Section Formula and Midpoint Formula for Class 10

Spe­cial cases#

• Mid­point: the ratio is $1 : 1$.
• Points of tri­sec­tion: the two points that divide $AB$ into three equal parts. They divide $AB$ in the ratios $1 : 2$ and $2 : 1$.
• Ratio $k : 1$: any ratio $m_1 : m_2$ can be writ­ten as $k : 1$ with $\displaystyle k = \frac{m_1}{m_2}$. This form uses one unknown instead of two.

The sec­tion for­mula#

State­ment#

The point $P(x, y)$ that divides the seg­ment join­ing $A(x_1, y_1)$ and $B(x_2, y_2)$ inter­nally in the ratio $m_1 : m_2$ is

$$\displaystyle P(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2},\ \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right)$$

A use­ful way to remem­ber it: each end point is mul­ti­plied by the part of the ratio next to the other end. So $x_2$ (from $B$) goes with $m_1$, and $x_1$ (from $A$) goes with $m_2$.

Why the for­mula is true#

Take the case where $x_1 \lt x \lt x_2$ and the seg­ment is not par­al­lel to an axis. Draw per­pen­dic­u­lars from $A$, $P$ and $B$ to the x-axis. Through $A$ draw a line par­al­lel to the x-axis meet­ing the per­pen­dic­u­lar from $P$ at $Q$. Through $P$ draw a line par­al­lel to the x-axis meet­ing the per­pen­dic­u­lar from $B$ at $C$. Then $Q = (x, y_1)$ and $C = (x_2, y)$.

In tri­an­gles $AQP$ and $PCB$:

• $\angle AQP = \angle PCB = 90^\circ$, since each is made by a hor­i­zon­tal line and a ver­ti­cal line;
• $\angle PAQ = \angle BPC$, since $AQ \parallel PC$ and $AB$ is a trans­ver­sal (cor­re­spond­ing angles).

So $\triangle AQP \sim \triangle PCB$ by the AA cri­te­rion. Cor­re­spond­ing sides of sim­i­lar tri­an­gles are pro­por­tional:

$$\displaystyle \frac{AP}{PB} = \frac{AQ}{PC} = \frac{QP}{CB}$$

Now $AQ = x - x_1$ and $PC = x_2 - x$, so

$$\displaystyle \begin{aligned} \frac{m_1}{m_2} &= \frac{x - x_1}{x_2 - x} \\ m_1 x_2 - m_1 x &= m_2 x - m_2 x_1 \\ m_1 x_2 + m_2 x_1 &= (m_1 + m_2)\, x \\ x &= \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \end{aligned}$$

The ver­ti­cal sides give the $y$-coor­di­nate in the same way. If $y$ increases from $A$ to $B$, then $QP = y - y_1$ and $CB = y_2 - y$. If $y$ decreases, as in the fig­ure below, then $QP = y_1 - y$ and $CB = y - y_2$. In both cases the ratio equals $\displaystyle \frac{y - y_1}{y_2 - y}$, which gives $\displaystyle y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$. The other arrange­ments of the points give the same result.

The mid­point for­mula#

Put $m_1 = m_2$ in the sec­tion for­mula. The mid­point of the seg­ment join­ing $A(x_1, y_1)$ and $B(x_2, y_2)$ is

$$\displaystyle M = \left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2} \right)$$

Each coor­di­nate of the mid­point is the aver­age of the cor­re­spond­ing coor­di­nates of the end points.

