NCERT Class 10 Mathematics — Coordinate Geometry, Exercise 7.1. All 10 questions solved.

Exercise 7.1 has one tool and asks for it ten different ways. The tool is the distance formula, which is Pythagoras’ theorem written in coordinates. The distance between $P(x_1, y_1)$ and $Q(x_2, y_2)$ is

$$PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$

and the distance of $P(x, y)$ from the origin is $\sqrt{x^2 + y^2}$. The questions then use it for three jobs:

• Collinearity. Three points lie on one line exactly when the two shorter distances add up to the longest.
• Naming shapes. A triangle or quadrilateral is identified by comparing the lengths of its sides and, for quadrilaterals, its diagonals.
• Equidistance. “Equidistant from A and B” means $PA = PB$, which is solved by squaring both sides.

Key insight. Work with squared distances whenever you compare or equate them. $PA = PB$ is the same statement as $PA^2 = PB^2$, and squaring removes the square roots. It also makes the $x^2$ and $y^2$ terms cancel, so the equation turns out to be linear. That is why questions 7, 9 and 10 look quadratic but are not. The same habit spares you surds when checking whether four sides are equal in questions 5 and 6.

Question 1

Find the distance between the following pairs of points:

(i) $(2, 3)$, $(4, 1)$    (ii) $(-5, 7)$, $(-1, 3)$    (iii) $(a, b)$, $(-a, -b)$

Solution. Subtract coordinates carefully, especially where a negative is being subtracted, square the differences and add.

(i) $\sqrt{(4 – 2)^2 + (1 – 3)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$

(ii) $\sqrt{(-1 – (-5))^2 + (3 – 7)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2}$

(iii) $\sqrt{(-a – a)^2 + (-b – b)^2} = \sqrt{4a^2 + 4b^2} = 2\sqrt{a^2 + b^2}$

In (iii) the two points are reflections of each other through the origin, so the answer is twice the distance of $(a, b)$ from the origin, which is exactly what $2\sqrt{a^2 + b^2}$ says.

Question 2

Find the distance between the points $(0, 0)$ and $(36, 15)$. Can you now find the distance between the two towns A and B discussed in Section 7.2.

Solution. The distance from the origin is

$$\sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39$$

In Section 7.2, town B is $36$ km east and $15$ km north of town A. Put A at the origin, take east along the $x$-axis and north along the $y$-axis, and let one unit stand for $1$ km. Then B is the point $(36, 15)$, and the distance between the towns is the $39$ units just found, that is, $39$ km.

Question 3

Determine if the points $(1, 5)$, $(2, 3)$ and $(-2, -11)$ are collinear.

Solution. Call the points $A(1, 5)$, $B(2, 3)$ and $C(-2, -11)$. If they lie on one line, one of them sits between the other two, and the two short distances add up to the long one.

$$AB = \sqrt{1^2 + (-2)^2} = \sqrt{5}$$

$$BC = \sqrt{(-4)^2 + (-14)^2} = \sqrt{212} = 2\sqrt{53}$$

$$AC = \sqrt{(-3)^2 + (-16)^2} = \sqrt{265}$$

The longest is $AC$ (about $16.28$), and $AB + BC$ is about $2.24 + 14.56 = 16.80$, which is already different. To be sure without decimals, square $AB + BC$:

$$\bigl(\sqrt{5} + \sqrt{212}\bigr)^2 = 217 + 2\sqrt{1060}$$

For this to equal $265$ we would need $\sqrt{1060} = 24$, that is $1060 = 576$, which is false. So $AB + BC \ne AC$, and the points are not collinear.

Question 4

Check whether $(5, -2)$, $(6, 4)$ and $(7, -2)$ are the vertices of an isosceles triangle.

Solution. An isosceles triangle needs two equal sides. Call the points $A(5, -2)$, $B(6, 4)$ and $C(7, -2)$.

$$AB = \sqrt{1^2 + 6^2} = \sqrt{37}, \qquad BC = \sqrt{1^2 + (-6)^2} = \sqrt{37}, \qquad AC = \sqrt{2^2 + 0^2} = 2$$

$AB = BC$, and the points do form a triangle, since $2 < \sqrt{37} + \sqrt{37}$ and the three are not on one line. The picture agrees: $A$ and $C$ share the $y$-coordinate $-2$, and $B$ sits directly above their midpoint $(6, -2)$.

