Orienting Yourself: The Use of Coordinates Class 9 Questions And Answers PDF Free Download

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NCERT Solutions for Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates: Exercise Questions

Below you will find NCERT Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates question answers explained step by step.

Fig. 1.3 shows Reiaan’s room with points OABC marking its corners. The x- and y-axes are marked in the figure. Point O is the origin.

Referring to Fig. 1.3, answer the following questions:

Question (i) If $D_1 R_1$ represents the door to Reiaan's room, how far is the door from the left wall (the y-axis) of the room? How far is the door from the x -axis?

$\textbf{Answer:}$

Since the door $D_1 R_1$ lies on the $x$-axis,
Distance of the door from the left wall, i.e., $y$-axis $=8$ units (since $D_1$ is at $x=8$ ).
Distance of the door from the $x$-axis $=0$ units, because the door lies on the $x$-axis.

Question (ii) What are the coordinates of $D_1$?

$\textbf{Answer:}$

From the figure, $D_1$ is on the x-axis. ie., $x=8$

So, the coordinates of $D_1=(8,0)$

Question (iii) If $\mathrm{R}_1$ is the point $(11.5,0)$, how wide is the door? Do you think this is a comfortable width for the room door? If a person in a wheelchair wants to enter the room, will he/she be able to do so easily?

$\textbf{Answer:}$
Given: $R_1=(11.5,0), D_1=(8,0)$

Width of the door $=11.5-8=3.5 \text { units }$

A width of 3.5 units is quite comfortable for a room door, and a person using a wheelchair should also be able to enter easily through that door.

Question (iv) If $\mathrm{B}_1(0,1.5)$ and $\mathrm{B}_2(0,4)$ represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?

$\textbf{Answer:}$

Given: $B_1=(0,1.5), B_2=(0,4)$
Bathroom door width $=4-1.5=2.5 \text { units }$
Since $2.5<3.5$, the bathroom door is narrower than the room door.

Think and Reflect [Page number: 5]

Question 1. What are the standard widths for a room door? Look around your home and in school.

$\textbf{Answer:}$

The standard width for a room door is 30 to 36 inches. That means 0.76 to 0.9 metres. Sometimes it can be smaller, and sometimes it can be wider.

Question 2. Are the doors in your school suitable for people in wheelchairs?

$\textbf{Answer:}$

Yes, the doors in our school are wider in shape and suitable for people with wheelchairs. Every school should have wider doors so that people with wheelchairs can enter freely into class.

Think and Reflect [Page number: 7]

Question 1. What is the x-coordinate of a point on the y-axis?

$\textbf{Answer:}$

The x-coordinate of a point on the y-axis is 0.

Question 2. Is there a similar generalisation for a point on the x-axis?

$\textbf{Answer:}$

Yes, the y-coordinate of a point on the x-axis is 0.

Question 3. Does point Q (y, x) ever coincide with point P (x, y)? Justify your answer.

$\textbf{Answer:}$

The point Q (y, x) can coincide with point P (x, y) when x = y.

For example, points (3, 3) and (3, 3) can coincide, but points (2, 5) and (5, 2) can not coincide.

Question 4. If x ≠ y, then (x, y) ≠ (y, x); and (x, y) = (y, x) if and only if x = y. Is this claim true?

$\textbf{Answer:}$

Yes, the claim is true.

If $x \neq y$, then swapping the coordinates changes the point.

So, $(x, y) \neq(y, x)$

If $x=y$, then: $(x, y)=(y, x)$, because both coordinates are the same.

On a graph sheet, mark the x-axis and y-axis and the origin O. Mark points from (– 7, 0) to (13, 0) on the x-axis and from (0, – 15) to (0, 12) on the y-axis. (Use the scale 1 cm = 1 unit.)
Using Fig. 1.5, answer the given questions.

Question 1. Place Reiaan's rectangular study table with three of its feet at the points $(8,9),(11,9)$ and $(11,7)$.
(i) Where will the fourth foot of the table be?
(ii) Is this a good spot for the table?
(iii) What is the width of the table? The length? Can you make out the height of the table?

$\textbf{Answer:}$

(i)
Given: Three of the tables' feet are at the points $(8,9),(11,9)$ and $(11,7)$.
Since the table is rectangular, the fourth corner will be: $(8, 7)$

(ii)

Yes. The table is placed near the upper-right part of the room and is near the bed. Also, it does not block the bed, wardrobe, or doors. It is in an ideal place to read as well.