The cen­troid of a tri­an­gle#

A median joins a ver­tex of a tri­an­gle to the mid­point of the oppo­site side. The three medi­ans meet at the cen­troid $G$, which divides each median in the ratio $2 : 1$ from the ver­tex. For $A(x_1, y_1)$, $B(x_2, y_2)$ and $C(x_3, y_3)$, let $D$ be the mid­point of $BC$:

$$\displaystyle D = \left( \frac{x_2 + x_3}{2},\ \frac{y_2 + y_3}{2} \right)$$

Divid­ing $AD$ in the ratio $2 : 1$:

$$\displaystyle x = \frac{2 \cdot \dfrac{x_2 + x_3}{2} + 1 \cdot x_1}{2 + 1} = \frac{x_1 + x_2 + x_3}{3}$$

and sim­i­larly for $y$. The same point comes out from any of the three medi­ans, so

$$\displaystyle G = \left( \frac{x_1 + x_2 + x_3}{3},\ \frac{y_1 + y_2 + y_3}{3} \right)$$

Method in steps#

Find­ing the point when the ratio is known#

1. Label the end points $A(x_1, y_1)$ and $B(x_2, y_2)$ in the order the ratio is given.
2. Write $m_1$ (next to $A$) and $m_2$ (next to $B$).
3. Sub­sti­tute into the sec­tion for­mula, cross­ing the ratio: $m_1$ with $B$ and $m_2$ with $A$.
4. Sim­plify each coor­di­nate sep­a­rately.

Find­ing the ratio when the point is known#

1. Let the ratio be $k : 1$.
2. Write one coor­di­nate of the point using the sec­tion for­mula with $m_1 = k$ and $m_2 = 1$.
3. Equate it to the known coor­di­nate and solve for $k$.
4. Write the ratio as $k : 1$ and clear any frac­tions. If $k$ is neg­a­tive, the divi­sion is exter­nal, which is out­side this syl­labus.
5. Use $k$ in the other coor­di­nate to find the miss­ing value, or to check the point.

When a seg­ment is cut by the x-axis, the point has $y = 0$. When it is cut by the y-axis, the point has $x = 0$.

Worked exam­ples#

Exam­ple 1: point of divi­sion and mid­point#

(a) Find the point that divides the seg­ment join­ing $A(-2, 3)$ and $B(8, -2)$ inter­nally in the ratio $2 : 3$.

Here $m_1 = 2$ and $m_2 = 3$.

$$\displaystyle \begin{aligned} x &= \frac{2 \times 8 + 3 \times (-2)}{2 + 3} = \frac{16 - 6}{5} = 2 \\ y &= \frac{2 \times (-2) + 3 \times 3}{2 + 3} = \frac{-4 + 9}{5} = 1 \end{aligned}$$

The point is $P(2, 1)$, as in the fig­ure above.

(b) Find the mid­point of the seg­ment join­ing $(-3, 7)$ and $(5, -1)$.

$$\displaystyle M = \left( \frac{-3 + 5}{2},\ \frac{7 + (-1)}{2} \right) = (1, 3)$$

Exam­ple 2: find­ing the ratio using k : 1#

A point $P$ with x-coor­di­nate $-1$ lies on the seg­ment join­ing $A(-4, 6)$ and $B(5, -6)$. Find the ratio in which $P$ divides $AB$, and find $P$.

Let the ratio be $k : 1$. Then

$$\displaystyle \begin{aligned} \frac{5k + (-4)}{k + 1} &= -1 \\ 5k - 4 &= -k - 1 \\ 6k &= 3 \\ k &= \frac{1}{2} \end{aligned}$$

So the ratio is $\displaystyle \frac{1}{2} : 1$, that is $1 : 2$. Now the y-coor­di­nate:

$$\displaystyle y = \frac{\frac{1}{2} \times (-6) + 1 \times 6}{\frac{1}{2} + 1} = \frac{-3 + 6}{\frac{3}{2}} = 2$$

The point is $P(-1, 2)$, and it divides $AB$ in the ratio $1 : 2$.

Exam­ple 3: divi­sion by the x-axis#

In what ratio does the x-axis divide the seg­ment join­ing $A(3, -4)$ and $B(-2, 6)$? Find the point of divi­sion.

