Chapter 1 Orienting Yourself: The Use of Coordinates (Class 9 - Latest Maths NCERT (Ganita Manjari I) Solutions)

Given:

From the provided coordinate map (Fig. $1.3$) of the room:

1. The left wall of the room lies along the $y$-axis (where $x = 0$).

2. The bottom wall of the room lies along the $x$-axis (where $y = 0$).

3. The coordinates of the corner points are $O(0, 0)$, $A(12, 0)$, $B(12, 10)$, and $C(0, 10)$.

4. The room door is represented by the segment $D_1R_1$ on the $x$-axis.

5. The coordinates of $R_1$ are $(11.5, 0)$.

6. The bathroom door is represented by the segment $B_1B_2$ on the $y$-axis, where $B_1(0, 1.5)$ and $B_2(0, 4)$.

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(i) To find: Distance of the door from the left wall and the $x$-axis.

Solution:

By observing the grid on the $x$-axis in Fig. $1.3$, the point $D_1$ is located at the marking $8$. Therefore, the $x$-coordinate of $D_1$ is $8$.

The distance of any point from the $y$-axis (left wall) is given by its $x$-coordinate.

Since the door $D_1R_1$ lies exactly on the $x$-axis, its distance from the $x$-axis is zero.

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(ii) To find: Coordinates of $D_1$.

Solution:

The point $D_1$ lies on the $x$-axis, which means its $y$-coordinate is $0$. As determined from the grid in the previous step, its $x$-coordinate is $8$.

Therefore, the coordinates of $D_1$ are $(8, 0)$.

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(iii) To find: Width of the door and its accessibility.

Solution:

The width of the door is the distance between points $D_1(8, 0)$ and $R_1(11.5, 0)$.

$\text{Width} = |x_{R1} - x_{D1}|$

In Indian residential standards, $1\text{ unit}$ often corresponds to $1\text{ foot}$. A width of $3.5\text{ units}$ is very comfortable for a room door.

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(iv) To compare: Width of the bathroom door versus the room door.

Solution:

The bathroom door is represented by points $B_1(0, 1.5)$ and $B_2(0, 4)$ on the $y$-axis.

$\text{Width of bathroom door} = |y_{B2} - y_{B1}|$

Comparing the two widths:

Width of room door $= 3.5\text{ units}$

Width of bathroom door $= 2.5\text{ units}$

Since $2.5 < 3.5$, the bathroom door is narrower than the room door.

Given:

Three feet of a rectangular table are located at coordinates $A(8, 9)$, $B(11, 9)$, and $C(11, 7)$.

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To Find:

(i) Coordinates of the fourth foot $D$.

(ii) Suitability of the location in the bedroom.

(iii) Dimensions of the table (Width, Length, and Height).

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Solution:

(i) In a rectangle, opposite sides are equal and parallel. The vertices are usually given in cyclic order.

Looking at the coordinates of $A(8, 9)$ and $B(11, 9)$, we see they have the same $y$-coordinate, forming a horizontal side.

Looking at $B(11, 9)$ and $C(11, 7)$, they have the same $x$-coordinate, forming a vertical side.

The fourth foot $D$ must have the same $x$-coordinate as $A$ and the same $y$-coordinate as $C$ to complete the rectangle.

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Visual Representation:

The following diagram illustrates the placement of the four feet of the study table within the bedroom coordinate system.

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(ii) Is this a good spot?

According to the floor plan (Fig 1.5), the bedroom floor extends up to $x = 12$ and $y = 10$. The table coordinates $(8, 7)$ to $(11, 9)$ place it in the top-right corner of the room. Since it does not block the wardrobe (located at $y = 0$ to $2$), the bed (located at $y = 5$ to $8$), or the entrance, it is a very good and efficient spot for the table.

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(iii) Dimensions:

The Width is the horizontal distance between the feet:

$\text{Width} = |11 - 8| = 3 \text{ units}$

The Length is the vertical distance between the feet:

$\text{Length} = |9 - 7| = 2 \text{ units}$

Height: From a two-dimensional (2D) plan view, we only see the floor coordinates ($x$ and $y$). We cannot determine the height of the table from this graph as the vertical dimension ($z$-axis) is not shown.

Given:

The wardrobe is placed between $x = 3$ and $x = 7$. Its left vertical edge (represented by $W_1 W_4$) lies on the line $x = 3$.

