A Weighted Average of Two Points Lands You Anywhere Along the Line Between Them

The distance formula told you how far apart two points are. It said nothing about what lies between them — and that is the gap this part fills.

The question is easy to state. Given two points, where is the point that sits a quarter of the way from the first to the second? A third of the way? Two-fifths? The answer is a weighted average of the two sets of coordinates, and the weights are exactly the two numbers in the ratio, used crosswise.

The simplest case makes the pattern obvious. The mid-point of $(3,4)$ and $(-1,2)$ is

$$\left(\frac{3+(-1)}{2},\frac{4+2}{2}\right)=(1,3)$$

**Average the $x$-coordinates and average the $y$-coordinates. The section formula is that same idea with unequal weights, so that the point can sit anywhere along the segment rather than only halfway.

From one formula come four kinds of question, and the last two are the ones that carry the most marks:

- Given the ratio, find the point — a direct substitution
- Given the point, find the ratio — set up one equation in the unknown ratio
- Find the ratio in which an axis divides a segment** — use the fact that a point on the $x$-axis has $y=0$
- Complete a parallelogram or find a point of trisection — use the mid-point in reverse

The fourth of these rests on one geometric fact that is worth stating now. The diagonals of a parallelogram bisect each other, so the mid-point of one diagonal is the mid-point of the other. That single sentence turns "find the fourth vertex" into two one-line equations.

This page covers the second part of the CBSE Class 10 Maths chapter on coordinate geometry: the section formula, the mid-point formula, finding a dividing ratio, trisection, and completing a parallelogram.

**Because the mid-point divides the segment in the ratio $1:1$, and equal weights turn a weighted average into an ordinary average.**

Substituting $m_1=m_2=1$ into the section formula:

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

so the mid-point formula is

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

**Average the $x$-coordinates, average the $y$-coordinates. There is nothing to cross over here, which is why the mid-point formula almost never goes wrong and the section formula often does.

Worked example 1.** Find the mid-point of $(3,4)$ and $(-1,2)$.

$$M=\left(\frac{3-1}{2},\frac{4+2}{2}\right)=(1,3)$$

Worked example 2 — the centre of a circle. The end points of a diameter of a circle are $(2,-3)$ and $(-2,5)$. Find the centre.

The centre is the mid-point of any diameter:

$$\left(\frac{2-2}{2},\frac{-3+5}{2}\right)=(0,1)$$

**The centre is $(0,1)$.

Check with the distance formula.** From $(0,1)$ to $(2,-3)$ is $\sqrt{4+16}=\sqrt{20}$, and to $(-2,5)$ is $\sqrt{4+16}=\sqrt{20}$. Equal, as radii must be.

Worked example 3 — working backwards to the other end. The mid-point of a segment is $(2,-3)$ and one end is $(4,1)$. Find the other end.

Let the other end be $(x,y)$. From the mid-point formula,

$$\frac{4+x}{2}=2⟹4+x=4⟹x=0$$
$$\frac{1+y}{2}=-3⟹1+y=-6⟹y=-7$$

**The other end is $(0,-7)$.

Check**: the mid-point of $(4,1)$ and $(0,-7)$ is $\left(\frac{4}{2},\frac{-6}{2}\right)=(2,-3)$. Correct.

A faster route to the same answer. Doubling the mid-point and subtracting the known end gives the other end directly:

$$(x,y)=(2\times 2-4,2\times (-3)-1)=(0,-7)$$

That shortcut is worth knowing, and it follows straight from rearranging the mid-point formula.

Worked example 4 — with unknowns in the coordinates. The mid-point of the segment joining $A(2a,4)$ and $B(-2,3b)$ is $(1,2a+1)$. Find $a$ and $b$.

The mid-point is

$$\left(\frac{2a-2}{2},\frac{4+3b}{2}\right)=\left(a-1,\frac{4+3b}{2}\right)$$

Equating the $x$-coordinates:

$$a-1=1⟹a=2$$

Equating the $y$-coordinates, and using $a=2$ so that $2a+1=5$:

$$\frac{4+3b}{2}=5⟹4+3b=10⟹b=2$$

**So $a=2$ and $b=2$.

Check**: the points become $A(4,4)$ and $B(-2,6)$, whose mid-point is $(1,5)$; and $2a+1=5$. Both coordinates match.

The order to work in matters here. The $x$-equation contained only $a$, so it could be solved on its own; the $y$-equation contained both $a$ and $b$. Solve the equation with one unknown first, and the second becomes a one-line substitution rather than a simultaneous pair.

**Let the ratio be $k:1$, substitute into the section formula, and use whichever coordinate you know to solve for $k$. Using $k:1$ instead of $m_1:m_2$ replaces two unknowns with one.

