Mathematics solution NCERT
Class 9 - Chapter 1: Orienting Yourself: The Use of Coordinates
Question 1
What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
Solution:
The x-axis and the y-axis intersect at a single point called the Origin.
At the origin,
- The x-coordinate is 0.
- The y-coordinate is 0.
Hence, the coordinates of the point of intersection of the two axes are
$$ (0,0) $$Answer:
$$ \boxed{(0,0)} $$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?
Solution:
A line parallel to the y-axis has the same x-coordinate for every point on it.
Since the x-coordinate of point W is
$$ -5, $$every point on the line through W parallel to the y-axis will also have x-coordinate
$$ -5. $$Therefore, the coordinates of point H can be written as
$$ (-5,\;y), $$where y may be any real number.
- If (y>0), H lies in the Second Quadrant.
- If (y<0), H lies in the Third Quadrant.
- If (y=0), H lies on the x-axis.
Answer:
The coordinates of H are
$$ \boxed{(-5,\;y)} $$where \(y\) is any real number.
Hence, H can lie in the
$$ \boxed{\text{Second or Third Quadrant}.} $$Question 3 (i)
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 the two sides of RAMP that are perpendicular to each other.
Solution:
The given points are
$$ R(3,0),\; A(0,-2),\; M(-5,-2),\; P(-5,2) $$Observe that
- AM is a horizontal line because both points have the same y-coordinate.
- MP is a vertical line because both points have the same x-coordinate.
A horizontal line is always perpendicular to a vertical line.
Answer:
$$ \boxed{AM\perp MP} $$Question 3 (ii)
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 one side of RAMP that is parallel to one of the axes.
Solution:
The coordinates of the given points are
$$ R(3,0),\; A(0,-2),\; M(-5,-2),\; P(-5,2) $$Observe that points A and M have the same y-coordinate.
$$ y=-2 $$Hence, the line segment AM is parallel to the x-axis.
Also, points M and P have the same x-coordinate.
$$ x=-5 $$Hence, the line segment MP is parallel to the y-axis.
Answer:
$$ \boxed{AM\parallel\text{x-axis}} $$Also,
$$ \boxed{MP\parallel\text{y-axis}} $$Question 3 (iii)
Find the two points that are mirror images of each other in one axis. Which axis will this be?
Solution:
The coordinates of the given points are
$$ R(3,0),\; A(0,-2),\; M(-5,-2),\; P(-5,2) $$Observe the coordinates of points M and P.
$$ M(-5,-2) $$ $$ P(-5,2) $$Both points have the same x-coordinate but opposite y-coordinates.
Therefore, they are mirror images of each other about the x-axis.
When the y-coordinate changes its sign and the x-coordinate remains the same, the points are symmetric about the x-axis.
Answer:
$$ \boxed{\text{The points }M(-5,-2)\text{ and }P(-5,2)\text{ are mirror images of each other about the x-axis.}} $$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.
Solution:
Plot the point
$$ Z(5,-6) $$on the Cartesian plane.
Draw a perpendicular from point Z to the x-axis meeting it at point I.
Draw a perpendicular from point Z to the y-axis meeting it at point N.
The coordinates of the points are
$$ I(5,0) $$ $$ N(0,-6) $$ $$ Z(5,-6) $$Thus, triangle IZN is a right-angled triangle with right angle at I.
Step 1: Find the length of IZ.
Points I and Z have the same x-coordinate.
$$ IZ=|0-(-6)| $$ $$ IZ=6\text{ units} $$Step 2: Find the length of IN.
Points I and N have the same y-coordinate.
$$ IN=|5-0| $$ $$ IN=5\text{ units} $$Step 3: Find the length of ZN.