(iii)

Horizontal distance = Length = 11 - 8 = 3 ft

Vertical distance = Width = 9 - 7 = 2 ft

Height cannot be determined from the floor plan as the vertical dimensions are not given.

Question 2. If the bathroom door has a hinge at $\mathrm{B}_1$ and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?

$\textbf{Answer:}$

According to Figure 1.5,
$B_1=(0,1.5)$ and $B_2=(0,4)$
So, width = $4-1.5=2.5$ ft

The distance of the wardrobe from the wall is 3 ft, as we can see in the picture.

So, when the bathroom door opens into the bedroom, it will not hit the wardrobe.

If the door is made wider, it may need a different opening direction or a sliding design.

Question 3. Look at Reiaan's bathroom.
(i) What are the coordinates of the four corners $\mathrm{O}, \mathrm{F}, \mathrm{R}$, and $\mathrm{P}$ of the bathroom?
(ii) What is the shape of the showering area SHWR in Reiaan's bathroom? Write the coordinates of the four corners.
(iii) Mark off a $3 \mathrm{ft} \times 2 \mathrm{ft}$ space for the washbasin and a $2 \mathrm{ft} \times 3 \mathrm{ft}$ space for the toilet. Write the coordinates of the corners of these spaces.

$\textbf{Answer:}$

(i)

From the figure, we get:
O = (0, 0)
F = (0, 9)
R = (-6, 9)
P = (-6, 0)

(ii)

The shape of the showering area is a trapezium, as one pair of opposite sides is parallel.
From the figure, we get:
S = (-6, 6)
H = (-3, 6)
W = (-2, 9)
R = (-6, 9)

(iii)

Washbasin space $(3 \mathrm{ft} \times 2 \mathrm{ft})$
One possible set of corners:
(-6,0),(-3,0),(-3,2),(-6,2)

Toilet space ( $2 \mathrm{ft} \times 3 \mathrm{ft}$ )
One possible set of corners:
(-6,2),(-4,2),(-4,5), (-6,5)

Question 4. Other rooms in the house:
(i) Reiaan's room door leads from the dining room which has the length 18 ft and width 15 ft. The length of the dining room extends from point P to point A. Sketch the dining room and mark the coordinates of its corners.
(ii) Place a rectangular $5 \mathrm{ft} \times 3 \mathrm{ft}$ dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.

$\textbf{Answer:}$

(i)

The two points of Reiaan's dining room are P(-6, 0) and A(12, 0).
Let the point on the P's side be M, and A's side be N.
It is given that the length of the dining room is 18 ft and the width is 15 ft.
So, the coordinates of the other two points below are:
M: (-6, -15)
N: (12, -15)

(ii)

The dining room extends
In the x-axis, -6 to 12
In the y-axis, 0 to -15
So, the centre of the dining room is $=(\frac{-6+12}2),(\frac{0-15}2)=(3,-7.5)$
To place a 5 ft × 3 ft table at the centre, we have to take the length 5 ft and the width 3 ft.
Half length = $\frac52=2.5$ and half width $=\frac32=1.5$
So, the coordinates of the four feet of the table are:

$\begin{aligned} & (3-2.5,-7.5-1.5)=(0.5,-9) \\ & (3+2.5,-7.5-1.5)=(5.5,-9) \\ & (3+2.5,-7.5+1.5)=(5.5,-6) \\ & (3-2.5,-7.5+1.5)=(0.5,-6)\end{aligned}$

Think and Reflect [Page number: 9]

Question 1. In moving from A (3, 4) to D (7, 1), what distance has been covered along the x-axis? What about the distance along the y-axis?

$\textbf{Answer:}$

The distance Covered along x-axis = 7 - 3 = 4 units

The distance covered along y-axis = 4 - 1 = 3 units

Question 2. Can these distances help you find the distance AD?

$\textbf{Answer:}$

Yes. These distances form the two perpendicular sides of a right triangle.

Using the Pythagoras Theorem:

$A D=\sqrt{4^2+3^2}=\sqrt{25}=5$ units

Think and Reflect [Page number: 11]

Question 1. What has remained the same and what has changed with this reflection?

$\textbf{Answer:}$
In the reflection of $\triangle$ ADM in the y -axis:

• The shape and size of the triangle remain the same.
• The distances between points remain the same.
• The y-coordinates remain unchanged.
• The x-coordinates change their signs:

Here,
$(3,4) \rightarrow(-3,4)$
$(7,1) \rightarrow(-7,1)$
$(9,6) \rightarrow(-9,6)$

So, reflection in the $y$-axis changes only the sign of the $x$-coordinate.