Let the ratio be $k : 1$. The point lies on the x-axis, so its y-coor­di­nate is $0$:

$$\displaystyle \begin{aligned} \frac{6k + (-4)}{k + 1} &= 0 \\ 6k - 4 &= 0 \\ k &= \frac{2}{3} \end{aligned}$$

The ratio is $2 : 3$. Using $m_1 = 2$ and $m_2 = 3$ for the x-coor­di­nate:

$$\displaystyle x = \frac{2 \times (-2) + 3 \times 3}{5} = \frac{-4 + 9}{5} = 1$$

The x-axis divides $AB$ in the ratio $2 : 3$ at the point $(1, 0)$.

Exam­ple 4: points of tri­sec­tion#

Find the points of tri­sec­tion of the seg­ment join­ing $A(-3, -5)$ and $B(6, 4)$.

Let $P$ and $Q$ be the points with $AP = PQ = QB$. Then $P$ divides $AB$ in $1 : 2$, and $Q$ divides $AB$ in $2 : 1$.

$$\displaystyle \begin{aligned} P &= \left( \frac{1 \times 6 + 2 \times (-3)}{3},\ \frac{1 \times 4 + 2 \times (-5)}{3} \right) = \left( \frac{0}{3},\ \frac{-6}{3} \right) = (0, -2) \\ Q &= \left( \frac{2 \times 6 + 1 \times (-3)}{3},\ \frac{2 \times 4 + 1 \times (-5)}{3} \right) = \left( \frac{9}{3},\ \frac{3}{3} \right) = (3, 1) \end{aligned}$$

Check: $Q$ should be the mid­point of $PB$, and $\displaystyle \left( \frac{0 + 6}{2}, \frac{-2 + 4}{2} \right) = (3, 1)$. It is.

Exam­ple 5: the fourth ver­tex of a par­al­lel­o­gram#

Three ver­tices of par­al­lel­o­gram $ABCD$, taken in order, are $A(1, -2)$, $B(4, 1)$ and $C(3, 5)$. Find $D$.

The diag­o­nals of a par­al­lel­o­gram bisect each other, so $AC$ and $BD$ have the same mid­point. Let $D = (x, y)$.

$$\displaystyle \text{Midpoint of } AC = \left( \frac{1 + 3}{2},\ \frac{-2 + 5}{2} \right) = \left( 2,\ \frac{3}{2} \right)$$

$$\displaystyle \text{Midpoint of } BD = \left( \frac{4 + x}{2},\ \frac{1 + y}{2} \right)$$

Equat­ing them:

$$\displaystyle \frac{4 + x}{2} = 2 \Rightarrow x = 0, \qquad \frac{1 + y}{2} = \frac{3}{2} \Rightarrow y = 2$$

So $D = (0, 2)$. A quick short­cut that fol­lows from this: $D = A + C - B$, coor­di­nate by coor­di­nate.

Exam­ple 6: the cen­troid through a median#

Find the cen­troid of the tri­an­gle with ver­tices $A(-1, 4)$, $B(5, 2)$ and $C(2, -3)$, first using a median and then using the for­mula.

The mid­point of $BC$ is

$$\displaystyle D = \left( \frac{5 + 2}{2},\ \frac{2 + (-3)}{2} \right) = \left( \frac{7}{2},\ -\frac{1}{2} \right)$$

The cen­troid divides $AD$ in the ratio $2 : 1$:

$$\displaystyle \begin{aligned} x &= \frac{2 \times \frac{7}{2} + 1 \times (-1)}{3} = \frac{7 - 1}{3} = 2 \\ y &= \frac{2 \times \left(-\frac{1}{2}\right) + 1 \times 4}{3} = \frac{-1 + 4}{3} = 1 \end{aligned}$$

By the for­mula: $\displaystyle G = \left( \frac{-1 + 5 + 2}{3}, \frac{4 + 2 - 3}{3} \right) = (2, 1)$. Both meth­ods give $G(2, 1)$.