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To Find:

Whether the door hits the wardrobe during its swing and suggestions for a wider door design.

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Solution:

First, we determine the width of the bathroom door based on the given coordinates:

When the door is hinged at $B_1(0, 1.5)$ and opens into the bedroom, it creates a circular swing path with a radius equal to the door width.

The maximum horizontal distance the door can reach into the bedroom is its radius ($2.5 \text{ units}$).

Comparing the values from (ii) and the wardrobe position, we see that $2.5 < 3$.

Therefore, the door will not hit the wardrobe as there is a clearance of $0.5 \text{ units}$ between the fully opened door and the wardrobe's edge.

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Suggestions for a Wider Door:

If the door is made wider (for example, $3.5 \text{ units}$), the swing radius would exceed the distance to the wardrobe ($3.5 > 3$), causing a collision. In such a case, the following changes are suggested:

1. Inward Opening: Modify the design so the door opens inward into the bathroom. This is standard in most Indian flats to keep the bedroom area unobstructed.

2. Sliding Mechanism: Install a sliding door that moves along the wall (the $y$-axis). This saves the entire space required for a "swing" and allows for a much wider entrance if needed.

3. Relocate Wardrobe: Shift the wardrobe further to the right, starting from $x = 4$ or $x = 5$, to provide enough clearance for a larger door swing.

4. Hinge Reversal: Move the hinge to $B_2(0, 4)$. The door would then swing towards the bed area instead of the wardrobe area, though this might interfere with access to the bed.

To Find:

(i) Coordinates of points $O$, $F$, $R$, and $P$.

(ii) Shape and coordinates of the showering area $SHWR$.

(iii) Coordinates for the washbasin and toilet areas.

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Solution:

(i) Coordinates of the Bathroom Corners:

By observing Fig 1.5 and using the updated coordinates provided:

Point $O$ is the origin: $O(0, 0)$

Point $F$ lies on the $y$-axis: $F(0, 9)$

Point $R$ is the top-left corner: $R(-6, 9)$

Point $P$ lies on the $x$-axis at the same distance as $R$: $P(-6, 0)$

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(ii) Showering Area $SHWR$:

The showering area is defined by the corners $S, H, W, \text{ and } R$. From the graph:

The coordinates are:

$R = (-6, 9)$

$W = (-2, 9)$

$H = (-2, 6)$

$S = (-6, 6)$

Now, let's calculate the dimensions:

Since the adjacent sides are perpendicular and unequal (3 units and 4 units), the shape of the showering area $SHWR$ is a Rectangle.

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(iii) Washbasin and Toilet Spaces:

Washbasin ($3\text{ ft} \times 2\text{ ft}$):

Corners: $(-6, 0.5)$, $(-5, 0.5)$, $(-5, 2)$, and $(-6, 2)$.

Toilet ($2\text{ ft} \times 3\text{ ft}$):

Placing it between the washbasin and the showering area:

Corners: $(-6, 3)$, $(-4.5, 3)$, $(-4.5, 4)$, and $(-6, 4)$.

Given:

Dining room length = $18 \text{ units}$ (from $P(-6, 0)$ to $A(12, 0)$).

Dining room width = $15 \text{ units}$ extending downwards (negative $y$ direction).

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Solution:

(i) The dining room corners are determined by extending $15 \text{ units}$ down from the $x$-axis.

$P = (-6, 0)$

$A = (12, 0)$

$A' = (12, -15)$

$P' = (-6, -15)$

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(ii) Centre of the dining room:

$x_{\text{centre}} = \frac{-6 + 12}{2} = \frac{6}{2} = 3$

$y_{\text{centre}} = \frac{0 + (-15)}{2} = -7.5$

The table is $5 \text{ ft} \times 3 \text{ ft}$. For it to be centered:

Half-length = $2.5 \text{ units}$; Half-width = $1.5 \text{ units}$.

$x \text{-coordinates} = 3 \pm 2.5 \Rightarrow (0.5, 5.5)$

$y \text{-coordinates} = -7.5 \pm 1.5 \Rightarrow (-6, -9)$

The coordinates of the feet of the table are:

$(0.5, -6)$, $(5.5, -6)$, $(5.5, -9)$, and $(0.5, -9)$.

