NCERT Solutions for Class 9 Maths Chapter 1 — End-of-Chapter Exercises#

The two axes meet at the origin $O$. The origin is 0 units from both axes.

∴ The x-coordinate is 0 and the y-coordinate is 0, so the point is $O(0, 0)$.

A line parallel to the y-axis is vertical, so every point on it has the same $x$-coordinate. The line passes through $W$, whose $x$-coordinate is $-5$.

∴ $H = (-5, y)$, where $y$ can be any number.

$H$ is to the left of the y-axis, because its $x$-coordinate is negative. So:

• if $y > 0$, $H$ lies in Quadrant II;
• if $y < 0$, $H$ lies in Quadrant III;
• if $y = 0$, $H$ lies on the x-axis, at $(-5, 0)$, which is not in any quadrant.

∴ $H$ can lie in Quadrant II or Quadrant III.

• $AM$ joins $(0, -2)$ and $(-5, -2)$. Both points have $y = -2$, so $AM$ is horizontal.
• $MP$ joins $(-5, -2)$ and $(-5, 2)$. Both points have $x = -5$, so $MP$ is vertical.

A horizontal line and a vertical line meet at a right angle.

∴ $AM \perp MP$.

• $AM$ is parallel to the x-axis, because both its endpoints have $y = -2$.
• $MP$ is parallel to the y-axis, because both its endpoints have $x = -5$.

Either one answers the question.

$M(-5, -2)$ and $P(-5, 2)$ have the same $x$-coordinate and $y$-coordinates that are equal in size but opposite in sign. $M$ is 2 units below the x-axis and $P$ is 2 units above it, on the same vertical line.

∴ $M$ and $P$ are mirror images in the x-axis.

Plot $R(3, 0)$, $A(0, -2)$, $M(-5, -2)$ and $P(-5, 2)$, and join them in the order $R \to A \to M \to P \to R$.

• $AM$ is horizontal and $MP$ is vertical, so they meet at a right angle at $M$ ✓
• $AM$ runs along $y = -2$, parallel to the x-axis ✓
• $M$ and $P$ are the same distance below and above the x-axis ✓

A convenient choice is to put the right angle at $Z$, with the two shorter sides parallel to the axes:

$$I = (5, 0), \quad Z = (5, -6), \quad N = (0, -6)$$

$IZ$ is vertical, because $I$ and $Z$ have the same $x$-coordinate:

$$IZ = 0 - (-6) = 6 \text{ units}$$

$ZN$ is horizontal, because $Z$ and $N$ have the same $y$-coordinate:

$$ZN = 5 - 0 = 5 \text{ units}$$

$IN$ is the hypotenuse. By the Baudhāyana–Pythagoras theorem:

$$IN^2 = IZ^2 + ZN^2$$

$$IN^2 = 6^2 + 5^2$$

$$IN^2 = 61$$

$$\Rightarrow IN = \sqrt{61} \text{ units}$$

∴ The sides are 5 units, 6 units and $\sqrt{61}$ units (about 7.81 units). Other choices of $I$ and $N$ give other correct answers.

Without negative numbers, both coordinates could only be 0 or positive ($x \geq 0$ and $y \geq 0$). Such a system could describe only the points on or to the right of the y-axis and on or above the x-axis, that is, only Quadrant I and the two half-axes around it.

Points to the left of the y-axis (Quadrants II and III) or below the x-axis (Quadrants III and IV) would have no coordinates.

∴ No, such a system would not let us locate all the points on a 2-D plane.

Find the three distances $MA$, $AG$ and $MG$. If the points lie on one straight line, the middle point divides the longest segment into the other two, so the two shorter distances add up to the longest one. If they do not, the three points form a triangle.