Question 2. Would these observations be the same if ΔADM is reflected in the x-axis (instead of the y-axis)?

$\textbf{Answer:}$

Yes, the observations would be similar if the triangle is reflected in the x-axis, but with one difference:

• The shape, size, and distances would remain the same.
• This time, the x-coordinates would remain unchanged.
• The y-coordinates would change their signs.

For example: $(3,4) \rightarrow(3,-4)$
So:

• Reflection in the y-axis changes the sign of x.
• Reflection in the x-axis changes the sign of y.

Question 1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes?

$\textbf{Answer:}$

The two coordinate axes, x and y, only intersect at the origin.

So, the coordinate is (0, 0).

Question 2. Point W has x-coordinate equal to – 5. Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in?

$\textbf{Answer:}$

A line parallel to the $y$-axis has the same $x$-coordinate for all points on it.
So point $H$ will have coordinates: $(-5, y)$, where $y$ = any real number

• If $y>0, H$ lies in Quadrant II.
• If $y<0, H$ lies in Quadrant III.
• If $y=0, H$ lies in x-axis

Question 3. Consider the points R (3, 0), A (0, – 2), M (– 5, – 2) and P (– 5, 2). If they are joined in the same order, predict:

(i) Two sides of RAMP that are perpendicular to each other.

(ii) One side of RAMP that is parallel to one of the axes.

(iii) Two points that are mirror images of each other in one axis. Which axis will this be? Now plot the points and verify your predictions.

$\textbf{Answer:}$

(i)
AM by joining A (0, – 2) and M (– 5, – 2), is horizontal.
MP by joining M (– 5, – 2) and P (– 5, 2) is vertical.
So, it is clear that $MP\perp AM$
$\therefore$ The two sides of RAMP are perpendicular to each other.

(ii)
Since points A and M have the same y -coordinate, -2, AM is parallel to the $x$-axis.
Also, MP is parallel to the y-axis, as M and P have the same x-coordinate, -5.

(iii)
If two points have the same x-coordinates and opposite y-coordinates, then they are mirror images in the x-axis.
Here, M and P have the same x-coordinate, -5 and opposite y-coordinates, 2 and -2. So, they are mirror images in the x-axis.

Question 4. Plot point Z (5, – 6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides. (Comment: Answers may differ from person to person.)

$\textbf{Answer:}$

Question 5. What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a 2-D plane?

$\textbf{Answer:}$

Without negative numbers:

• We could only mark points in Quadrant I and on the positive x and y axes.
• Points in Quadrants II, III, and IV could not be located.

$\therefore$ No, that system would not allow us to locate all points on a 2-D plane because negative coordinates are needed for points in other quadrants.

*Question 6. Are the points M (– 3, – 4), A (0, 0) and G (6, 8) on the same straight line? Suggest a method to check this without plotting and joining the points.

$\textbf{Answer:}$

We know that the condition for collinearity of three points is that if the slopes between consecutive pairs of points are equal, then the points lie on the same straight line.
So, for three points $(x_1,y_1),(x_2,y_2),$ and $(x_3,y_3)$ is collinear
If $\frac{y_2-y_1}{x_2-x_1}=\frac{y_3-y_2}{x_3-x_2}$.

Here, points are: M (– 3, – 4), A (0, 0) and G (6, 8)
Slope of $M A$: $\frac{0-(-4)}{0-(-3)}=\frac{4}{3}$
Slope of $A G$: $\frac{8-0}{6-0}=\frac{8}{6}=\frac{4}{3}$
Since the slopes are equal, the points lie on the same straight line.

*Question 7. Use your method (from Problem 6) to check if the points R (– 5, – 1), B (– 2, – 5) and C (4, – 12) are on the same straight line. Now plot both sets of points and check your answers.

$\textbf{Answer:}$

We know that for three points $(x_1,y_1),(x_2,y_2),$ and $(x_3,y_3)$ is collinear
If $\frac{y_2-y_1}{x_2-x_1}=\frac{y_3-y_2}{x_3-x_2}$.

Here, points are: R (– 5, – 1), B (– 2, – 5) and C (4, – 12)
Slope of $R B$ :$\frac{-5-(-1)}{-2-(-5)}=-\frac{4}{3}$
Slope of $B C$ :$\frac{-12-(-5)}{4-(-2)}=-\frac{7}{6}$
Since:
$-\frac{4}{3} \neq-\frac{7}{6}$
Since the slopes are not equal, the points do not lie on the same straight line.