To Find:

The $x$-coordinate and $y$-coordinate of the point of intersection of the $x$-axis and $y$-axis.

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Solution:

The point where the horizontal axis ($x$-axis) and the vertical axis ($y$-axis) intersect is called the Origin.

At this point, the distance from both the axes is zero.

Therefore, the coordinates of the point of intersection are $(0, 0)$.

Given:

Point $W$ has an $x$-coordinate of $-5$. A line passes through $W$ and is parallel to the $y$-axis.

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Solution:

Any line parallel to the $y$-axis has a constant $x$-coordinate for all points lying on it. Since the line passes through $W$ where $x = -5$, the equation of the line is:

(i) Coordinates of $H$:

Since $H$ lies on this line, its $x$-coordinate must be $-5$. The $y$-coordinate can be any real number $y$.

Therefore, the coordinates of $H$ are $(-5, y)$.

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(ii) Quadrants for $H$:

The $x$-coordinate is negative ($x < 0$).

If $y > 0$, the point $H$ lies in the Second Quadrant (II).

If $y < 0$, the point $H$ lies in the Third Quadrant (III).

If $y = 0$, the point $H$ lies on the negative $x$-axis.

Given:

Points $R(3, 0)$, $A(0, -2)$, $M(-5, -2)$, and $P(-5, 2)$.

To Predict and Verify:

(i) Perpendicular sides of quadrilateral $RAMP$.

(ii) Sides parallel to the coordinate axes.

(iii) Points that are mirror images across an axis.

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Solution:

(i) Perpendicular Sides:

We analyze the coordinates of the segments formed by joining the points in order: $RA$, $AM$, $MP$, and $PR$.

Since a horizontal line and a vertical line are always perpendicular, the sides $AM$ and $MP$ are perpendicular to each other.

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(ii) Side parallel to an axis:

From the observations in (i):

1. Side $AM$ is parallel to the $x$-axis (as it is a horizontal line where $y = -2$).

2. Side $MP$ is parallel to the $y$-axis (as it is a vertical line where $x = -5$).

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(iii) Mirror Images:

Consider the points $M(-5, -2)$ and $P(-5, 2)$.

The $x$-coordinates are the same, while the $y$-coordinates are equal in magnitude but opposite in sign ($2$ and $-2$).

Therefore, $M$ and $P$ are mirror images of each other in the $x$-axis.

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Verification through Plotting:

The points are plotted on the Cartesian plane below. Joining them in order shows the shape of the quadrilateral and confirms the geometric properties predicted above.

By looking at the graph, it is verified that $\angle AMP = 90^\circ$, side $AM$ is horizontal, side $MP$ is vertical, and point $P$ is the reflection of $M$ across the $x$-axis.

Given:

A point $Z$ with coordinates $(5, -6)$ in the Fourth Quadrant of the Cartesian plane.

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To Construct:

A right-angled triangle $IZN$ using $Z$ as one of the vertices and determining the lengths of its sides.

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Construction and Solution:

To construct a right-angled triangle easily, we can choose the other two vertices such that the sides $IZ$ and $ZN$ are parallel to the coordinate axes. This ensures that the angle between them is exactly $90^\circ$.

1. Let us choose Point $I$ on the $x$-axis directly above $Z$. Since it is on the $x$-axis, its $y$-coordinate is $0$.

2. Let us choose Point $N$ on the $y$-axis directly to the left of $Z$. Since it is on the $y$-axis, its $x$-coordinate is $0$.

3. Now, we calculate the lengths of the three sides using the distance formula or simple subtraction for lines parallel to the axes.

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Lengths of the Sides:

Side $IZ$ (Vertical side):

$IZ = 6 \text{ units}$

Side $ZN$ (Horizontal side):

$ZN = 5 \text{ units}$

Side $IN$ (Hypotenuse):

By Pythagoras' Theorem:

$IN = \sqrt{IZ^2 + ZN^2}$

$IN = \sqrt{6^2 + 5^2}$

$IN = \sqrt{36 + 25}$

$IN = \sqrt{61} \approx 7.81 \text{ units}$

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Graphical Verification:

The points $I(5,0)$, $Z(5,-6)$, and $N(0,-6)$ form a right-angled triangle with the right angle at vertex $Z$.

Therefore, the lengths of the three sides of $\triangle IZN$ are $6 \text{ units}$, $5 \text{ units}$, and $\sqrt{61} \text{ units}$.