Question 5

In a classroom, $4$ friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Fig. 7.8 is a grid of columns $0$ to $10$ and rows $0$ to $10$. A sits at column 3, row 4; B at column 6, row 7; C at column 9, row 4; D at column 6, row 1. Taking columns as $x$ and rows as $y$:

Solution. The coordinates are $A(3, 4)$, $B(6, 7)$, $C(9, 4)$ and $D(6, 1)$. A square needs four equal sides and two equal diagonals. Four equal sides alone only make a rhombus.

$$AB = \sqrt{3^2 + 3^2} = 3\sqrt{2}, \quad BC = \sqrt{3^2 + (-3)^2} = 3\sqrt{2}, \quad CD = \sqrt{(-3)^2 + (-3)^2} = 3\sqrt{2}, \quad DA = \sqrt{(-3)^2 + 3^2} = 3\sqrt{2}$$

$$AC = \sqrt{6^2 + 0^2} = 6, \qquad BD = \sqrt{0^2 + (-6)^2} = 6$$

All four sides are equal and the diagonals are equal, so ABCD is a square. As a check, $AB^2 + BC^2 = 18 + 18 = 36 = AC^2$, so the angle at B is a right angle, as it must be.

Question 6

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

(i) $(-1, -2)$, $(1, 0)$, $(-1, 2)$, $(-3, 0)$

(ii) $(-3, 5)$, $(3, 1)$, $(0, 3)$, $(-1, -4)$

(iii) $(4, 5)$, $(7, 6)$, $(4, 3)$, $(1, 2)$

Solution. Take the points in order as $A$, $B$, $C$, $D$, find all four sides and both diagonals, and then match the pattern:

• all sides equal and diagonals equal: square
• all sides equal, diagonals unequal: rhombus
• opposite sides equal and diagonals equal: rectangle
• opposite sides equal, diagonals unequal: parallelogram

(i) $A(-1, -2)$, $B(1, 0)$, $C(-1, 2)$, $D(-3, 0)$:

$$AB = \sqrt{2^2 + 2^2} = 2\sqrt{2}, \quad BC = \sqrt{(-2)^2 + 2^2} = 2\sqrt{2}, \quad CD = \sqrt{(-2)^2 + (-2)^2} = 2\sqrt{2}, \quad DA = \sqrt{2^2 + (-2)^2} = 2\sqrt{2}$$

$$AC = \sqrt{0^2 + 4^2} = 4, \qquad BD = \sqrt{(-4)^2 + 0^2} = 4$$

Equal sides and equal diagonals: a square.

(ii) $A(-3, 5)$, $B(3, 1)$, $C(0, 3)$, $D(-1, -4)$. Before naming a shape, check that one exists. Look at $A$, $B$ and $C$:

$$AB = \sqrt{6^2 + (-4)^2} = \sqrt{52} = 2\sqrt{13}, \qquad AC = \sqrt{3^2 + (-2)^2} = \sqrt{13}, \qquad CB = \sqrt{3^2 + (-2)^2} = \sqrt{13}$$

$AC + CB = 2\sqrt{13} = AB$, so $C$ lies on the segment $AB$. In fact it is the midpoint. Three of the four points are on one straight line, so the four points cannot be the corners of a quadrilateral: joining them gives a triangle with an extra point on one side. There is no quadrilateral.

(iii) $A(4, 5)$, $B(7, 6)$, $C(4, 3)$, $D(1, 2)$:

$$AB = \sqrt{3^2 + 1^2} = \sqrt{10}, \quad BC = \sqrt{(-3)^2 + (-3)^2} = 3\sqrt{2}, \quad CD = \sqrt{(-3)^2 + (-1)^2} = \sqrt{10}, \quad DA = \sqrt{3^2 + 3^2} = 3\sqrt{2}$$

$$AC = \sqrt{0^2 + (-2)^2} = 2, \qquad BD = \sqrt{(-6)^2 + (-4)^2} = 2\sqrt{13}$$

Opposite sides are equal ($AB = CD$ and $BC = DA$), but the adjacent sides differ, so it is not a rhombus. The diagonals are unequal, so it is not a rectangle. It is a parallelogram.

Question 7

Find the point on the $x$-axis which is equidistant from $(2, -5)$ and $(-2, 9)$.

Solution. Every point on the $x$-axis has $y$-coordinate $0$, so the point is $P(x, 0)$ and only $x$ is unknown. Equidistance means equal squared distances:

$$(x – 2)^2 + (0 + 5)^2 = (x + 2)^2 + (0 – 9)^2$$

$$x^2 – 4x + 4 + 25 = x^2 + 4x + 4 + 81$$

The $x^2$ terms cancel, as the key insight promised, leaving $-8x = 56$, so $x = -7$.