*Question 8. Using the origin as one vertex, plot the vertices of:

(i) A right-angled isosceles triangle.

(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.

$\textbf{Answer:}$

(i)

Set of vertices of $\triangle$OAB is: O(0,0), A(4,0) and B(0,4)
Here, the two perpendicular sides are equal.

(ii)

Set of vertices is: $\mathrm{O}=(0,0), \mathrm{P}=(-3,-4), \mathrm{Q}=(3,-4)$

Here,

• P lies in Quadrant III
• Q lies in Quadrant IV
• O P=O Q=5 units

So $\triangle$OPQ is an isosceles triangle.

*Question 9. The following table shows the coordinates of points S, M and T. In each case, state whether M is the midpoint of segment ST. Justify your answer.

When M is the mid-point of ST, can you find any connection between the coordinates of M, S and T?

$\textbf{Answer:}$

We know that If the two points are: $\left(x_1, y_1\right) \text { and }\left(x_2, y_2\right)$

then their midpoint is: $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

*Question 10. Use the connection you found to find the coordinates of B given that M (–7, 1) is the midpoint of A (3, – 4) and B (x, y).

$\textbf{Answer:}$

We know that If the two points are: $\left(x_1, y_1\right) \text { and }\left(x_2, y_2\right)$
then their midpoint is: $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
Given that,
M (–7, 1) is the midpoint of A (3, – 4) and B (x, y).
Using the formula, we get,
$\frac{3+x}{2}=-7$
$\therefore x=-14$
Also,
$\frac{3+x}{2}=-7$
$\therefore y=6$

So, the coordinates of B are $(-14,6)$.

*Question 11. Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for the case when the points are A (4, 7) and B (16, –2).

$\textbf{Answer:}$

Given:
$A(4,7), B(16,-2)$
Difference in $x$-coordinates: $16-4=12$
Difference in $y$-coordinates: $-2-7=-9$
For trisection: $\frac{12}{3}=4$ and $\frac{-9}{3}=-3$
That means, for each step, it:

• moves 4 units in the x-direction
• moves 3 units downward in the y-direction

Point $P$ is one-third from $A$ :
$P=(4+4,7-3)=(8,4)$
Point $Q$ is two-thirds from $A$ :
$Q=(4+8,7-6)=(12,1)$

So the trisection points are: $(8,4) \text { and }(12,1)$

*Question 12.

(i) Given the points A (1, – 8), B (– 4, 7) and C (–7, – 4), show that they lie on a circle K whose center is the origin O (0, 0). What is the radius of circle K?

(ii) Given the points D (–5, 6) and E (0, 9), check whether D and E lie within the circle, on the circle, or outside the circle K.

$\textbf{Answer:}$

(i)

Given: Points A (1, – 8), B (– 4, 7) and C (–7, – 4)

We know that distance of points (x, y) from the origin is $\sqrt{x^2+y^2}$

So, the distance of point A(1, -8) from the centre O:

$O A=\sqrt{1^2+(-8)^2}=\sqrt{1+64}=\sqrt{65}$

The distance of point B(-4,7) from the centre O:

$O B=\sqrt{(-4)^2+7^2}=\sqrt{16+49}=\sqrt{65}$

The distance of point C(-7,-4) from the centre O:

$O C=\sqrt{(-7)^2+(-4)^2}=\sqrt{49+16}=\sqrt{65}$

All distances are equal.

So all three points lie on the same circle centred at the origin.

$\therefore$ The radius of the circle, K with centre (0, 0) is $\sqrt{65}$.

(ii)

Given: Points D (–5, 6) and E (0, 9)
We already know that the radius of the circle K with centre (0, 0) is $\sqrt{65}$.
So, the distance of point D(-5,6) from the centre O:
$O D=\sqrt{25+36}=\sqrt{61}$
Since $\sqrt{61}<\sqrt{65}$, point $D$ lies inside the circle.

The distance of point E(0,9) from the centre O:
$O E=\sqrt{0+81}=9$
Since$9>\sqrt{65}$, the point $E$ lies outside the circle.

*Question 13. The midpoints of the sides of triangle ABC are the points D, E, and F. Given that the coordinates of D, E, and F are (5, 1), (6, 5), and (0, 3), respectively, find the coordinates of A, B and C.