We find $MA$:

$$= \sqrt{(0 - (-3))^2 + (0 - (-4))^2}$$

$$= \sqrt{9 + 16}$$

$$= \sqrt{25}$$

$$= 5$$

We find $AG$:

$$= \sqrt{(6 - 0)^2 + (8 - 0)^2}$$

$$= \sqrt{36 + 64}$$

$$= \sqrt{100}$$

$$= 10$$

We find $MG$:

$$= \sqrt{(6 - (-3))^2 + (8 - (-4))^2}$$

$$= \sqrt{81 + 144}$$

$$= \sqrt{225}$$

$$= 15$$

Comparing the two shorter distances with the longest:

$$\Rightarrow MA + AG = 5 + 10 = 15 = MG$$

∴ $MA + AG = MG$, so M, A and G lie on the same straight line, with $A$ between $M$ and $G$.

We find $RB$:

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

$$= \sqrt{9 + 16}$$

$$= 5$$

We find $BC$:

$$= \sqrt{(4 - (-2))^2 + (-12 - (-5))^2}$$

$$= \sqrt{36 + 49}$$

$$= \sqrt{85}$$

We find $RC$:

$$= \sqrt{(4 - (-5))^2 + (-12 - (-1))^2}$$

$$= \sqrt{81 + 121}$$

$$= \sqrt{202}$$

The longest distance is $RC$. The values are very close ($5 + \sqrt{85} \approx 14.22$ and $\sqrt{202} \approx 14.21$), so we compare them exactly by squaring:

$$(RB + BC)^2 = (5 + \sqrt{85})^2$$

$$(RB + BC)^2 = 25 + 10\sqrt{85} + 85$$

$$(RB + BC)^2 = 110 + 10\sqrt{85}$$$$RC^2 = 202$$

For these to be equal we would need $10\sqrt{85} = 92$, that is, $\sqrt{85} = 9.2$. But $9.2^2 = 84.64$, which is less than 85:

$$\Rightarrow \sqrt{85} > 9.2$$

$$\Rightarrow RB + BC > RC$$

∴ $RB + BC \neq RC$, so R, B and C are not on the same straight line. They form a very thin triangle, so on a graph they look almost in line. Plotting both sets shows $M$, $A$, $G$ exactly on one line, and $B$ slightly off the line $RC$.

Put the right angle at the origin and the two equal sides along the axes. For example:

$$O = (0, 0), \quad P = (4, 0), \quad Q = (0, 4)$$$$OP = 4 - 0 = 4$$

$$OQ = 4 - 0 = 4$$

The axes are perpendicular, so $\angle POQ = 90°$.

∴ $OP = OQ$ and $\angle POQ = 90°$, so $\triangle OPQ$ is a right-angled isosceles triangle.

We choose two points that are mirror images in the y-axis, one in each quadrant, so that they are the same distance from $O$:

$$O = (0, 0), \quad A = (-3, -4) \text{ (Quadrant III)}, \quad B = (3, -4) \text{ (Quadrant IV)}$$

We find $OA$:

$$= \sqrt{(-3)^2 + (-4)^2}$$

$$= \sqrt{25}$$

$$= 5$$

We find $OB$:

$$= \sqrt{3^2 + (-4)^2}$$

$$= \sqrt{25}$$

$$= 5$$

∴ $OA = OB$, so $\triangle OAB$ is isosceles.

For each row, we find the midpoint of $ST$ with the midpoint formula and compare it with $M$.

S	M	T	Midpoint of ST	Is M the midpoint?
$(-3, 0)$	$(0, 0)$	$(3, 0)$	$\left(\dfrac{-3 + 3}{2}, \dfrac{0 + 0}{2}\right) = (0, 0)$	Yes
$(2, 3)$	$(3, 4)$	$(4, 5)$	$\left(\dfrac{2 + 4}{2}, \dfrac{3 + 5}{2}\right) = (3, 4)$	Yes
$(0, 0)$	$(0, 5)$	$(0, -10)$	$\left(\dfrac{0 + 0}{2}, \dfrac{0 + (-10)}{2}\right) = (0, -5)$	No, $(0, 5) \neq (0, -5)$
$(-8, 7)$	$(0, -2)$	$(6, -3)$	$\left(\dfrac{-8 + 6}{2}, \dfrac{7 + (-3)}{2}\right) = (-1, 2)$	No, $(0, -2) \neq (-1, 2)$

Connection: when $M$ is the midpoint of $ST$, the $x$-coordinate of $M$ is the average of the $x$-coordinates of $S$ and $T$, and the $y$-coordinate of $M$ is the average of their $y$-coordinates.