Worked example 1 — a given point.** Find the ratio in which $(-4,6)$ divides the segment joining $A(-6,10)$ and $B(3,-8)$.

Let the ratio be $k:1$. Then the $x$-coordinate of the dividing point is

$$\frac{k(3)+1(-6)}{k+1}=-4$$
$$3k-6=-4k-4$$
$$7k=2⟹k=\frac{2}{7}$$

**So the ratio is $\frac{2}{7}:1$, which is $2:7$.

Check with the $y$-coordinate**, which was not used:

$$\frac{\frac{2}{7}(-8)+10}{\frac{2}{7}+1}=\frac{-\frac{16}{7}+\frac{70}{7}}{\frac{9}{7}}=\frac{\frac{54}{7}}{\frac{9}{7}}=6$$

**The $y$-coordinate comes out as $6$, exactly as given.** That is a genuine independent check, because only the $x$-coordinate was used to find $k$ — and it is the step that proves the point really does lie on the segment. If the second coordinate disagrees, the point is not on the line at all.

**Worked example 2 — the $x$-axis.** In what ratio does the $x$-axis divide the segment joining $(1,-5)$ and $(-4,5)$? Find the point of division too.

**The key observation: any point on the $x$-axis has $y=0$.** So set the $y$-coordinate of the section formula to zero with ratio $k:1$:

$$\frac{k(5)+1(-5)}{k+1}=0⟹5k-5=0⟹k=1$$

**The ratio is $1:1$**, so the $x$-axis crosses the segment at its mid-point. The point of division is therefore

$$\left(\frac{1-4}{2},\frac{-5+5}{2}\right)=\left(-\frac{3}{2},0\right)$$

Check that it makes sense. One end has $y=-5$ and the other $y=+5$, symmetric about zero, so the crossing must be halfway. The answer agrees with the geometry, and a ratio other than $1:1$ would have been a sign of error.

**Worked example 3 — the $y$-axis.** In what ratio does the $y$-axis divide the segment joining $(5,-6)$ and $(-1,-4)$? Find the point.

**Any point on the $y$-axis has $x=0$:**

$$\frac{k(-1)+1(5)}{k+1}=0⟹-k+5=0⟹k=5$$

**The ratio is $5:1$.** The $y$-coordinate is then

$$\frac{5(-4)+1(-6)}{6}=\frac{-20-6}{6}=-\frac{26}{6}=-\frac{13}{3}$$

**The point of division is $\left(0,-\frac{13}{3}\right)$.

Check by the fraction-of-the-way reasoning.** A ratio of $5:1$ puts the point five-sixths of the way from $(5,-6)$ to $(-1,-4)$. The $x$-coordinate travels from $5$ to $-1$, a change of $-6$, and five-sixths of that is $-5$, landing on $0$. Correct.

Worked example 4 — a ratio with a letter in it. Find the ratio in which the point $\left(\frac{3}{4},\frac{5}{12}\right)$ divides the segment joining $\left(\frac{1}{2},\frac{3}{2}\right)$ and $(2,-5)$.

Using the $x$-coordinate with ratio $k:1$:

$$\frac{2k+\frac{1}{2}}{k+1}=\frac{3}{4}$$
$$4\left(2k+\frac{1}{2}\right)=3(k+1)⟹8k+2=3k+3$$
$$5k=1⟹k=\frac{1}{5}$$

**The ratio is $1:5$.

Check with the $y$-coordinate:**

$$\frac{\frac{1}{5}(-5)+\frac{3}{2}}{\frac{1}{5}+1}=\frac{-1+\frac{3}{2}}{\frac{6}{5}}=\frac{\frac{1}{2}}{\frac{6}{5}}=\frac{1}{2}\times \frac{5}{6}=\frac{5}{12}$$

**Exactly the given $y$-coordinate. Fractions in the data did not change the method at all.

The two things to take from this section.** Use $k:1$ so there is only one unknown, and always verify with the coordinate you did not use. That verification costs three lines and does something no other check can: it confirms the point lies on the segment rather than merely having the right value in one coordinate.

**Trisection needs the section formula twice, in the ratios $1:2$ and $2:1$. A parallelogram's missing vertex needs only the fact that the diagonals bisect each other.

Worked example 1 — points of trisection.** Find the points of trisection of the segment joining $A(4,-1)$ and $B(-2,-3)$.