$\textbf{Answer:}$
Let:
$
A\left(x_1, y_1\right), B\left(x_2, y_2\right), $ and $ C\left(x_3, y_3\right)
$
Given:
$
D(5,1), E(6,5), F(0,3)
$
Since $D$ is midpoint of $B C$ :

$
\begin{aligned}
& \frac{x_2+x_3}{2}=5 \\
& ⇒x_2+x_3=10.....(1) \\
& \frac{y_2+y_3}{2}=1 \\
&⇒ y_2+y_3=2...........(2)
\end{aligned}
$
Since $E$ is midpoint of $C A$ :
$
\begin{aligned}
& x_3+x_1=12.......(3) \\
& y_3+y_1=10.........(4)
\end{aligned}
$
Since $F$ is midpoint of $A B$ :
$
\begin{aligned}
& x_1+x_2=0........(5) \\
& y_1+y_2=6....(6)
\end{aligned}
$

Adding equations 1 and 5, we get,
$x_1+2x_2+x_3=10$
$⇒12+2x_2=10[\because x_3+x_1=12.......(3)]$
$\therefore x_2=-1$
Using this value, we get $x_1=1$ and $x_3=11$

Adding equations 2 and 6, we get,
$y_1+2y_2+y_3=8$
$⇒10+2y_2=8[\because y_3+y_1=10.........(4)]$
$\therefore y_2=-1$
Using this value, we get $y_1=7$ and $y_3=3$

Hence:$A=(-1,-1), B=(11,3), $ and $C=(1,7)$

Question 14. A city has two main roads which cross each other at the centre of the city. These two roads are along the North–South (N–S) direction and East–West (E–W) direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 10 streets in each direction.

(i) Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines.

(ii) There are street intersections in the model. Each street intersection is formed by two streets — one running in the N–S direction and another in the E–W direction. Each street intersection is referred to in the following manner: If the second street running in the N–S direction and 5th street in the E–W direction meet at some crossing, then we call this street intersection (2, 5).

Using this convention, find:

(a) How many street intersections can be referred to as (4, 3)?

(b) How many street intersections can be referred to as (3, 4)?

$\textbf{Answer:}$

(i)

Given:
10 vertical lines (N–S roads)

10 horizontal lines (E–W roads)

Distance between adjacent roads: 1 cm = 200 m

(ii)
(a)
One vertical road and one horizontal road intersect at only one point.

That point is (4, 3).

(b)
Similarly, only one intersection exists.

That point is (3, 4).

Question 15. A computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre at the point A (100, 150). Another circular icon of radius 100 pixels is drawn with its centre at the point B (250, 230).

Determine:

(i) whether any part of either circle lies outside the screen.

(ii) whether the two circles intersect each other.

$\textbf{Answer:}$

(i)

For Circle 1:
Given: Centre: (100,150) and Radius: 80
On Left: $100-80=20>0$
On Right: $100+80=180<800$
On Bottom: $150-80=70>0$
On Top: $150+80=230<600$
So Circle A lies completely inside the screen.

For Circle 2:
Given: Centre: (250, 230) and Radius: 100
On Left: $250-100=150>0$
On Right: $250+100=350<800$
On Bottom: $230-100=130>0$
On Top: $230+100=330<600$
So Circle B also lies completely inside the screen.

So, no part of either circle lies outside the screen.

(ii)

Distance between the centres of the two circles

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

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

$=\sqrt{22500+6400} $

$=\sqrt{28900}$

$=170$

Sum of radii of both circles = 80+100=180

Since: 170<180

Hence, two circles intersect each other.

Question 16. Plot the points A (2, 1), B (–1, 2), C (–2, –1), and D (1, –2) in the coordinate plane. Is ABCD a square? Can you explain why? What is the area of this square?

$\textbf{Answer:}$

Yes, ABCD is a square.

$A(2,1), B(-1,2), C(-2,-1), D(1,-2)$

We know, by the Baudhāyana-Pythagoras Theorem, the distance between points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$.

$A B=\sqrt{(-1-2)^2+(2-1)^2}=\sqrt{(-3)^2+1^2}=\sqrt{10}$

$B C=\sqrt{(-2+1)^2+(-1-2)^2}=\sqrt{(-1)^2+(-3)^2}=\sqrt{10}$

$C D=\sqrt{(1+2)^2+(-2+1)^2}=\sqrt{3^2+(-1)^2}=\sqrt{10}$

$D A=\sqrt{(2-1)^2+(1+2)^2}=\sqrt{1^2+3^2}=\sqrt{10}$
All four sides are equal.