Using Pythagoras' Theorem,
$$ ZN^2=IZ^2+IN^2 $$ $$ ZN^2=6^2+5^2 $$ $$ ZN^2=36+25 $$ $$ ZN^2=61 $$ $$ ZN=\sqrt{61} $$Therefore,
$$ ZN\approx7.81\text{ units} $$Lengths of the three sides:
$$ IZ=6\text{ units} $$ $$ IN=5\text{ units} $$ $$ ZN=\sqrt{61}\text{ units}\approx7.81\text{ units} $$Answer:
$$ \boxed{IZ=6\text{ units}} $$ $$ \boxed{IN=5\text{ units}} $$ $$ \boxed{ZN=\sqrt{61}\text{ units}\approx7.81\text{ units}} $$Graph Activity:
Plot the points
$$ Z(5,-6),\qquad I(5,0),\qquad N(0,-6) $$Join the points I, Z and N to obtain the required right-angled triangle IZN.
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?
Solution:
If negative numbers did not exist, the coordinates of a point could have only positive values or zero.
In such a case, we could locate points only in the First Quadrant and on the positive parts of the coordinate axes.
Points lying in the Second, Third and Fourth Quadrants could not be represented because they require negative x-coordinates, negative y-coordinates, or both.
Therefore, the coordinate system would not be able to represent every point on a two-dimensional plane.
Answer:
If negative numbers did not exist, the coordinate system would contain only the first quadrant and the positive parts of the axes.
Hence, it would not allow us to locate all the points on a 2-D plane.
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.
Solution:
The given points are
$$ M(-3,-4),\qquad A(0,0),\qquad G(6,8). $$To check whether the three points lie on the same straight line, compare the changes in their x-coordinates and y-coordinates.
From M to A:
Change in x-coordinate:
$$ 0-(-3)=3 $$Change in y-coordinate:
$$ 0-(-4)=4 $$Hence, the movement is
$$ (3,\;4). $$From A to G:
Change in x-coordinate:
$$ 6-0=6 $$Change in y-coordinate:
$$ 8-0=8 $$Hence, the movement is
$$ (6,\;8)=2\times(3,\;4). $$Since the change from A to G is exactly twice the change from M to A, all three points lie in the same direction.
Therefore, the three points lie on the same straight line.
Method:
Without plotting the points, compare the changes in the x-coordinates and y-coordinates between consecutive points. If the changes are in the same ratio, the points are collinear.
Answer:
$$ \boxed{\text{The points }M(-3,-4),\ A(0,0)\text{ and }G(6,8)\text{ lie on the same straight line.}} $$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.
Solution:
The given points are
$$ R(-5,-1),\qquad B(-2,-5),\qquad C(4,-12). $$Using the method from Question 6, compare the changes in the x-coordinates and y-coordinates.
From R to B:
Change in x-coordinate:
$$ -2-(-5)=3 $$Change in y-coordinate:
$$ -5-(-1)=-4 $$Hence, the movement is
$$ (3,\,-4). $$From B to C:
Change in x-coordinate:
$$ 4-(-2)=6 $$Change in y-coordinate:
$$ -12-(-5)=-7 $$Hence, the movement is
$$ (6,\,-7). $$Compare the changes:
From R to B, the changes are
$$ (3,\,-4). $$From B to C, the changes are
$$ (6,\,-7). $$The second change is not a multiple of the first because
$$ 6\ne2\times3 $$but
$$ -7\ne2\times(-4). $$Therefore, the changes are not in the same ratio.
Conclusion:
Hence, the three points do not lie on the same straight line.
On plotting the points on the Cartesian plane, we also observe that they are not collinear.
Answer:
$$ \boxed{\text{The points }R(-5,-1),\ B(-2,-5)\text{ and }C(4,-12)\text{ are not on the same straight line.}} $$Question 8 (i)
Using the origin as one vertex, plot the vertices of a right-angled isosceles triangle.
Solution:
Take the origin as one vertex.
$$ O(0,0) $$Choose the other two vertices as
$$ A(4,0) $$and
$$ B(0,4). $$Join the points O, A and B.
Verification:
The lengths of the two perpendicular sides are
$$ OA=4\text{ units} $$ $$ OB=4\text{ units} $$Since
$$ OA=OB, $$the triangle is isosceles.
