NCERT Class 10 Mathematics — Coordinate Geometry, Exercise 7.2. All 10 questions solved.

Exercise 7.1 measured segments. Exercise 7.2 cuts them. The tool is the section formula. The point $P$ that divides the segment from $A(x_1, y_1)$ to $B(x_2, y_2)$ internally in the ratio $m_1 : m_2$, so that $\frac{AP}{PB} = \frac{m_1}{m_2}$, is

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

Two special cases do much of the work:

• Midpoint (ratio $1 : 1$): $\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$.
• Unknown ratio. Write it as $k : 1$. Then there is a single unknown, and one coordinate equation finds it.

Key insight. In the section formula each weight goes on the far endpoint. $m_1$, the share next to $A$, multiplies $B$’s coordinates, and $m_2$ multiplies $A$’s. That feels backwards, but it has to be so: if $m_1$ is small, $P$ is near $A$, so $A$ must carry the larger weight. Every question below is this formula run forwards (ratio known, find the point) or backwards (point known, find the ratio or a missing vertex).

Question 1

Find the coordinates of the point which divides the join of $(-1, 7)$ and $(4, -3)$ in the ratio $2 : 3$.

Solution. Here $A(-1, 7)$, $B(4, -3)$, $m_1 = 2$, $m_2 = 3$. The $2$ multiplies $B$’s coordinates and the $3$ multiplies $A$’s:

$$x = \frac{2(4) + 3(-1)}{2 + 3} = \frac{5}{5} = 1, \qquad y = \frac{2(-3) + 3(7)}{2 + 3} = \frac{15}{5} = 3$$

A quick sense check: the ratio $2 : 3$ puts the point two-fifths of the way from $A$ to $B$. The $x$-coordinate moves $\frac{2}{5}$ of the way from $-1$ to $4$, which is $-1 + 2 = 1$.

Question 2

Find the coordinates of the points of trisection of the line segment joining $(4, -1)$ and $(-2, -3)$.

Solution. Trisection points cut the segment into three equal parts, so there are two of them. With $A(4, -1)$ and $B(-2, -3)$, the first, $P$, divides $AB$ in the ratio $1 : 2$ and the second, $Q$, in the ratio $2 : 1$.

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

$$Q = \left(\frac{2(-2) + 1(4)}{3},\; \frac{2(-3) + 1(-1)}{3}\right) = \left(0,\; -\frac{7}{3}\right)$$

Check: $Q$ should be the midpoint of $P$ and $B$, and $\left(\frac{2 – 2}{2}, \frac{-5/3 – 3}{2}\right) = \left(0, -\frac{7}{3}\right)$.

Question 3

To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of $1$ m each. $100$ flower pots have been placed at a distance of $1$ m from each other along AD, as shown in Fig. 7.12. Niharika runs $\frac{1}{4}$th the distance AD on the $2$nd line and posts a green flag. Preet runs $\frac{1}{5}$th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

In Fig. 7.12, A is the bottom-left corner of the ground, B bottom right, C top right and D top left. The chalk lines, numbered 1 to 10, run parallel to AD, $1$ m apart, starting from the side AD. The flower pots stand along AD, the first $1$ m from A and each one $1$ m beyond the last, so the hundredth is at D and $AD = 100$ m.

Solution. Make the ground a coordinate plane: $A$ is the origin, $AB$ is the $x$-axis and $AD$ is the $y$-axis, with $1$ m as the unit. A point on the $k$th line at a distance $y$ metres from $AB$ is then $(k, y)$.

Niharika runs $\frac{1}{4} \times 100 = 25$ m along the 2nd line, so the green flag is at $G(2, 25)$. Preet runs $\frac{1}{5} \times 100 = 20$ m along the 8th line, so the red flag is at $R(8, 20)$.

$$GR = \sqrt{(8 – 2)^2 + (20 – 25)^2} = \sqrt{36 + 25} = \sqrt{61} \text{ m}$$

“Exactly halfway” means the midpoint of $GR$:

$$M = \left(\frac{2 + 8}{2},\; \frac{25 + 20}{2}\right) = (5,\; 22.5)$$

An $x$-coordinate of $5$ means the 5th line, and a $y$-coordinate of $22.5$ means $22.5$ m along it from $AB$.