Solution:

If a coordinate system did not have negative numbers, it would consist only of the First Quadrant ($x \geq 0$ and $y \geq 0$). Both the $x$-axis and $y$-axis would be rays starting from the origin $(0, 0)$ rather than infinite lines.

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Would it allow us to locate all points?

No, this system would not allow us to locate all points on a 2-D plane. A complete 2-D Cartesian plane extends infinitely in all directions. Without negative numbers, we would be unable to represent:

1. Points to the left of the $y$-axis (where $x < 0$).

2. Points below the $x$-axis (where $y < 0$).

In other words, we would lose the ability to describe positions in the Second, Third, and Fourth Quadrants.

Given:

Three points $M(-3, -4)$, $A(0, 0)$, and $G(6, 8)$.

To Find:

Determine if the points are collinear (lie on the same straight line) using the Distance Formula.

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Solution:

The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

For three points to be collinear, the sum of the distances between the two shorter segments must be equal to the distance of the longest segment.

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1. Calculating distance $MA$ (between $M(-3, -4)$ and $A(0, 0)$):

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

$MA = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25}$

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2. Calculating distance $AG$ (between $A(0, 0)$ and $G(6, 8)$):

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

$AG = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100}$

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3. Calculating distance $MG$ (between $M(-3, -4)$ and $G(6, 8)$):

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

$MG = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225}$

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Conclusion:

Now, let us check if the sum of the two shorter distances ($MA$ and $AG$) equals the longest distance ($MG$):

$MA + AG = 5 + 10 = 15$

Since the sum of the lengths of two segments is equal to the length of the third segment, the points $M$, $A$, and $G$ are collinear and lie on the same straight line.

Given:

Three points $R(-5, -1)$, $B(-2, -5)$, and $C(4, -12)$.

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To Find:

Determine if the points are collinear using the Distance Formula.

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Solution:

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

We calculate the lengths of the segments $RB$, $BC$, and $RC$.

1. Distance $RB$ (between $R(-5, -1)$ and $B(-2, -5)$):

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

$RB = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16}$

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2. Distance $BC$ (between $B(-2, -5)$ and $C(4, -12)$):

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

$BC = \sqrt{(6)^2 + (-7)^2} = \sqrt{36 + 49}$

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3. Distance $RC$ (between $R(-5, -1)$ and $C(4, -12)$):

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

$RC = \sqrt{(9)^2 + (-11)^2} = \sqrt{81 + 121}$

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Conclusion:

For the points to be on the same straight line, the sum of the two shorter segments must be equal to the longest segment.

Shortest distances: $RB = 5$ and $BC \approx 9.22$.

$RB + BC = 5 + 9.22 = 14.22$

Longest distance: $RC = 14.21$

Since the sum of the segments is not exactly equal to the third segment (even if very close), the points $R$, $B$, and $C$ are not collinear. Plotting the points will show a very slight bend at point $B$, forming a triangle rather than a single line.

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Verification through Plotting:

To Plot:

(i) A right-angled isosceles triangle with one vertex at the origin.

(ii) An isosceles triangle with the origin as a vertex and other vertices in Quadrant $III$ and Quadrant $IV$.

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Solution:

(i) Right-angled isosceles triangle:

To form a right-angled isosceles triangle using the origin $O(0, 0)$ as the vertex where the right angle is formed, we can choose two points on the coordinate axes that are at an equal distance from the origin.

Let the distance be $a = 5$ units.

The vertices are:

1. $O (0, 0)$ (The origin)

2. $A (5, 0)$ (On the positive $x$-axis)

3. $B (0, 5)$ (On the positive $y$-axis)

Since $OA$ lies on the $x$-axis and $OB$ lies on the $y$-axis, the angle between them is $90^\circ$.

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(ii) Isosceles triangle with vertices in Quadrant $III$ and $IV$:

Let the origin $O(0, 0)$ be the first vertex. For the triangle to be isosceles with vertices in Quadrant $III$ (where both $x$ and $y$ are negative) and Quadrant $IV$ (where $x$ is positive and $y$ is negative), we can choose points that are symmetric about the $y$-axis.

