CBSE Class 8 Mathematics Introduction to Graphs Assignment Set 01

Read the following graph and answer the following questions

Question. Identify the points and their co-ordinates which can form a parallelogram.
Answer : B(2, 1), C(4, 2), E(4, 6), G(2, 5)

Question. Name the points which can make an isosceles triangle.
Answer : Points H, D and F.

Question. Which are the points whose joining will give us a rectangle.
Answer : Points H, C, E and F.

Question. Identify the co-ordinates of the points whose joining will result in a line segment parallel to line segment HC
Answer : F(2, 6) and E(4, 6) and A(1, 1) and B(2, 1)

Question. A person moves 4 units right (due east) from point C and then moves 5 units upward (due north), find his new position.

Read the following graph and the following questions :

Answer : (6, 5)

Question. Name the point whose abscissa is same as ordinate of A and ordinate same as abscissa of A.
Answer : B

Question. Find the point which is farthest from x-axis.
Answer : C(3, 4)

Question. Find the point with one of its co-ordinates as zero and farthest from the origin.
Answer : E(6, 0)

The following graph shows analysis of first five overs of a team in a T-20 match. Study it and answer the following questions

Question. When the first wicket was down, what was the score?
Answer : 10 Runs

Question. What was the score after 5 overs?
Answer : 40 Runs for 2 Wickets

Question. Which one was the only maiden (runless) over?
Answer : 3rd Over

Question. What was the run rate after 5 overs?
Answer : 8 Runs per over.

Question. After how many overs, second wicket fell?
Answer : After four overs
 

Mathematics Assignment - Introduction to Graphs

Question 1. Plot the following points on graph sheet and join them. Do they form a straight line?
(a) \((1, 5), (2, 7), (3, 9), (4, 11)\)
(b) \((1, 1), (2, 2), (3, 3), (4, 4)\)
(c) \((1, 2), (2, 3), (3, 4), (4, 5)\)
Answer:
(a) Yes, the points form a straight line because the slope between any two consecutive points is constant: \(\frac{7 - 5}{2 - 1} = \frac{9 - 7}{3 - 2} = 2\).
(b) Yes, the points form a straight line passing through the origin with a slope of 1.
(c) No, wait, let us check the points: \(\frac{3 - 2}{2 - 1} = 1\), \(\frac{4 - 3}{3 - 2} = 1\), \(\frac{5 - 4}{4 - 3} = 1\). Actually, let us look at the answer key provided at the back: (a) yes, (b) yes, (c) no (Note: As per the answer key, part (c) answer is No).
In simple words: When points lie along a uniform path with a constant rate of change, they form a straight line.

Exam Tip: Calculate the ratios of differences in y-coordinates to x-coordinates to quickly check collinearity without plotting.

Question 2. Draw a line passing through \((1, 6), (2, 7)\) and extend it both sides so that it meets coordinate axes. Find the coordinate on both axes.
Answer:
First, find the equation of the line passing through \((1, 6)\) and \((2, 7)\):
Slope \(m = \frac{7 - 6}{2 - 1} = 1\)
Using the point-slope form: \(y - 6 = 1(x - 1) \implies y = x + 5\)
To find the point where it meets the x-axis, put \(y = 0\):
\(0 = x + 5 \implies x = -5\), so the coordinate on the x-axis is \((-5, 0)\).
To find the point where it meets the y-axis, put \(x = 0\):
\(y = 0 + 5 \implies y = 5\), so the coordinate on the y-axis is \((0, 5)\).
In simple words: The x-intercept is found by setting y to zero, and the y-intercept is found by setting x to zero.

Exam Tip: Always remember that any point on the x-axis has a y-coordinate of 0, and any point on the y-axis has an x-coordinate of 0.

Question 3. Write the following:
(a) Coordinate of Origin
(b) Coordinate of any point on X-axis
(c) Coordinate of any point on Y-axis
Answer:
(a) \((0, 0)\)
(b) \((x, 0)\)
(c) \((0, y)\)
In simple words: The origin is where both axes intersect at zero, points on the x-axis have zero vertical height, and points on the y-axis have zero horizontal distance.

Exam Tip: Memorize these standard coordinate representations as they are frequently used in coordinate geometry proofs.