Trisection means dividing the segment into three equal parts, so there are two points. The first, $P$, is one-third of the way from $A$, which is the ratio $1:2$:

$$P_x=\frac{1(-2)+2(4)}{3}=\frac{-2+8}{3}=2$$
$$P_y=\frac{1(-3)+2(-1)}{3}=\frac{-3-2}{3}=-\frac{5}{3}$$

The second, $Q$, is two-thirds of the way, which is the ratio $2:1$:

$$Q_x=\frac{2(-2)+1(4)}{3}=\frac{-4+4}{3}=0$$
$$Q_y=\frac{2(-3)+1(-1)}{3}=\frac{-6-1}{3}=-\frac{7}{3}$$

**The points of trisection are $\left(2,-\frac{5}{3}\right)$ and $\left(0,-\frac{7}{3}\right)$.

Check with a mid-point.** Since the three parts are equal, $Q$ must be the mid-point of $P$ and $B$:

$$\left(\frac{2+(-2)}{2},\frac{-\frac{5}{3}+(-3)}{2}\right)=\left(0,\frac{-\frac{5}{3}-\frac{9}{3}}{2}\right)=\left(0,\frac{-\frac{14}{3}}{2}\right)=\left(0,-\frac{7}{3}\right)$$

**Exactly $Q$.** And by the same reasoning $P$ must be the mid-point of $A$ and $Q$, which is a second check available for free.

The mistake this question is designed to catch. Using the ratios $1:3$ and $2:3$. **Trisection points divide the segment $1:2$ and $2:1$, because the ratio compares the two pieces**, not a piece with the whole. The three pieces are each one-third, so the first point has one piece behind it and two ahead — hence $1:2$.

Worked example 2 — the fourth vertex of a parallelogram. Three vertices of a parallelogram $ABCD$ are $A(1,2)$, $B(4,3)$ and $C(6,6)$. Find $D$.

The diagonals of a parallelogram bisect each other, so the mid-point of $AC$ equals the mid-point of $BD$.

Mid-point of $AC$:

$$\left(\frac{1+6}{2},\frac{2+6}{2}\right)=\left(\frac{7}{2},4\right)$$

Let $D=(x,y)$. Mid-point of $BD$:

$$\left(\frac{4+x}{2},\frac{3+y}{2}\right)$$

Equating the two:

$$\frac{4+x}{2}=\frac{7}{2}⟹x=3$$
$$\frac{3+y}{2}=4⟹y=5$$

**So $D=(3,5)$.

Check with the sides.** In a parallelogram $AB$ and $DC$ must be equal and parallel. Going from $A(1,2)$ to $B(4,3)$ means moving $3$ right and $1$ up; going from $D(3,5)$ to $C(6,6)$ means moving $3$ right and $1$ up. Identical, so $ABCD$ really is a parallelogram.

Which diagonal you use is decided by the lettering. In $ABCD$ the diagonals are $AC$ and $BD$, because the vertices are named in order around the figure. **Pairing $A$ with $B$ instead would give a side, not a diagonal**, and the resulting equations would place $D$ somewhere meaningless.

Worked example 3 — two unknown coordinates. If $(1,2)$, $(4,y)$, $(x,6)$ and $(3,5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$.

The diagonals are the first-and-third and the second-and-fourth points. Equating their mid-points:

$$\left(\frac{1+x}{2},\frac{2+6}{2}\right)=\left(\frac{4+3}{2},\frac{y+5}{2}\right)$$

From the $x$-coordinates:

$$\frac{1+x}{2}=\frac{7}{2}⟹x=6$$

From the $y$-coordinates:

$$4=\frac{y+5}{2}⟹y+5=8⟹y=3$$

**So $x=6$ and $y=3$**, and the four vertices are $(1,2)$, $(4,3)$, $(6,6)$ and $(3,5)$ — the same parallelogram as the previous example, which is a reassuring consistency.

Worked example 4 — three equal parts of a different kind. Find the coordinates of the points that divide the segment joining $(-2,0)$ and $(0,8)$ into four equal parts.

Four equal parts means three points, at the ratios $1:3$, $1:1$ and $3:1$:

$$P_1=\left(\frac{1(0)+3(-2)}{4},\frac{1(8)+3(0)}{4}\right)=\left(-\frac{3}{2},2\right)$$
$$P_2=\left(\frac{-2+0}{2},\frac{0+8}{2}\right)=(-1,4)$$
$$P_3=\left(\frac{3(0)+1(-2)}{4},\frac{3(8)+1(0)}{4}\right)=\left(-\frac{1}{2},6\right)$$

Check the pattern. The $y$-coordinates run $0,2,4,6,8$ — equal steps of $2$ — and the $x$-coordinates run $-2,-\frac{3}{2},-1,-\frac{1}{2},0$, equal steps of $\frac{1}{2}$. Both coordinates form arithmetic progressions, which is exactly what equal division along a straight line must produce, and it is the fastest way to verify any answer of this kind.

Notice how the number of points relates to the number of parts. **Dividing into $n$ equal parts needs $n-1$ points**, at the ratios $1:(n-1)$, $2:(n-2)$, and so on. Trisection is the case $n=3$ with two points; four parts needs three points. Counting the points wrongly is the first thing to check if an answer feels short.