:n.\) The point of division is

\[P=\left( \dfrac{mx_2 - nx_1}{m-n}, \dfrac{my_2 - ny_1}{m-n} \right) .\]

This proof of this result is similar to the proof in internal divisions, by drawing two similar right triangles. The yellow and orange triangles have their sides in the ratio \(m : n\).

From the figure, we see that point \(P\) is at a distance of \(\frac{m}{m-n} \times AB\) away from point \(A:\)

\[\begin{align} x & = x_1 + \frac{m}{m - n} (x_2 - x_1) \\ & = \frac{(m - n)x_1 + mx_2 - mx_1 }{m-n} \\ & = \frac{mx_2 - nx_1}{m-n}. \qquad (3) \end{align}\]

Similarly, solving for \(y\) gives

\[y=\frac { m{ y }_{ 2 }-n{ y }_{ 1 } }{ m-n }. \qquad (4)\]

Therefore, from \((3)\) and \((4)\)

\[ P(x, y) = \left( \dfrac { m{ x }_{ 2 }-n{ x }_{ 1 } }{ m-n }, \dfrac { m{ y }_{ 2 }-n{ y }_{ 1 } }{ m-n } \right).\ _\square\]