Let the vertices be:

1. $O (0, 0)$

2. $P (-4, -4)$ (Located in Quadrant $III$)

3. $Q (4, -4)$ (Located in Quadrant $IV$)

To verify if the triangle is isosceles, we find the lengths of $OP$ and $OQ$ using the distance formula:

$OP = \sqrt{(-4 - 0)^2 + (-4 - 0)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} $$ = \sqrt{32} \text{ units}$

$OQ = \sqrt{(4 - 0)^2 + (-4 - 0)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} $$ = \sqrt{32} \text{ units}$

Given:

The coordinates of points $S$, $M$, and $T$ for four different scenarios as provided in the table.

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To Find:

Determine whether $M$ is the midpoint of segment $ST$ using the Distance Formula and establish the connection between the coordinates of $S, M,$ and $T$.

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Solution:

To verify if $M$ is the midpoint of segment $ST$ using the distance formula, we must check two conditions:

1. The distance $SM$ must be equal to the distance $MT$.

2. The points must be collinear such that $SM + MT = ST$.

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

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Case (i): $S(-3, 0), M(0, 0), T(3, 0)$

$SM = \sqrt{(0 - (-3))^2 + (0 - 0)^2} = \sqrt{3^2 + 0} = 3 \text{ units}$

$MT = \sqrt{(3 - 0)^2 + (0 - 0)^2} = \sqrt{3^2 + 0} = 3 \text{ units}$

$ST = \sqrt{(3 - (-3))^2 + (0 - 0)^2} = \sqrt{6^2 + 0} = 6 \text{ units}$

Conclusion: Yes, $M$ is the midpoint of $ST$.

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Case (ii): $S(2, 3), M(3, 4), T(4, 5)$

$SM = \sqrt{(3 - 2)^2 + (4 - 3)^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \text{ units}$

$MT = \sqrt{(4 - 3)^2 + (5 - 4)^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \text{ units}$

$ST = \sqrt{(4 - 2)^2 + (5 - 3)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \text{ units}$

Conclusion: Yes, $M$ is the midpoint of $ST$.

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Case (iii): $S(0, 0), M(0, 5), T(0, -10)$

$SM = \sqrt{(0 - 0)^2 + (5 - 0)^2} = \sqrt{0 + 25} = 5 \text{ units}$

$MT = \sqrt{(0 - 0)^2 + (-10 - 5)^2} = \sqrt{0 + (-15)^2} = 15 \text{ units}$

Conclusion: No, $M$ is not the midpoint of $ST$.

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Case (iv): $S(-8, 7), M(0, -2), T(6, -3)$

$SM = \sqrt{(0 - (-8))^2 + (-2 - 7)^2} = \sqrt{8^2 + (-9)^2} = \sqrt{64 + 81} $$ = \sqrt{145} \text{ units}$

$MT = \sqrt{(6 - 0)^2 + (-3 - (-2))^2} = \sqrt{6^2 + (-1)^2} = \sqrt{36 + 1} $$ = \sqrt{37} \text{ units}$

Conclusion: No, $M$ is not the midpoint of $ST$.

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Final Table Results:

Endpoints $S$ and $T$	Point $M$	Is $M$ Midpoint?	Reason (Distances)
$(-3, 0), (3, 0)$	$(0, 0)$	Yes	$SM = 3, MT = 3$
$(2, 3), (4, 5)$	$(3, 4)$	Yes	$SM = \sqrt{2}, MT = \sqrt{2}$
$(0, 0), (0, -10)$	$(0, 5)$	No	$SM = 5, MT = 15$
$(-8, 7), (6, -3)$	$(0, -2)$	No	$SM = \sqrt{145}, MT = \sqrt{37}$

Given:

Coordinates of point $A = (3, -4)$

Coordinates of the midpoint $M = (-7, 1)$

Coordinates of point $B = (x, y)$

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To Find:

The values of $x$ and $y$ using the Distance Formula.