Now, we will check whether the adjacent sides are perpendicular.
We know that when two lines are perpendicular, the product of their slopes is −1.
Slope of AB = $=\frac{2-1}{-1-2}=-\frac{1}{3}$
Slope of BC = $\frac{-1-2}{-2+1}=3$
Now, $-\frac{1}{3}\times3=-1$
So, the sides AB and BC are perpendicular.
$\therefore$ ABCD is a square.

Side length of the square is $\sqrt{10}$ unit
Area of a square = $\text{Side}^2=(\sqrt{10})^2=10$ square units

Class 9 Maths NCERT Chapter 1 Solutions: Extra Questions

Question 1: Write the coordinates of a point whose ordinate is 4 and which lies on the y-axis.

$\textbf{Answer:}$

Any point on the y-axis has its abscissa (x-coordinate) equal to 0.

Given that the ordinate (y-coordinate) is 4, the point is: (0, 4)

Question 2:

A slope of line AB is $-\frac{2}{3}$. Co-ordinates of points A and B are $(x,–3)$ and $(5, 2)$, respectively. What is the value of $x$?

$\textbf{Answer:}$

Given: A slope of line AB is $-\frac{2}{3}$. Co-ordinates of points A and B are $(x,–3)$ and $(5, 2)$, respectively.
Slope of line(m) $=\frac{y_{2}–y_{1}}{x_{2}–x_{1}}$
$⇒-\frac{2}{3}= \frac{2–(–3)}{5–x}$
$⇒-\frac{2}{3}= \frac{2+3}{5–x}$
$⇒-\frac{2}{3}= \frac{5}{5–x}$
$⇒-10+2x=15$
$\therefore x=12.5$
Hence, the correct answer is 12.5.

Question 3:

What is the $y$-intersect of the linear equation $59x + 14y – 112 = 0$?

$\textbf{Answer:}$

Given: The linear equation is $59x + 14y – 112 = 0$.
To find the y-intercept of the linear equation $59x + 14y – 112 = 0$, put $x = 0$ and solve for $y$.
$59x + 14y – 112 = 0$
⇒ $0 + 14y – 112 = 0$
⇒ $14y = 112$
$\therefore$ $y = 8$
Hence, the correct answer is 8.

Question 4:

What is the reflection of the point $(4,7)$ over the line $y=–1$?

$\textbf{Answer:}$

Given: The point is $(4,7)$ and the line is $y=–1$.
The formula for the reflection of the point $(x_1, y_1)$ over the line $ax+by+c=0$ is $\frac{x–x_1}{a}=\frac{y–y_1}{b}=\frac{–2(ax_1+by_1+c)}{a^2+b^2}$.
Here are the values of $a=0$, $b=1$, and $c=1$.
$(x–4)=0$ and $(y–7)=\frac{–2(7+1)}{1}$
⇒ $x=4$ and $y=–16+7$
⇒ $x=4$ and $y=–9$
Hence, the correct answer is $(4,–9)$.

Question 5:

What is the equation of the line if its slope is $\frac{–2}{5}$ and it passes through the point $(1,–3)$?

$\textbf{Answer:}$

Given: Slope = $-\frac{2}{5}$
Passing through the point $(1,–3)$.
The equation of the line with slope $m$ is $y=mx+c$
Putting the values of $m$ and $(x,y)$,
⇒ $ –3= -\frac{2}{5}×1+c$
⇒ $c= \frac{2}{5}-3$
⇒ $c=-\frac{13}{5}$
So, the equation is $y=-\frac{2}{5}x+(-\frac{13}{5})$
⇒ $5y=–2x–13$
⇒ $2x+5y=–13$
Hence, the correct answer is $2x+5y=–13$.

Orienting Yourself: The Use of Coordinates Class 9 Chapter 1: Topics

Topics you will learn in NCERT Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates include:

• 1.1 Introduction
• 1.2 Settling In
• 1.3 The 2-D Cartesian Coordinate System
• 1.4 Distance Between Two Points in the 2-D Plane

NCERT Solutions for Class 9 Maths Chapter Wise

We at Careers360 compiled all the NCERT Class 9 Maths solutions in one place for easy student reference. The following links will allow you to access them.

NCERT Class 9 Books & Syllabus

Students can also refer to the NCERT Books and the latest Syllabus for Class 9 provided by Careers360 using the following links.

• NCERT Books Class 9 Maths
• NCERT Books for Class 9 Science
• NCERT Syllabus Class 9 Maths
• NCERT Books Class 9
• NCERT Syllabus Class 9