The $x$-coordinate of $M$ is the average of the $x$-coordinates of $A$ and $B$, so:

$$\dfrac{3 + x}{2} = -7$$

$$\Rightarrow 3 + x = -14$$

$$\Rightarrow x = -17$$

The $y$-coordinate of $M$ is the average of the $y$-coordinates of $A$ and $B$, so:

$$\dfrac{-4 + y}{2} = 1$$

$$\Rightarrow -4 + y = 2$$

$$\Rightarrow y = 6$$

∴ $B = (-17, 6)$.

Check: $\left(\dfrac{3 + (-17)}{2}, \dfrac{-4 + 6}{2}\right) = (-7, 1) = M$ ✓

$P$ and $Q$ cut $AB$ into three equal parts: $AP = PQ = QB$. So:

• $P$ is the midpoint of $A$ and $Q$ (because $AP = PQ$);
• $Q$ is the midpoint of $P$ and $B$ (because $PQ = QB$).

Using the midpoint formula on the $x$-coordinates (the $y$-coordinates work the same way):

$$x_P = \dfrac{x_A + x_Q}{2} \quad \cdots (1)$$

$$x_Q = \dfrac{x_P + x_B}{2} \quad \cdots (2)$$

Substituting equation (2) in equation (1):

$$x_P = \dfrac{x_A + x_Q}{2}$$

$$x_P = \dfrac{1}{2}\left(x_A + \dfrac{x_P + x_B}{2}\right)$$

$$\Rightarrow 4x_P = 2x_A + x_P + x_B$$

$$\Rightarrow 3x_P = 2x_A + x_B$$

$$\Rightarrow x_P = \dfrac{2x_A + x_B}{3}$$

In the same way, $x_Q = \dfrac{x_A + 2x_B}{3}$. So:

$$P = \left(\dfrac{2x_A + x_B}{3}, \dfrac{2y_A + y_B}{3}\right), \qquad Q = \left(\dfrac{x_A + 2x_B}{3}, \dfrac{y_A + 2y_B}{3}\right)$$

$$x_P = \dfrac{2(4) + 16}{3} = \dfrac{24}{3} = 8$$

$$y_P = \dfrac{2(7) + (-2)}{3} = \dfrac{12}{3} = 4$$

$$x_Q = \dfrac{4 + 2(16)}{3} = \dfrac{36}{3} = 12$$

$$y_Q = \dfrac{7 + 2(-2)}{3} = \dfrac{3}{3} = 1$$

∴ $P = (8, 4)$ and $Q = (12, 1)$.

Check: the midpoint of $A(4, 7)$ and $Q(12, 1)$ is $\left(\dfrac{4 + 12}{2}, \dfrac{7 + 1}{2}\right) = (8, 4) = P$ ✓, and the midpoint of $P(8, 4)$ and $B(16, -2)$ is $\left(\dfrac{8 + 16}{2}, \dfrac{4 + (-2)}{2}\right) = (12, 1) = Q$ ✓

A point lies on a circle with centre $O$ when its distance from $O$ equals the radius. We find each distance from $O$.

We find $OA$:

$$= \sqrt{1^2 + (-8)^2}$$

$$= \sqrt{1 + 64}$$

$$= \sqrt{65}$$

We find $OB$:

$$= \sqrt{(-4)^2 + 7^2}$$

$$= \sqrt{16 + 49}$$

$$= \sqrt{65}$$

We find $OC$:

$$= \sqrt{(-7)^2 + (-4)^2}$$

$$= \sqrt{49 + 16}$$

$$= \sqrt{65}$$$$\Rightarrow OA = OB = OC = \sqrt{65}$$

∴ $A$, $B$ and $C$ are all the same distance from $O$, so they lie on circle $K$, and the radius of circle K is $\sqrt{65}$ units (about 8.06 units).