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Solution:

If $M$ is the midpoint of segment $AB$, then it must satisfy two conditions based on distances:

1. The distance $AM$ is equal to the distance $MB$.

2. The distance $AB$ is equal to twice the distance $AM$ (since $A, M, B$ are collinear).

The distance formula is:

Step 1: Calculate the distance $AM$:

$AM = \sqrt{(-7 - 3)^2 + (1 - (-4))^2}$

$AM = \sqrt{(-10)^2 + (5)^2}$

$AM = \sqrt{100 + 25} = \sqrt{125}$

Step 2: Set $AM^2 = MB^2$:

$(\sqrt{125})^2 = (x - (-7))^2 + (y - 1)^2$

$125 = x^2 + 14x + 49 + y^2 - 2y + 1$

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Step 3: Set $AB^2 = (2 \times AM)^2$:

$AB = 2\sqrt{125} = \sqrt{4 \times 125} = \sqrt{500}$

$(x - 3)^2 + (y - (-4))^2 = 500$

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Step 4: Solve the equations:

Subtracting equation (iii) from equation (v):

$(x^2 + y^2 - 6x + 8y) - (x^2 + y^2 + 14x - 2y) = 475 - 75$

$-20x + 10y = 400$

Dividing the entire equation by 10:

Substituting $y = 2x + 40$ back into equation (i):

$x^2 + (2x + 40)^2 + 14x - 2(2x + 40) = 75$

$x^2 + 4x^2 + 160x + 1600 + 14x - 4x - 80 = 75$

$5x^2 + 170x + 1520 = 75$

$5x^2 + 170x + 1445 = 0$

Dividing by 5:

$x^2 + 34x + 289 = 0$

This is a perfect square: $(x + 17)^2 = 0$

Therefore, $x = -17$.

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Step 5: Find the value of $y$:

Substitute $x = -17$ in equation (iii):

$y = 2(-17) + 40$

$y = -34 + 40$

Therefore, $y = 6$.

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Final Answer:

Using the distance formula, the coordinates of point $B$ are calculated to be $(-17, 6)$.

Given:

Endpoints of the line segment: $A(4, 7)$ and $B(16, -2)$.

Points of trisection: $P$ and $Q$, where $P$ is closer to $A$ and $Q$ is closer to $B$.

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To Find:

The coordinates of points $P(x_1, y_1)$ and $Q(x_2, y_2)$ using midpoint logic.

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Solution:

Points of trisection divide the line segment $AB$ into three equal parts. Therefore, $AP = PQ = QB$.

Based on this property, we can derive two midpoint relationships:

1. Since $AP = PQ$, point $P$ is the midpoint of the segment $AQ$.

2. Since $PQ = QB$, point $Q$ is the midpoint of the segment $PB$.

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Step 1: Finding $x$-coordinates:

Using the midpoint formula for $P$ being the midpoint of $AQ$:

$\Rightarrow 2x_1 - x_2 = 4$

Using the midpoint formula for $Q$ being the midpoint of $PB$:

$\Rightarrow -x_1 + 2x_2 = 16$

Multiplying equation (ii) by 2, we get:

$-2x_1 + 4x_2 = 32$

Adding this to equation (i):

$(2x_1 - x_2) + (-2x_1 + 4x_2) = 4 + 32$

$3x_2 = 36$

$x_2 = \frac{36}{3} = 12$

Substituting $x_2 = 12$ in equation (i):

$2x_1 - 12 = 4$

$2x_1 = 16 \implies x_1 = 8$

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Step 2: Finding $y$-coordinates:

Using the midpoint formula for $P$ being the midpoint of $AQ$:

$\Rightarrow 2y_1 - y_2 = 7$

Using the midpoint formula for $Q$ being the midpoint of $PB$:

$\Rightarrow -y_1 + 2y_2 = -2$

Multiplying equation (iv) by 2, we get:

$-2y_1 + 4y_2 = -4$

Adding this to equation (iii):

$(2y_1 - y_2) + (-2y_1 + 4y_2) = 7 - 4$

$3y_2 = 3 \implies y_2 = 1$

Substituting $y_2 = 1$ in equation (iii):

$2y_1 - 1 = 7$

$2y_1 = 8 \implies y_1 = 4$

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Final Answer:

The coordinates of the points of trisection are $P(8, 4)$ and $Q(12, 1)$.

Given:

1. Center of the circle $K$ is the origin, $O(0, 0)$.

2. Points to be verified on the circle: $A(1, -8)$, $B(-4, 7)$, and $C(-7, -4)$.

3. Points to be checked for position: $D(-5, 6)$ and $E(0, 9)$.

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To Find/Verify:

(i) Show that $A$, $B$, and $C$ lie on circle $K$ and find the radius.

(ii) Determine if $D$ and $E$ lie inside, on, or outside circle $K$.