Question 4

Find the ratio in which the line segment joining the points $(-3, 10)$ and $(6, -8)$ is divided by $(-1, 6)$.

Solution. The ratio is unknown, so call it $k : 1$. With $A(-3, 10)$ and $B(6, -8)$, the dividing point is

$$\left(\frac{6k – 3}{k + 1},\; \frac{-8k + 10}{k + 1}\right)$$

Its $x$-coordinate must be $-1$:

$$\frac{6k – 3}{k + 1} = -1 \;\Rightarrow\; 6k – 3 = -k – 1 \;\Rightarrow\; 7k = 2 \;\Rightarrow\; k = \frac{2}{7}$$

The $y$-coordinate must agree, otherwise $(-1, 6)$ would not lie on the segment at all. With $k = \frac{2}{7}$:

$$\frac{-8 \cdot \frac{2}{7} + 10}{\frac{2}{7} + 1} = \frac{\frac{54}{7}}{\frac{9}{7}} = 6 \checkmark$$

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

Question 5

Find the ratio in which the line segment joining A$(1, -5)$ and B$(-4, 5)$ is divided by the $x$-axis. Also find the coordinates of the point of division.

Solution. Let the $x$-axis cut $AB$ in the ratio $k : 1$. The point of division is

$$\left(\frac{-4k + 1}{k + 1},\; \frac{5k – 5}{k + 1}\right)$$

What we know about this point is that it lies on the $x$-axis, so its $y$-coordinate is $0$. That gives $5k – 5 = 0$, so $k = 1$. The ratio is $1 : 1$ and the point is the midpoint:

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

This makes sense: $A$ is $5$ units below the axis and $B$ is $5$ units above it, so the axis cuts the segment exactly in half.

Question 6

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

Solution. The diagonals of a parallelogram bisect each other, so both diagonals have the same midpoint. “Taken in order” fixes which vertices are opposite. With $A(1, 2)$, $B(4, y)$, $C(x, 6)$ and $D(3, 5)$, the diagonals are $AC$ and $BD$.

$$\text{Midpoint of } AC = \left(\frac{1 + x}{2},\; \frac{2 + 6}{2}\right) = \left(\frac{1 + x}{2},\; 4\right)$$

$$\text{Midpoint of } BD = \left(\frac{4 + 3}{2},\; \frac{y + 5}{2}\right) = \left(\frac{7}{2},\; \frac{y + 5}{2}\right)$$

Equating coordinates: $1 + x = 7$, so $x = 6$, and $y + 5 = 8$, so $y = 3$. Both diagonals then meet at $\left(\frac{7}{2}, 4\right)$.

Question 7

Find the coordinates of a point A, where AB is the diameter of a circle whose centre is $(2, -3)$ and B is $(1, 4)$.

Solution. The centre of a circle is the midpoint of every diameter. With $A(x, y)$ and $B(1, 4)$:

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

So $x = 3$ and $y = -10$. In words, to get from B to the centre you move $1$ right and $7$ down, and A is one more such step beyond the centre: $(2 + 1, -3 – 7)$.

Question 8

If A and B are $(-2, -2)$ and $(2, -4)$, respectively, find the coordinates of P such that $\text{AP} = \dfrac{3}{7}\,\text{AB}$ and P lies on the line segment AB.

Solution. The section formula needs the ratio $AP : PB$, not $AP : AB$. Since $P$ is on the segment, $PB = AB – AP = \frac{4}{7}AB$, so

$$AP : PB = \frac{3}{7} : \frac{4}{7} = 3 : 4$$

With $m_1 = 3$ on $B(2, -4)$ and $m_2 = 4$ on $A(-2, -2)$:

$$P = \left(\frac{3(2) + 4(-2)}{7},\; \frac{3(-4) + 4(-2)}{7}\right) = \left(-\frac{2}{7},\; -\frac{20}{7}\right)$$

Question 9

Find the coordinates of the points which divide the line segment joining A$(-2, 2)$ and B$(2, 8)$ into four equal parts.