As a check, both distances are $\sqrt{81 + 25} = \sqrt{106}$ and $\sqrt{25 + 81} = \sqrt{106}$.

Question 8

Find the values of $y$ for which the distance between the points P$(2, -3)$ and Q$(10, y)$ is $10$ units.

Solution. Set the squared distance equal to $10^2$:

$$(10 – 2)^2 + (y + 3)^2 = 100 \;\Rightarrow\; (y + 3)^2 = 36 \;\Rightarrow\; y + 3 = \pm 6$$

So $y = 3$ or $y = -9$. Two answers make sense geometrically. Q must lie on the vertical line $x = 10$, and a circle of radius $10$ about P meets that line twice, at points $6$ units above and below P’s level $y = -3$.

Question 9

If Q$(0, 1)$ is equidistant from P$(5, -3)$ and R$(x, 6)$, find the values of $x$. Also find the distances QR and PR.

Solution. First the fixed distance:

$$QP^2 = (5 – 0)^2 + (-3 – 1)^2 = 25 + 16 = 41$$

Setting $QR^2$ equal to it:

$$(x – 0)^2 + (6 – 1)^2 = 41 \;\Rightarrow\; x^2 = 16 \;\Rightarrow\; x = \pm 4$$

Either way $QR = QP = \sqrt{41}$, since that was the condition. PR does depend on which R is taken:

• For $R(4, 6)$: $PR = \sqrt{(4 – 5)^2 + (6 + 3)^2} = \sqrt{1 + 81} = \sqrt{82}$.
• For $R(-4, 6)$: $PR = \sqrt{(-4 – 5)^2 + (6 + 3)^2} = \sqrt{81 + 81} = 9\sqrt{2}$.

Question 10

Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the point $(3, 6)$ and $(-3, 4)$.

Solution. Equate squared distances from $(x, y)$ to the two points:

$$(x – 3)^2 + (y – 6)^2 = (x + 3)^2 + (y – 4)^2$$

$$x^2 – 6x + 9 + y^2 – 12y + 36 = x^2 + 6x + 9 + y^2 – 8y + 16$$

The squares cancel, leaving $-12x – 4y + 20 = 0$. Dividing by $-4$:

$$3x + y – 5 = 0$$

This is the equation of a line, and it has to be. The points equidistant from two fixed points form the perpendicular bisector of the segment joining them. As a check, the midpoint of $(3, 6)$ and $(-3, 4)$ is $(0, 5)$, and $3(0) + 5 – 5 = 0$.

Common mistakes

• Sign slips when subtracting negatives, question 1(ii). $-1 – (-5)$ is $+4$, not $-6$. Question 1(iii) has the same trap: $-a – a = -2a$, whose square is $4a^2$.
• Calling a shape a square on the strength of four equal sides, questions 5 and 6(i). A rhombus also has four equal sides. It is the equal diagonals that make it a square, which is why both questions need $AC$ and $BD$ as well.
• Naming a quadrilateral in question 6(ii) without checking that one exists. Four points only make a quadrilateral if no three of them are collinear. Here $(0, 3)$ is the midpoint of the segment from $(-3, 5)$ to $(3, 1)$, so the answer is “no quadrilateral”.
• Calling question 6(iii) a rectangle. Equal opposite sides give a parallelogram. A rectangle also needs equal diagonals, and here they are $2$ and $2\sqrt{13}$.
• Taking the point in question 7 as $(0, y)$. A point on the $x$-axis has $y = 0$, so it is $(x, 0)$. The $y$-axis version gives a different, wrong answer.
• Deciding collinearity from rounded decimals, question 3. Here the decimals are far enough apart to be convincing, but in general “close” does not mean “equal”. Squaring $\sqrt{5} + \sqrt{212}$ settles it exactly.
• Keeping only $x = 4$ in question 9. $x^2 = 16$ has two roots, and they give two different values of PR. Dropping $x = -4$ loses the answer $9\sqrt{2}$.

Practise next

• Exercise 7.2 — the section formula, which locates a point dividing a segment in a given ratio instead of measuring the segment’s length.
• Class 11 Straight Lines, Exercise 9.1 — slope, which gives a second and often quicker test for collinearity and for perpendicular sides.
• Class 11 Introduction to Three Dimensional Geometry, Exercise 11.2 — the distance formula extended to points in space, used for the same collinearity and triangle questions.