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Solution:

(i) Verification of points $A$, $B$, and $C$

A point $(x, y)$ lies on a circle if its distance from the center $(h, k)$ is equal to the radius $r$. For a circle centered at the origin $O(0, 0)$, the distance $d$ is given by:

Calculating distance $OA$:

Calculating distance $OB$:

Calculating distance $OC$:

Since $OA = OB = OC = \sqrt{65}$, all three points are equidistant from the origin. Thus, they lie on the same circle $K$.

The radius of the circle $K$ is $r = \sqrt{65}$ units.

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(ii) Position of points $D$ and $E$

We compare the distance of each point from the origin with the radius $r = \sqrt{65}$.

For Point $D(-5, 6)$:

$OD = \sqrt{(-5)^2 + (6)^2} = \sqrt{25 + 36} = \sqrt{61}$

Since $OD < r$, point $D$ lies within (inside) the circle $K$.

For Point $E(0, 9)$:

$OE = \sqrt{(0)^2 + (9)^2} = \sqrt{0 + 81} = \sqrt{81} = 9$

Since $OE > r$, point $E$ lies outside the circle $K$.

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Summary Table:

Point	Distance from $O(0,0)$	Comparison with $r = \sqrt{65}$	Position
$A(1, -8)$	$\sqrt{65}$	$Equal$	On the circle
$B(-4, 7)$	$\sqrt{65}$	$Equal$	On the circle
$C(-7, -4)$	$\sqrt{65}$	$Equal$	On the circle
$D(-5, 6)$	$\sqrt{61}$	$Less$ $than$	Inside the circle
$E(0, 9)$	$9$ ($\sqrt{81}$)	$Greater$ $than$	Outside the circle

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Given:

Midpoints of the sides of $\triangle ABC$ are:

1. $D(5, 1)$ (Midpoint of $BC$)

2. $E(6, 5)$ (Midpoint of $AC$)

3. $F(0, 3)$ (Midpoint of $AB$)

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To Find:

The coordinates of the vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.

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Solution:

According to the midpoint formula, the coordinates of the midpoint of a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ are $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.

For the $x$-coordinates:

Adding equations (i), (ii), and (iii):

$2(x_1 + x_2 + x_3) = 0 + 10 + 12$

$2(x_1 + x_2 + x_3) = 22$

Now, we find individual $x$-coordinates by subtracting (i), (ii), and (iii) from (iv):

$x_1 = (x_1 + x_2 + x_3) - (x_2 + x_3) = 11 - 10 = 1$

$x_2 = (x_1 + x_2 + x_3) - (x_1 + x_3) = 11 - 12 = -1$

$x_3 = (x_1 + x_2 + x_3) - (x_1 + x_2) = 11 - 0 = 11$

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For the $y$-coordinates:

Adding equations (v), (vi), and (vii):

$2(y_1 + y_2 + y_3) = 6 + 2 + 10$

$2(y_1 + y_2 + y_3) = 18$

Now, we find individual $y$-coordinates by subtracting (v), (vi), and (vii) from (viii):

$y_1 = (y_1 + y_2 + y_3) - (y_2 + y_3) = 9 - 2 = 7$

$y_2 = (y_1 + y_2 + y_3) - (y_1 + y_3) = 9 - 10 = -1$

$y_3 = (y_1 + y_2 + y_3) - (y_1 + y_2) = 9 - 6 = 3$

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Final Result:

The coordinates of the vertices of the triangle are:

$A = (1, 7)$

$B = (-1, -1)$

$C = (11, 3)$

Given:

Two main roads cross at the city centre: North–South ($N$–$S$) and East–West ($E$–$W$).

Total streets in each direction = 10.

Distance between parallel streets = $200\text{ m}$.

Scale for model = $1\text{ cm} = 200\text{ m}$.

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To Find:

(i) Draw a city model using the given scale.

(ii) Determine how many street intersections are referred to as $(4, 3)$ and $(3, 4)$.

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Solution:

(i) City Model Construction:

In our model, the $N$–$S$ roads can be represented as vertical lines (parallel to the $y$-axis) and the $E$–$W$ roads as horizontal lines (parallel to the $x$-axis). Since the distance between each street is $200\text{ m}$ and our scale is $1\text{ cm} = 200\text{ m}$, each street will be drawn at a distance of $1\text{ cm}$ from the previous one.