We compare each distance from $O$ with the radius $\sqrt{65}$.

We find $OD$:

$$= \sqrt{(-5)^2 + 6^2}$$

$$= \sqrt{25 + 36}$$

$$= \sqrt{61}$$$$\Rightarrow \sqrt{61} < \sqrt{65}$$

∴ $OD$ < radius, so D lies inside circle K.

We find $OE$:

$$= \sqrt{0^2 + 9^2}$$

$$= 9$$$$\Rightarrow 9 = \sqrt{81} > \sqrt{65}$$

∴ $OE$ > radius, so E lies outside circle K.

Let $D$ be the midpoint of $BC$, $E$ the midpoint of $CA$, and $F$ the midpoint of $AB$. Let $A = (x_1, y_1)$, $B = (x_2, y_2)$ and $C = (x_3, y_3)$.

x-coordinates. The $x$-coordinate of each midpoint is the average of two $x$-coordinates.

$D(5, 1)$ is the midpoint of $BC$, so:

$$\dfrac{x_2 + x_3}{2} = 5$$

$$\Rightarrow x_2 + x_3 = 10 \quad \cdots (1)$$

$E(6, 5)$ is the midpoint of $CA$, so:

$$\dfrac{x_3 + x_1}{2} = 6$$

$$\Rightarrow x_3 + x_1 = 12 \quad \cdots (2)$$

$F(0, 3)$ is the midpoint of $AB$, so:

$$\dfrac{x_1 + x_2}{2} = 0$$

$$\Rightarrow x_1 + x_2 = 0 \quad \cdots (3)$$

Adding equation (1), equation (2) and equation (3):

$$\begin{array}{rcl} x_2 + x_3 &=& 10 \\ x_3 + x_1 &=& 12 \\ x_1 + x_2 &=& 0 \\ \hline 2(x_1 + x_2 + x_3) &=& 22 \end{array}$$

$$\Rightarrow x_1 + x_2 + x_3 = 11 \quad \cdots (4)$$

Subtracting equation (1) from equation (4):

$$\begin{array}{rcl} x_1 + x_2 + x_3 &=& 11 \\ x_2 + x_3 &=& 10 \\ \hline x_1 &=& 1 \end{array}$$

Subtracting equation (2) from equation (4):

$$\begin{array}{rcl} x_1 + x_2 + x_3 &=& 11 \\ x_3 + x_1 &=& 12 \\ \hline x_2 &=& -1 \end{array}$$

Subtracting equation (3) from equation (4):

$$\begin{array}{rcl} x_1 + x_2 + x_3 &=& 11 \\ x_1 + x_2 &=& 0 \\ \hline x_3 &=& 11 \end{array}$$

y-coordinates. In the same way, from the $y$-coordinates of $D$, $E$ and $F$:

$$y_2 + y_3 = 2 \quad \cdots (5)$$

$$y_3 + y_1 = 10 \quad \cdots (6)$$

$$y_1 + y_2 = 6 \quad \cdots (7)$$

Adding equation (5), equation (6) and equation (7):

$$\begin{array}{rcl} y_2 + y_3 &=& 2 \\ y_3 + y_1 &=& 10 \\ y_1 + y_2 &=& 6 \\ \hline 2(y_1 + y_2 + y_3) &=& 18 \end{array}$$

$$\Rightarrow y_1 + y_2 + y_3 = 9 \quad \cdots (8)$$

Subtracting equation (5), equation (6) and equation (7) from equation (8) one at a time:

$$y_1 = 9 - 2 = 7$$

$$y_2 = 9 - 10 = -1$$

$$y_3 = 9 - 6 = 3$$

∴ $A = (1, 7)$, $B = (-1, -1)$ and $C = (11, 3)$.