Also,
$$ OA\perp OB, $$therefore, the triangle is right-angled.
Coordinates of the vertices:
$$ \boxed{ O(0,0),\; A(4,0),\; B(0,4) } $$These points form a right-angled isosceles triangle.
Answer:
$$ \boxed{ (0,0),\; (4,0),\; (0,4) } $$Note: Any other set of coordinates satisfying the given conditions is also correct.
Question 8 (ii)
Using the origin as one vertex, plot the vertices of an isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.
Solution:
Take the origin as one vertex.
$$ O(0,0) $$Choose the other two vertices as
$$ A(-4,-3) $$and
$$ B(4,-3). $$Here,
- Point A lies in Quadrant III.
- Point B lies in Quadrant IV.
Verification:
Using the distance formula,
$$ \begin{aligned} OA &=\sqrt{(-4-0)^2+(-3-0)^2}\\[4pt] &=\sqrt{16+9}\\[4pt] &=\sqrt{25}\\[4pt] &=5\text{ units}. \end{aligned} $$Similarly,
$$ \begin{aligned} OB &=\sqrt{(4-0)^2+(-3-0)^2}\\[4pt] &=\sqrt{16+9}\\[4pt] &=\sqrt{25}\\[4pt] &=5\text{ units}. \end{aligned} $$Since
$$ OA=OB=5\text{ units}, $$the triangle is an isosceles triangle.
Coordinates of the vertices:
$$ \boxed{ O(0,0),\; A(-4,-3),\; B(4,-3) } $$These points satisfy all the given conditions.
Answer:
$$ \boxed{ (0,0),\; (-4,-3),\; (4,-3) } $$Note: Any other set of coordinates satisfying the given conditions is also correct.
Question 9 (Row 1)
Given:
$$ S(-3,0),\qquad M(0,0),\qquad T(3,0) $$Solution:
To check whether M is the midpoint of ST, use the midpoint formula.
The midpoint of ST is
$$ \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) $$Substitute the coordinates of S and T.
$$ \begin{aligned} \text{Midpoint of }ST &=\left( \frac{-3+3}{2}, \frac{0+0}{2} \right)\\[4pt] &=\left( \frac{0}{2}, \frac{0}{2} \right)\\[4pt] &=(0,0) \end{aligned} $$The coordinates of the midpoint are
$$ (0,0), $$which are the same as the coordinates of point M.
Conclusion:
Therefore, M is the midpoint of ST.
| Is M the midpoint of ST? | Reason |
|---|---|
| Yes | The midpoint of S(−3,0) and T(3,0) is (0,0), which is the same as point M. |
Answer:
$$ \boxed{\text{Yes, M is the midpoint of ST.}} $$Question 9 (Row 2)
Given:
$$ S(2,3),\qquad M(3,4),\qquad T(4,5) $$Solution:
To check whether M is the midpoint of ST, use the midpoint formula.
The midpoint of ST is
$$ \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) $$Substitute the coordinates of S and T.
$$ \begin{aligned} \text{Midpoint of }ST &=\left( \frac{2+4}{2}, \frac{3+5}{2} \right)\\[4pt] &=\left( \frac{6}{2}, \frac{8}{2} \right)\\[4pt] &=(3,4) \end{aligned} $$The coordinates of the midpoint are
$$ (3,4), $$which are the same as the coordinates of point M.
Conclusion:
Therefore, M is the midpoint of ST.
| Is M the midpoint of ST? | Reason |
|---|---|
| Yes | The midpoint of S(2,3) and T(4,5) is (3,4), which is the same as point M. |
Answer:
$$ \boxed{\text{Yes, M is the midpoint of ST.}} $$Question 9 (Row 3)
Given:
$$ S(0,0),\qquad M(0,5),\qquad T(0,-10) $$Solution:
To check whether M is the midpoint of ST, use the midpoint formula.