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(ii) Identification of Intersections:

The city layout is based on a coordinate system where each intersection is a unique point defined by the crossing of a specific $N$–$S$ street and an $E$–$W$ street.

(a) For the intersection $(4, 3)$:

This refers to the intersection of the $4^{\text{th}}$ street in the $N$–$S$ direction and the $3^{\text{rd}}$ street in the $E$–$W$ direction. In any coordinate plane, a specific pair of coordinates $(x, y)$ identifies exactly one point.

Therefore, there is only one street intersection that can be referred to as $(4, 3)$.

(b) For the intersection $(3, 4)$:

This refers to the intersection of the $3^{\text{rd}}$ street in the $N$–$S$ direction and the $4^{\text{th}}$ street in the $E$–$W$ direction. This point is distinct from $(4, 3)$.

Therefore, there is only one street intersection that can be referred to as $(3, 4)$.

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Conclusion:

Both $(4, 3)$ and $(3, 4)$ are unique intersections in the city grid. Their coordinates are distinct, and each pair points to a single, specific location where two streets cross.

Given:

Screen Dimensions: Width = $800$ pixels, Height = $600$ pixels.

The screen boundaries are from $x = 0$ to $x = 800$ and from $y = 0$ to $y = 600$.

Circle A: Centre at $(100, 150)$ and Radius $r_1 = 80$ pixels.

Circle B: Centre at $(250, 230)$ and Radius $r_2 = 100$ pixels.

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To Find:

(i) Check if any part of either circle lies outside the screen boundary.

(ii) Check if the two circles intersect each other.

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Solution:

(i) Checking if circles are within the screen:

A circle stays inside the screen if its leftmost, rightmost, topmost, and bottom-most points are within the screen limits ($0 \le x \le 800$ and $0 \le y \le 600$).

For Circle A:

Horizontal range: From $100 - 80 = 20$ to $100 + 80 = 180$.

Vertical range: From $150 - 80 = 70$ to $150 + 80 = 230$.

Since 20 and 180 are between 0 and 800, and 70 and 230 are between 0 and 600, Circle A is completely inside the screen.

For Circle B:

Horizontal range: From $250 - 100 = 150$ to $250 + 100 = 350$.

Vertical range: From $230 - 100 = 130$ to $230 + 100 = 330$.

Since 150 and 350 are between 0 and 800, and 130 and 330 are between 0 and 600, Circle B is also completely inside the screen.

Conclusion: No part of either circle lies outside the screen.

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(ii) Checking if the circles intersect:

To check if two circles intersect, we compare the distance between their centres ($d$) with the sum of their radii ($r_1 + r_2$).

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

Calculating the distance $d$:

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

$d = \sqrt{(150)^2 + (80)^2}$

$d = \sqrt{22500 + 6400}$

$d = \sqrt{28900}$

Now, finding the sum of the radii:

We observe that $d < r_1 + r_2$ ($170 < 180$).

Conclusion: Since the distance between the centres is less than the sum of their radii, the two circles intersect each other.

Given:

The coordinates of the points are $A (2, 1)$, $B (-1, 2)$, $C (-2, -1)$, and $D (1, -2)$.

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To Find:

(i) Plot the points in the coordinate plane.

(ii) Verify if $ABCD$ is a square.

(iii) Calculate the area of the figure.

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Solution:

To determine if $ABCD$ is a square, we need to check two properties using the distance formula:

1. All four sides must be of equal length ($AB = BC = CD = DA$).

2. The diagonals must be of equal length ($AC = BD$).

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Step 1: Calculating the lengths of the four sides:

Since $AB = BC = CD = DA = \sqrt{10}$, the quadrilateral is a rhombus.

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Step 2: Calculating the lengths of the diagonals:

Since the diagonals are equal ($AC = BD$), the quadrilateral $ABCD$ is a square.

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Step 3: Calculating the Area:

The area of a square is calculated as $\text{Side} \times \text{Side}$.

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

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Alternate Solution:

We can also find the area using the length of the diagonals. The area of a square is given by:

$\text{Area} = \frac{1}{2} \times (\text{diagonal})^2$

$\text{Area} = \frac{1}{2} \times (\sqrt{20})^2$

$\text{Area} = \frac{1}{2} \times 20$

Area = $10 \text{ sq. units}$

This matches our previous result, confirming the calculation.