Com­mon mis­takes#

• Pair­ing the ratio the wrong way: writ­ing $m_1 x_1 + m_2 x_2$ instead of $m_1 x_2 + m_2 x_1$.
• Read­ing the ratio from the wrong end. "$P$ divides $AB$ in $1 : 2$" means $P$ is nearer to $A$.
• Giv­ing only one point of tri­sec­tion when the ques­tion asks for both.
• Using $y = 0$ for divi­sion by the y-axis. On the y-axis it is $x = 0$ that holds.
• Leav­ing the ratio as $\displaystyle \frac{2}{3} : 1$ instead of writ­ing $2 : 3$.
• For a par­al­lel­o­gram, equat­ing the mid­points of adja­cent ver­tices, such as $AB$ and $CD$. It is the diag­o­nals, $AC$ and $BD$, that bisect each other.
• Divid­ing by 2 instead of 3 in the cen­troid for­mula.

Try these#

1. Find the point that divides the seg­ment join­ing $(4, -3)$ and $(-6, 7)$ inter­nally in the ratio $3 : 2$. Answer: $(-2, 3)$.
2. In what ratio does the y-axis divide the seg­ment join­ing $(-4, 2)$ and $(6, -3)$? Find the point. Answer: $2 : 3$, at $(0, 0)$.
3. The mid­point of $AB$ is $(2, -1)$ and $A = (-3, 4)$. Find $B$. Answer: $B(7, -6)$.
4. Find the points of tri­sec­tion of the seg­ment join­ing $(2, -2)$ and $(-7, 4)$. Answer: $(-1, 0)$ and $(-4, 2)$.
5. $A(-2, 1)$, $B(3, 2)$ and $C(5, 6)$ are three ver­tices of par­al­lel­o­gram $ABCD$ in order. Find $D$. Answer: $D(0, 5)$.
6. Find the cen­troid of the tri­an­gle with ver­tices $(0, 6)$, $(8, 0)$ and $(-2, -3)$. Answer: $(2, 1)$.

Key terms#

Sec­tion for­mula

The coor­di­nates of the point divid­ing a seg­ment inter­nally in the ratio $m_1 : m_2$.

Inter­nal divi­sion

Divi­sion of a seg­ment by a point lying between its end points.

Mid­point

The point divid­ing a seg­ment in the ratio $1 : 1$.

Points of tri­sec­tion

The two points that divide a seg­ment into three equal parts.

Median

A seg­ment join­ing a ver­tex of a tri­an­gle to the mid­point of the oppo­site side.

Cen­troid

The com­mon point of the three medi­ans of a tri­an­gle; it divides each median in the ratio $2 : 1$ from the ver­tex.

Ratio k : 1

A way of writ­ing an unknown ratio with a sin­gle unknown $k$.

Com­mon ques­tions#

Why is the k : 1 form so use­ful?#

An unknown ratio $m_1 : m_2$ has two unknowns but only its quo­tient mat­ters. Writ­ing it as $k : 1$ leaves one unknown, which a sin­gle equa­tion can find.

How do I know which coor­di­nate to use when find­ing a ratio?#

Use the coor­di­nate you are given. If the point lies on the x-axis, use $y = 0$. If it lies on the y-axis, use $x = 0$. Then use the other coor­di­nate to find the full point.

What if k comes out neg­a­tive?#

A neg­a­tive $k$ means the point lies out­side the seg­ment, so the divi­sion is exter­nal. For a ques­tion about inter­nal divi­sion, recheck your work­ing.

Is the mid­point for­mula a sep­a­rate result?#

No. It is the sec­tion for­mula with $m_1 = m_2$, so the two parts of each coor­di­nate are sim­ply aver­aged.

Can I find the fourth ver­tex with­out the mid­point method?#

The short­cut $D = A + C - B$ gives the same answer, but it comes from the diag­o­nals bisect­ing each other. In an exam, show the mid­point work­ing.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.
3. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
4. Cox­eter, H. S. M. and Gre­itzer, S. L. Geom­e­try Revis­ited. Math­e­mat­i­cal Asso­ci­a­tion of Amer­ica.