Precalculus

Analytic Geometry

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Division of a Line Segment

Division of a Line Segment in a Given Ratio

Let's consider in space two different points A and B and a straight line defined by these points. Let M be any point of the indicated line other than B. The point M divides the line segment AB in a ratio λ which is defined by the formula

\[\lambda = \frac{AM}{MB}.\]

Suppose the points A and B have coordinates \(\left({x_1,y_1,z_1}\right)\) and \(\left({x_2,y_2,z_2}\right)\) respectively. If the lambda number λ is known, then the coordinates \(\left({x,y,z}\right)\) of point M are given by

\[x_0 = \frac{x_1 + \lambda x_2}{1 + \lambda},\;y_0 = \frac{y_1 + \lambda y_2}{1 + \lambda},\;z_0 = \frac{z_1 + \lambda z_2}{1 + \lambda},\]

where \(\lambda \ne -1.\)

For positive lambda values the point M lies between the points A and B (Figure 1), and for negative values it lies outside the line segment AB (Figure 2).

Dividing a line segment in the ratio lambda (lambda is positive)
Figure 1.
Dividing a line segment in the ratio lambda (lambda is negative)
Figure 2.

Division of a Line Segment in Half

It is obvious that if the point M divides the segment AB in half, then

\[AM = MB, \Rightarrow \lambda = \frac{AM}{MB} = 1.\]

In this case, the coordinates of the point M are

\[x_0 = \frac{x_1 + x_2}{2},\;y_0 = \frac{y_1 + y_2}{2},\;z_0 = \frac{z_1 + z_2}{2}.\]

Solved Problems

Example 1.

Divide a rod \(36\text{ cm}\) long in a ratio of \(3:4:5.\)

Solution.

Let's denote the division points as \(K\left({x_K}\right)\) and \(M\left({x_M}\right).\) The boundaries of the rod are defined by points \(A\left({0}\right)\) and \(B\left({36}\right).\)

The point K divides the rod into \(2\) parts in a ratio of \(\lambda_K = \frac{3}{4+5} = \frac{1}{3}.\) The coordinate of this point is

\[x_K = \frac{x_A + \lambda_K x_B}{1 + \lambda_K} = \frac{0 + \frac{1}{3}\cdot36}{1 + \frac{1}{3}} = \frac{12}{\frac{4}{3}} = 9\text{ cm};\]

The point M divides the rod in a ratio of \(\lambda_M = \frac{3 + 4}{5} = \frac{7}{5}.\) Its coordinate is

\[x_M = \frac{x_A + \lambda_M x_B}{1 + \lambda_M} = \frac{0 + \frac{7}{5}\cdot36}{1 + \frac{7}{5}} = \frac{\frac{252}{5}}{\frac{12}{5}} = \frac{252}{12} = 21\text{ cm}.\]

The points \(K\left({9}\right),\) \(M\left({21}\right)\) divide the rod into \(3\) parts with a length of \(9, 12,\) and \(15\text{ cm}.\)

Example 2.

The line segment bounded by points \(A\left({-3,1,-2}\right)\) and \(B\left({6,7,1}\right)\) is divided into three equal parts. Determine the coordinates of the division points.

Solution.

Let the division points be \(K\left({x_K,y_K,z_K}\right)\) and \(M\left({x_M,y_M,z_M}\right).\) If point K is located at a distance of \(\frac{1}{3}AB\) from point A, then \(\lambda_K = \frac{1}{2}\) for this point. Respectively, for the point M we obtain \(\lambda_M = 2.\)

Calculate the coordinates of point K:

\[x_K = \frac{x_A + \lambda_K x_B}{1 + \lambda_K} = \frac{-3 + \frac{1}{2}\cdot6}{1 + \frac{1}{2}} = 0;\]
\[y_K = \frac{y_A + \lambda_K y_B}{1 + \lambda_K} = \frac{1 + \frac{1}{2}\cdot7}{1 + \frac{1}{2}} = \frac{\frac{9}{2}}{\frac{3}{2}} = 3;\]
\[z_K = \frac{z_A + \lambda_K z_B}{1 + \lambda_K} = \frac{-2 + \frac{1}{2}\cdot1}{1 + \frac{1}{2}} = \frac{-\frac{3}{2}}{\frac{3}{2}} = -1.\]

Similarly, we find the coordinates of point M:

\[x_M = \frac{x_A + \lambda_M x_B}{1 + \lambda_M} = \frac{-3 + 2\cdot6}{1 + 2} = 3;\]
\[y_M = \frac{y_A + \lambda_M y_B}{1 + \lambda_M} = \frac{1 + 2\cdot7}{1 + 2} = \frac{\frac{9}{2}}{\frac{3}{2}} = 5;\]
\[z_M = \frac{z_A + \lambda_M z_B}{1 + \lambda_M} = \frac{-2 + 2\cdot1}{1 + 2} = 0.\]

So the division points have coordinates \(K\left({0,3,-1}\right),\) \(M\left({3,5,0}\right).\)

Example 3.

Points \(K\left({2,3}\right),\) \(L\left({6,4}\right)\) and \(M\left({5,2}\right)\) are the midpoints of the sides of a triangle. Find the coordinates of its vertices.

Solution.

Let \(A\left({x_A,y_A}\right),\) \(B\left({x_B,y_B}\right),\) \(C\left({x_C,y_C}\right)\) be the vertices of the triangle.

A triangle with given midpoints of sides
Figure 3.