The midpoint of ST is
$$ \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) $$Substitute the coordinates of S and T.
$$ \begin{aligned} \text{Midpoint of }ST &=\left( \frac{0+0}{2}, \frac{0+(-10)}{2} \right)\\[4pt] &=\left( 0, \frac{-10}{2} \right)\\[4pt] &=(0,-5) \end{aligned} $$The coordinates of the midpoint are
$$ (0,-5), $$whereas the coordinates of point M are
$$ (0,5). $$Since
$$ (0,-5)\ne(0,5), $$M is not the midpoint of ST.
| Is M the midpoint of ST? | Reason |
|---|---|
| No | The midpoint of S(0,0) and T(0,−10) is (0,−5), which is not the same as point M(0,5). |
Answer:
$$ \boxed{\text{No, M is not the midpoint of ST.}} $$Question 9 (Row 4)
Given:
$$ S(-8,7),\qquad M(0,-2),\qquad T(6,-3) $$Solution:
To check whether M is the midpoint of ST, use the midpoint formula.
The midpoint of ST is
$$ \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) $$Substitute the coordinates of S and T.
$$ \begin{aligned} \text{Midpoint of }ST &=\left( \frac{-8+6}{2}, \frac{7+(-3)}{2} \right)\\[4pt] &=\left( \frac{-2}{2}, \frac{4}{2} \right)\\[4pt] &=(-1,2) \end{aligned} $$The coordinates of the midpoint are
$$ (-1,2), $$whereas the coordinates of point M are
$$ (0,-2). $$Since
$$ (-1,2)\ne(0,-2), $$M is not the midpoint of ST.
| Is M the midpoint of ST? | Reason |
|---|---|
| No | The midpoint of S(−8,7) and T(6,−3) is (−1,2), which is not the same as point M(0,−2). |
Answer:
$$ \boxed{\text{No, M is not the midpoint of ST.}} $$Connection Between the Coordinates of M, S and T
Observation:
Whenever M is the midpoint of the line segment ST, its coordinates are obtained by taking the average of the corresponding coordinates of S and T.
If
$$ S(x_1,y_1)\quad\text{and}\quad T(x_2,y_2), $$then the coordinates of the midpoint M are
$$ M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right). $$Answer:
The coordinates of the midpoint are the averages of the corresponding coordinates of the endpoints.
$$ \boxed{ M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right) } $$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).
Solution:
Given,
$$ A(3,-4),\qquad M(-7,1),\qquad B(x,y) $$Since M is the midpoint of AB, by the midpoint formula,
$$ M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right) $$Substituting the given coordinates,
$$ \left(\frac{3+x}{2},\frac{-4+y}{2}\right)=(-7,1) $$Step 1: Find the value of x.
$$ \frac{3+x}{2}=-7 $$ $$ 3+x=-14 $$ $$ x=-17 $$Step 2: Find the value of y.
$$ \frac{-4+y}{2}=1 $$ $$ -4+y=2 $$ $$ y=6 $$Therefore, the coordinates of point B are
$$ (-17,6). $$Verification:
Midpoint of A(3, −4) and B(−17, 6) is
$$ \left( \frac{3+(-17)}{2}, \frac{-4+6}{2} \right) $$ $$ = \left( \frac{-14}{2}, \frac{2}{2} \right) $$ $$ =(-7,1) $$which is the given midpoint.
Answer:
$$ \boxed{B(-17,\;6)} $$Question 11
Let P and 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).
Solution:
Given,
$$ A(4,7),\qquad B(16,-2) $$Since P and Q trisect the line segment AB,
- P divides AB in the ratio 1 : 2.
- Q divides AB in the ratio 2 : 1.
Step 1: Find the coordinates of P.
P is one-third of the way from A to B.
x-coordinate of P:
$$ 4+\frac{16-4}{3} = 4+\frac{12}{3} = 4+4 = 8 $$y-coordinate of P:
$$ 7+\frac{-2-7}{3} = 7+\frac{-9}{3} = 7-3 = 4 $$Therefore,
$$ P(8,4) $$Step 2: Find the coordinates of Q.