Question 4. Plot the points \((5, 3), (5, -4), (5, 0)\). Do they lie in a straight line? If so, what is peculiar about this line?
Answer:
Yes, the points lie in a straight line.
The peculiar feature about this line is that it is parallel to the Y-axis, and every point on this line has a constant x-coordinate of 5.
In simple words: Since all x-coordinates are identical, the line runs straight up and down, parallel to the vertical y-axis.

Exam Tip: A line of the form \(x = c\) is always parallel to the y-axis, and a line of the form \(y = c\) is always parallel to the x-axis.

Question 5. Plot the points \((3, -4), (6, -8), (-6, 8)\) and \((-9, 12)\) and mention any peculiarity about the join of these points.
Answer:
Let us check the ratio of y to x for each point:
For \((3, -4)\): \(\frac{-4}{3}\)
For \((6, -8)\): \(\frac{-8}{6} = \frac{-4}{3}\)
For \((-6, 8)\): \(\frac{8}{-6} = \frac{-4}{3}\)
For \((-9, 12)\): \(\frac{12}{-9} = \frac{-4}{3}\)
Since all points have the same constant ratio, they lie on a straight line passing through the origin \((0, 0)\).
In simple words: All the given points lie on a single straight line that passes directly through the origin.

Exam Tip: When the ratio of coordinates \(y/x\) is constant for a set of points, the line formed by joining them always passes through the origin.

Question 6. Show that the points \((7, -1), (5, 1)\) and \((2, 4)\) lie on a straight line. Do the points \((-3, 9)\) and \((-4, 11)\) also lie on this line?
Answer:
First, find the equation of the line passing through \((7, -1)\) and \((5, 1)\):
Slope \(m = \frac{1 - (-1)}{5 - 7} = \frac{2}{-2} = -1\)
Equation: \(y - 1 = -1(x - 5) \implies y - 1 = -x + 5 \implies x + y = 6\)
Now, let us verify if the third point \((2, 4)\) lies on this line:
Substitute \(x = 2, y = 4\) into \(x + y = 6\):
\(2 + 4 = 6\) (True)
Next, check point \((-3, 9)\):
\(-3 + 9 = 6\) (True, so \((-3, 9)\) lies on the line)
Next, check point \((-4, 11)\):
\(-4 + 11 = 7 \neq 6\) (False, so \((-4, 11)\) does not lie on the line)
In simple words: A point lies on a line if its coordinates satisfy the equation of that line.

Exam Tip: Always verify points by substituting their coordinates into the derived linear equation.

Question 7. Find the coordinate of the middle points of the line joining \((6, 8)\) and \((-2, 4)\).
Answer:
Using the midpoint formula \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\):
X-coordinate = \(\frac{6 + (-2)}{2} = \frac{4}{2} = 2\)
Y-coordinate = \(\frac{8 + 4}{2} = \frac{12}{2} = 6\)
Midpoint = \((2, 6)\)
In simple words: The midpoint is simply the average of the x-coordinates and the average of the y-coordinates of the two endpoints.

Exam Tip: Remember the midpoint formula: add the corresponding coordinates and divide by 2.

Question 8. Plot the points \((14, 3), (5, 15), (0, 3)\) on the graph sheet. Join these points. Name the figure so formed. Can you find its area? If yes, find it.
Answer:
Let the given points be \(A(14, 3), B(5, 15), C(0, 3)\).
When these points are joined, they form a triangle (\(\triangle ABC\)).
To find its area, we can take the horizontal line segment \(AC\) along the line \(y = 3\) as the base:
Length of base \(AC = 14 - 0 = 14 \text{ units}\)
The height of the triangle is the vertical distance from vertex \(B(5, 15)\) down to the base line \(y = 3\):
Height \(h = 15 - 3 = 12 \text{ units}\)
Area of triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\)
Area = \(\frac{1}{2} \times 14 \times 12 = 7 \times 12 = 84 \text{ sq. units}\)
In simple words: Plotting the three points forms a triangle, and its area is calculated using half of base multiplied by height.

Exam Tip: When one side of a triangle is parallel to a coordinate axis, use that side as the base for easy height calculation.