Solution. Three points, $P$, $Q$ and $R$, cut $AB$ into four equal parts. They divide it in the ratios $1 : 3$, $1 : 1$ and $3 : 1$. The middle one is simply the midpoint, and the other two are midpoints of the halves.

$$Q = \left(\frac{-2 + 2}{2},\; \frac{2 + 8}{2}\right) = (0, 5)$$

$$P = \text{midpoint of } AQ = \left(\frac{-2 + 0}{2},\; \frac{2 + 5}{2}\right) = \left(-1,\; \frac{7}{2}\right)$$

$$R = \text{midpoint of } QB = \left(\frac{0 + 2}{2},\; \frac{5 + 8}{2}\right) = \left(1,\; \frac{13}{2}\right)$$

The section formula agrees. For $P$ with ratio $1 : 3$, $\left(\frac{1(2) + 3(-2)}{4}, \frac{1(8) + 3(2)}{4}\right) = \left(-1, \frac{7}{2}\right)$.

Question 10

Find the area of a rhombus if its vertices are $(3, 0)$, $(4, 5)$, $(-1, 4)$ and $(-2, -1)$ taken in order. [Hint: Area of a rhombus $= \frac{1}{2}$ (product of its diagonals)]

Solution. With the vertices in order as $A(3, 0)$, $B(4, 5)$, $C(-1, 4)$ and $D(-2, -1)$, the diagonals join opposite vertices, $AC$ and $BD$. This is the distance formula from Exercise 7.1 again:

$$AC = \sqrt{(-1 – 3)^2 + (4 – 0)^2} = \sqrt{32} = 4\sqrt{2}$$

$$BD = \sqrt{(-2 – 4)^2 + (-1 – 5)^2} = \sqrt{72} = 6\sqrt{2}$$

$$\text{Area} = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 48 = 24 \text{ square units}$$

It is worth confirming the figure really is a rhombus: each side works out to $\sqrt{26}$ (for instance $AB = \sqrt{1^2 + 5^2}$). The diagonals also share the midpoint $(1, 2)$, as the diagonals of any parallelogram must.

Common mistakes

• Putting the weights on the wrong endpoints in question 1. Writing $\frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$ gives $(2, 1)$, the point dividing the segment in $3 : 2$. That point is closer to $B$, not $A$. Each weight multiplies the far endpoint.
• Giving only one point in question 2. “Points of trisection” is plural. There are two, at ratios $1 : 2$ and $2 : 1$.
• Using $AP : AB$ as the ratio in question 8. $\frac{3}{7}$ compares $AP$ with the whole segment. The section formula needs $AP : PB = 3 : 4$. Using $3 : 7$ puts $P$ in the wrong place.
• Pairing the wrong vertices in question 6. “Taken in order” means the diagonals join the 1st and 3rd vertices and the 2nd and 4th. Equating the midpoints of $AB$ and $CD$ instead produces values that do not give a parallelogram.
• Working from the wrong coordinate in question 5. What is known about a point on the $x$-axis is that its $y$-coordinate is $0$, so the ratio comes from the $y$-equation. The $x$-coordinate is not known in advance.
• Mixing up the axes in question 3. The line number is the $x$-coordinate and the distance run is the $y$-coordinate. Writing the green flag as $(25, 2)$, or taking $AD$ as $10$ m because there are ten lines, spoils everything after it.
• Multiplying sides instead of diagonals in question 10. The hint’s formula uses the diagonals, $4\sqrt{2}$ and $6\sqrt{2}$, and the factor $\frac12$ matters. Without it you get $48$.

Practise next

• Exercise 7.1 — the distance formula, used again in questions 3 and 10 here, and the tool for classifying triangles and quadrilaterals.
• Class 11 Straight Lines, Exercise 9.1 — slopes and collinearity. It also uses the midpoint and section formula to find a missing vertex or a point on a line.
• Chapter 8, Exercise 8.1 — the next chapter, trigonometric ratios, where the right-angled triangles behind the distance formula come back.