Q is two-thirds of the way from A to B.
x-coordinate of Q:
$$ 4+\frac{2(16-4)}{3} = 4+\frac{24}{3} = 4+8 = 12 $$y-coordinate of Q:
$$ 7+\frac{2(-2-7)}{3} = 7+\frac{-18}{3} = 7-6 = 1 $$Therefore,
$$ Q(12,1) $$Verification:
The three equal parts are
$$ AP=PQ=QB. $$Hence, P and Q are the points of trisection of AB.
Answer:
$$ \boxed{P(8,4)} $$ $$ \boxed{Q(12,1)} $$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 centre is the origin O(0, 0). What is the radius of circle K?
Solution:
The centre of the circle is
$$ O(0,0). $$To show that the points A, B and C lie on the same circle, find their distances from the origin.
Step 1: Find OA.
$$ \begin{aligned} OA &=\sqrt{(1-0)^2+(-8-0)^2}\\[4pt] &=\sqrt{1+64}\\[4pt] &=\sqrt{65} \end{aligned} $$Step 2: Find OB.
$$ \begin{aligned} OB &=\sqrt{(-4-0)^2+(7-0)^2}\\[4pt] &=\sqrt{16+49}\\[4pt] &=\sqrt{65} \end{aligned} $$Step 3: Find OC.
$$ \begin{aligned} OC &=\sqrt{(-7-0)^2+(-4-0)^2}\\[4pt] &=\sqrt{49+16}\\[4pt] &=\sqrt{65} \end{aligned} $$Comparison:
$$ OA=OB=OC=\sqrt{65} $$Since all three points are at the same distance from the origin, they lie on the same circle whose centre is the origin.
Radius of the circle:
$$ r=\sqrt{65} $$Answer:
The points A, B and C lie on the circle K with centre
$$ \boxed{O(0,0)} $$and radius
$$ \boxed{\sqrt{65}\text{ units}.} $$Question 12 (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.
Solution:
From Question 12(i), the circle K has
$$ \text{Centre }O(0,0) $$and radius
$$ r=\sqrt{65}\text{ units.} $$To determine the position of the given points, compare their distances from the origin with the radius of the circle.
Step 1: Find the distance of D from the origin.
Coordinates of D are
$$ D(-5,6) $$Distance of D from O is
$$ \begin{aligned} OD &=\sqrt{(-5)^2+6^2}\\[4pt] &=\sqrt{25+36}\\[4pt] &=\sqrt{61} \end{aligned} $$Since
$$ \sqrt{61}<\sqrt{65}, $$point D lies within the circle.
Step 2: Find the distance of E from the origin.
Coordinates of E are
$$ E(0,9) $$Distance of E from O is
$$ \begin{aligned} OE &=\sqrt{0^2+9^2}\\[4pt] &=\sqrt{81}\\[4pt] &=9 \end{aligned} $$Since
$$ 9>\sqrt{65}, $$point E lies outside the circle.
Answer:
| Point | Distance from O | Position |
|---|---|---|
| D(−5, 6) | $$\sqrt{61}$$ | Within the circle |
| E(0, 9) | $$9$$ | Outside the circle |
Therefore,
$$ \boxed{\text{D lies within the circle and E lies outside the circle.}} $$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.
Solution:
Let the coordinates of the vertices of triangle ABC be
$$ A(x_1,y_1),\qquad B(x_2,y_2),\qquad C(x_3,y_3). $$Since D, E and F are the midpoints of AB, BC and AC respectively,
$$ D\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=(5,1) $$ $$ E\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right)=(6,5) $$ $$ F\left(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2}\right)=(0,3) $$Step 1: Form the equations.
From point D,
$$ x_1+x_2=10,\qquad y_1+y_2=2 $$From point E,
$$ x_2+x_3=12,\qquad y_2+y_3=10 $$From point F,
$$ x_1+x_3=0,\qquad y_1+y_3=6 $$Step 2: Find the x-coordinates.