If K is the midpoint of side AB, L is the midpoint of side BC and M is the midpoint of side AC, then we can write

\[\left\{ \begin{array}{l} x_K = \frac{x_A + x_B}{2}\\ x_L = \frac{x_B + x_C}{2}\\ x_M = \frac{x_A + x_C}{2}\\ y_K = \frac{y_A + y_B}{2}\\ y_L = \frac{y_B + y_C}{2}\\ y_M = \frac{y_A + y_C}{2} \end{array} \right.\]

From the first three equations of the system we can determine the coordinates \(x_A,\) \(x_B,\) \(x_C.\)

\[\left\{ \begin{array}{l} x_K = \frac{x_A + x_B}{2}\\ x_L = \frac{x_B + x_C}{2}\\ x_M = \frac{x_A + x_C}{2} \end{array} \right., \Rightarrow \left\{ \begin{array}{l} x_A + x_B = 2x_K\\ x_B + x_C = 2x_L\\ x_A + x_C = 2x_M \end{array} \right..\]

We express \(x_A\) from the first equation and substitute in the third equation:

\[\Rightarrow \left\{ \begin{array}{l} x_A = 2x_K - x_B\\ x_B + x_C = 2x_L\\ 2x_K - x_B + x_C = 2x_M \end{array} \right., \Rightarrow \left\{ \begin{array}{l} x_A = 2x_K - x_B\\ x_B + x_C = 2x_L\\ - x_B + x_C = 2x_M - 2x_K \end{array} \right..\]

Add the second and third equation to eliminate \(x_B:\)

\[\Rightarrow \left\{ \begin{array}{l} x_A = 2x_K - x_B\\ x_B + x_C = 2x_L\\ 2x_K + 2x_C = 2x_L + 2x_M \end{array} \right., \Rightarrow \left\{ \begin{array}{l} x_A = 2x_K - x_B\\ x_B + x_C = 2x_L\\ x_C = x_L + x_M - x_K \end{array} \right..\]

Now we know \(x_C\) and can find \(x_B\) and \(x_A:\)

\[\Rightarrow \left\{ \begin{array}{l} x_A = 2x_K - x_B\\ x_B + x_L + x_M - x_K = 2x_L\\ x_C = x_L + x_M - x_K \end{array} \right., \Rightarrow \left\{ \begin{array}{l} x_A = 2x_K - x_B\\ x_B = x_K + x_L - x_M\\ x_C = x_L + x_M - x_K \end{array} \right., \Rightarrow \left\{ \begin{array}{l} x_A = 2x_K - x_K - x_L + x_M\\ x_B = x_K + x_L - x_M\\ x_C = x_L + x_M - x_K \end{array} \right., \Rightarrow \left\{ \begin{array}{l} x_A = x_K + x_M - x_L\\ x_B = x_K + x_L - x_M\\ x_C = x_L + x_M - x_K \end{array} \right..\]

Similarly, one can determine the \(y-\)coordinates. We obtain the following final formulas:

\[\left\{ \begin{array}{l} x_A = x_K + x_M - x_L\\ x_B = x_K + x_L - x_M\\ x_C = x_L + x_M - x_K\\ y_A = y_K + y_M - y_L\\ y_B = y_K + y_L - y_M\\ y_C = y_L + y_M - y_K \end{array} \right..\]

Now you can easily calculate the coordinates of the vertices of the triangle:

\[\left\{ \begin{array}{l} x_A = 2 + 5 - 6\\ x_B = 2 + 6 - 5\\ x_C = 6 + 5 - 2\\ y_A = 3 + 2 - 4\\ y_B = 3 + 4 - 2\\ y_C = 4 + 2 - 3 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} x_A = 1\\ x_B = 3\\ x_C = 9\\ y_A = 1\\ y_B = 5\\ y_C = 3 \end{array} \right..\]

So the vertices of the triangle have the following coordinates: \(A\left({1,1}\right),\) \(B\left({3,5}\right),\) \(C\left({9,3}\right).\)

Example 4.

Given the coordinates of two vertices of a triangle \(A\left({2,-1}\right),\) \(B\left({-3,5}\right)\) and the point of intersection of medians \(M\left({1,1}\right),\) find the coordinates of the vertex C.

Solution.

In a triangle, the point of intersection of its medians (centroid) divides each median in the ratio \(2:1.\) Hence, we have

\[\frac{AM}{MK} = \lambda = 2,\]

where K is the midpoint of side BC.

A triangle with three medians
Figure 4.

We denote the coordinates of the points in the figure as follows:

\[A\left({x_A,y_A}\right), B\left({x_B,y_B}\right), C\left({x_C,y_C}\right), M\left({x_M,y_M}\right), K\left({x_K,y_K}\right).\]

Using the formulas for division of a line segment, we get

\[x_M=\frac{x_A + \lambda x_K}{1 + \lambda},\;y_M=\frac{y_A + \lambda y_K}{1 + \lambda}.\]

Solve these equations for \(x_K, y_K:\)

\[x_K=\frac{x_M\left({1+\lambda}\right) - x_A}{\lambda} = \frac{1\cdot3 - 2}{2} = \frac{1}{2};\]
\[y_K=\frac{y_M\left({1+\lambda}\right) - y_A}{\lambda} = \frac{1\cdot3 + 1}{2} = 2.\]

Find now the coordinates of the vertex C.

\[x_K=\frac{x_B + x_C}{2},\;y_K=\frac{y_B + y_C}{2}.\]

Hence

\[x_C = 2x_K - x_B = 2\cdot\frac{1}{2}+3=4,\;y_C = 2y_K - y_B = 2\cdot2 - 5 = -1.\]

Thus the vertex C has coordinates \(\left({4,-1}\right).\)