Check: the midpoint of $BC$ is $\left(\dfrac{-1 + 11}{2}, \dfrac{-1 + 3}{2}\right) = (5, 1) = D$ ✓

Draw the two main roads as two perpendicular lines crossing at the centre: the N–S road vertical and the E–W road horizontal. Since $200 \text{ m} = 1 \text{ cm}$, draw the other streets as lines 1 cm apart, parallel to the main roads, with 10 lines in each direction. The model looks like a square grid of lines, like graph paper.

(a) There is exactly one 4th N–S street and exactly one 3rd E–W street, and two crossing lines meet at exactly one point.

∴ Only one street intersection can be called $(4, 3)$.

(b) In the same way, the 3rd N–S street and the 4th E–W street meet at exactly one point.

∴ Only one street intersection can be called $(3, 4)$.

$(4, 3)$ and $(3, 4)$ are different intersections: the first number always names the N–S street and the second names the E–W street. Like coordinates, the order matters.

The screen covers $0 \leq x \leq 800$ and $0 \leq y \leq 600$. A circle stays on the screen if its leftmost, rightmost, lowest and highest points do.

Circle with centre $A(100, 150)$ and radius 80:

Edge of the circle	Value	On the screen?
Leftmost $x$	$100 - 80 = 20$	Yes ($\geq 0$)
Rightmost $x$	$100 + 80 = 180$	Yes ($\leq 800$)
Lowest $y$	$150 - 80 = 70$	Yes ($\geq 0$)
Highest $y$	$150 + 80 = 230$	Yes ($\leq 600$)

Circle with centre $B(250, 230)$ and radius 100:

Edge of the circle	Value	On the screen?
Leftmost $x$	$250 - 100 = 150$	Yes ($\geq 0$)
Rightmost $x$	$250 + 100 = 350$	Yes ($\leq 800$)
Lowest $y$	$230 - 100 = 130$	Yes ($\geq 0$)
Highest $y$	$230 + 100 = 330$	Yes ($\leq 600$)

∴ Both circles lie completely on the screen, so no part of either circle lies outside the screen.

We find the distance $AB$ between the centres:

$$= \sqrt{(250 - 100)^2 + (230 - 150)^2}$$

$$= \sqrt{150^2 + 80^2}$$

$$= \sqrt{22500 + 6400}$$

$$= \sqrt{28900}$$

$$= 170 \text{ pixels}$$

Two circles cross each other when the distance between their centres is less than the sum of the radii, but more than their difference:

$$\text{Sum of radii} = 80 + 100 = 180$$

$$\text{Difference of radii} = 100 - 80 = 20$$$$\Rightarrow 20 < 170 < 180$$

∴ The distance between the centres is less than the sum of the radii (so the circles are close enough to meet) and more than their difference (so neither circle is inside the other). The two circles intersect, crossing at two points.

Sides. We find $AB$:

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

$$= \sqrt{9 + 1}$$

$$= \sqrt{10}$$

We find $BC$:

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

$$= \sqrt{1 + 9}$$

$$= \sqrt{10}$$

We find $CD$:

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

$$= \sqrt{9 + 1}$$

$$= \sqrt{10}$$

We find $DA$:

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

$$= \sqrt{1 + 9}$$

$$= \sqrt{10}$$

Diagonals. We find $AC$:

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

$$= \sqrt{16 + 4}$$

$$= \sqrt{20}$$

We find $BD$:

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

$$= \sqrt{4 + 16}$$

$$= \sqrt{20}$$$$\Rightarrow AB = BC = CD = DA = \sqrt{10}, \quad AC = BD = \sqrt{20}$$

∴ All four sides are equal (so $ABCD$ is a rhombus) and the diagonals are equal (a rhombus with equal diagonals has right angles). So ABCD is a square.

Area.

$$\text{Area} = (\text{side})^2$$

$$\text{Area} = (\sqrt{10})^2$$

$$\text{Area} = 10 \text{ square units}$$

(Also, $AB^2 + BC^2 = 10 + 10 = 20 = AC^2$, so $\angle ABC = 90°$ by the converse of the Baudhāyana–Pythagoras theorem.)