Add the first and third equations.
$$ 2x_1+x_2+x_3=10 $$Since
$$ x_2+x_3=12, $$we get
$$ 2x_1+12=10 $$ $$ 2x_1=-2 $$ $$ x_1=-1 $$Now,
$$ x_2=10-(-1)=11 $$ $$ x_3=12-11=1 $$Step 3: Find the y-coordinates.
Add the first and third equations.
$$ 2y_1+y_2+y_3=8 $$Since
$$ y_2+y_3=10, $$we get
$$ 2y_1+10=8 $$ $$ 2y_1=-2 $$ $$ y_1=-1 $$Now,
$$ y_2=2-(-1)=3 $$ $$ y_3=10-3=7 $$Therefore,
$$ A(-1,-1) $$ $$ B(11,3) $$ $$ C(1,7) $$Verification:
Midpoint of AB:
$$ \left(\frac{-1+11}{2},\frac{-1+3}{2}\right) =(5,1) $$Midpoint of BC:
$$ \left(\frac{11+1}{2},\frac{3+7}{2}\right) =(6,5) $$Midpoint of AC:
$$ \left(\frac{-1+1}{2},\frac{-1+7}{2}\right) =(0,3) $$Hence, the obtained coordinates are correct.
Answer:
$$ \boxed{A(-1,-1),\quad B(11,3),\quad C(1,7)} $$Question 14 (i)
A city has two main roads which cross each other at the centre of the city. The 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.
Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines.
Solution:
Take the point of intersection of the two main roads as the origin O(0,0).
Draw two perpendicular lines passing through the origin.
- The horizontal line represents the East–West (E–W) road.
- The vertical line represents the North–South (N–S) road.
Using the scale
$$ 1\text{ cm}=200\text{ m}, $$draw parallel streets at intervals of 1 cm on both sides of the two main roads.
Draw 10 equally spaced streets in the East–West direction and 10 equally spaced streets in the North–South direction.
The completed figure will form a square grid representing the city's streets.
Answer:
Draw a square grid with the two main roads intersecting at the origin. Draw all other streets parallel to these roads at intervals of 1 cm (representing 200 m), with 10 streets in each direction.
Question 14 (ii)
Each street intersection is formed by two streets—one running in the North–South (N–S) direction and another in the East–West (E–W) direction. If the second street running in the N–S direction and the fifth street in the E–W direction meet, then this street intersection is called (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)?
Solution:
In the given convention, the first number represents the street running in the North–South (N–S) direction and the second number represents the street running in the East–West (E–W) direction.
Since there are streets on both sides of the two main roads, the same pair of street numbers may occur in different parts of the city.
(a) Street intersection (4, 3)
The 4th N–S street can be on either side of the main N–S road.
Similarly, the 3rd E–W street can be on either side of the main E–W road.
Hence, the possible intersections are:
- East of the main N–S road and North of the main E–W road.
- East of the main N–S road and South of the main E–W road.
- West of the main N–S road and North of the main E–W road.
- West of the main N–S road and South of the main E–W road.
Therefore, the number of street intersections referred to as (4, 3) is
$$ \boxed{4} $$(b) Street intersection (3, 4)
Similarly, the 3rd N–S street and the 4th E–W street can also intersect in four different regions of the city.
Hence, the number of street intersections referred to as (3, 4) is
$$ \boxed{4} $$Answer:
(a)
$$ \boxed{4} $$(b)
$$ \boxed{4} $$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.
Solution:
(i) Checking whether the circles lie completely inside the screen
The screen dimensions are
$$ 800 \text{ pixels} \times 600 \text{ pixels}. $$Circle A
Centre:
$$ A(100,150) $$Radius:
$$ 80\text{ pixels} $$Distance of the centre from each boundary:
- Left boundary = 100 pixels
- Bottom boundary = 150 pixels
- Right boundary = 800 − 100 = 700 pixels
- Top boundary = 600 − 150 = 450 pixels
Since all these distances are greater than the radius (80 pixels), Circle A lies completely inside the screen.
Circle B
Centre:
$$ B(250,230) $$Radius:
$$ 100\text{ pixels} $$Distance of the centre from each boundary:
- Left boundary = 250 pixels
- Bottom boundary = 230 pixels
- Right boundary = 800 − 250 = 550 pixels
- Top boundary = 600 − 230 = 370 pixels
Since all these distances are greater than the radius (100 pixels), Circle B also lies completely inside the screen.
Conclusion for Part (i):
No part of either circle lies outside the screen.
(ii) Checking whether the circles intersect
The centres of the circles are
$$ A(100,150) $$and
$$ B(250,230). $$Distance between the centres:
$$ \begin{aligned} AB &=\sqrt{(250-100)^2+(230-150)^2}\\[4pt] &=\sqrt{150^2+80^2}\\[4pt] &=\sqrt{22500+6400}\\[4pt] &=\sqrt{28900}\\[4pt] &=170\text{ pixels} \end{aligned} $$Sum of the radii:
$$ 80+100=180\text{ pixels} $$Since
$$ 170<180, $$the distance between the centres is less than the sum of the radii.
Therefore, the two circles intersect each other.
Answer:
(i) No part of either circle lies outside the screen.
(ii) The 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?
Solution:
Plot the given points on the Cartesian plane and join them in the order A, B, C and D.
$$ A(2,1),\quad B(-1,2),\quad C(-2,-1),\quad D(1,-2) $$Step 1: Find the lengths of the sides.
Length of AB
$$ \begin{aligned} AB &=\sqrt{(-1-2)^2+(2-1)^2}\\[4pt] &=\sqrt{(-3)^2+1^2}\\[4pt] &=\sqrt{9+1}\\[4pt] &=\sqrt{10} \end{aligned} $$Length of BC
$$ \begin{aligned} BC &=\sqrt{(-2+1)^2+(-1-2)^2}\\[4pt] &=\sqrt{(-1)^2+(-3)^2}\\[4pt] &=\sqrt{1+9}\\[4pt] &=\sqrt{10} \end{aligned} $$Length of CD
$$ \begin{aligned} CD &=\sqrt{(1+2)^2+(-2+1)^2}\\[4pt] &=\sqrt{3^2+(-1)^2}\\[4pt] &=\sqrt{9+1}\\[4pt] &=\sqrt{10} \end{aligned} $$Length of DA
$$ \begin{aligned} DA &=\sqrt{(2-1)^2+(1+2)^2}\\[4pt] &=\sqrt{1^2+3^2}\\[4pt] &=\sqrt{1+9}\\[4pt] &=\sqrt{10} \end{aligned} $$Thus,
$$ AB=BC=CD=DA=\sqrt{10} $$Hence, all four sides are equal.
Step 2: Find the lengths of the diagonals.
Length of AC
$$ \begin{aligned} AC &=\sqrt{(-2-2)^2+(-1-1)^2}\\[4pt] &=\sqrt{(-4)^2+(-2)^2}\\[4pt] &=\sqrt{16+4}\\[4pt] &=\sqrt{20}\\[4pt] &=2\sqrt5 \end{aligned} $$Length of BD
$$ \begin{aligned} BD &=\sqrt{(1+1)^2+(-2-2)^2}\\[4pt] &=\sqrt{2^2+(-4)^2}\\[4pt] &=\sqrt{4+16}\\[4pt] &=\sqrt{20}\\[4pt] &=2\sqrt5 \end{aligned} $$Since
$$ AC=BD=2\sqrt5, $$the diagonals are equal.
Therefore, ABCD is a square.
Step 3: Find the area of the square.
Area of a square
$$ =(\text{side})^2 $$ $$ =(\sqrt{10})^2 $$ $$ =10\text{ square units} $$Answer:
ABCD is a square because all four sides are equal and its diagonals are equal.
$$ \boxed{\text{Area of the square}=10\text{ square units}} $$