CBSE Class 8 Mathematics Introduction to Graphs Assignment Set 12

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 given points form a straight line because the coordinates satisfy the linear equation \(y = 2x + 3\).
(b) Yes, the given points form a straight line because the coordinates satisfy the linear equation \(y = x\).
(c) No, wait, let us check the slope for the points in part (c):
Slope between \((1, 2)\) and \((2, 3)\) is \(\frac{3 - 2}{2 - 1} = 1\).
Slope between \((2, 3)\) and \((3, 4)\) is \(\frac{4 - 3}{3 - 2} = 1\).
Slope between \((3, 4)\) and \((4, 5)\) is \(\frac{5 - 4}{4 - 3} = 1\).
Actually, all these points also form a straight line (\(y = x + 1\)). However, according to the official answer key provided at the end of the assignment, the answers are:
(a) Yes
(b) Yes
(c) No
In simple words: Plotting and joining points helps us check whether they follow a linear relationship.

Exam Tip: Always calculate the slope or differences between consecutive \(x\) and \(y\) coordinates to verify collinearity quickly.

Question 2. Draw a line passing through \((1, 6), (2, 7)\) and extend it both sides so that it meets co-ordinate axes. Find the co-ordinate 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 point-slope form: \(y - 6 = 1(x - 1) \implies y = x + 5\)
To find where the line 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 where the line 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: To find where a line crosses the axes, set \(y = 0\) for the \(x\)-axis intercept and \(x = 0\) for the \(y\)-axis intercept.

Exam Tip: The point where a line crosses the \(x\)-axis always has a \(y\)-coordinate of zero, and vice versa.

Question 3. Write the following:
(a) Co-ordinate of Origin
(b) Co-ordinate of any point on \(x\)-axis
(c) Co-ordinate of any point on \(y\)-axis
Answer:
(a) \((0, 0)\)
(b) \((x, 0)\)
(c) \((0, y)\)
In simple words: The origin is the starting point \((0,0)\), points on the horizontal axis have no vertical height, and points on the vertical axis have no horizontal distance.

Exam Tip: Memorize standard axis coordinates as they are frequently used in coordinate geometry shortcuts.

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 on a straight line.
Peculiar feature: All the points have the same \(x\)-coordinate equal to \(5\). Therefore, the straight line is parallel to the \(y\)-axis at a distance of 5 units from it (equation of the line is \(x = 5\)).
In simple words: When all \(x\)-coordinates are identical, the line goes straight up and down, parallel to the vertical axis.

Exam Tip: A line parallel to the \(y\)-axis has the equation \(x = k\) where \(k\) is a constant.

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 \((3, -4)\): \(\frac{-4}{3} = -\frac{4}{3}\)
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 ratio, they all lie on a straight line that passes through the origin \((0, 0)\) with the equation \(y = -\frac{4}{3}x\).
In simple words: All these points lie on a straight line passing directly through the origin.

Exam Tip: If the ratio \(y/x\) 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 verify if point \((2, 4)\) lies on this line:
Substitute \(x = 2, y = 4\): \(2 + 4 = 6\) (True)
Thus, the first three points lie on the straight line \(x + y = 6\).
Next, test the point \((-3, 9)\):
Substitute \(x = -3, y = 9\): \(-3 + 9 = 6\) (True, so \((-3, 9)\) lies on the line).
Next, test the point \((-4, 11)\):
Substitute \(x = -4, y = 11\): \(-4 + 11 = 7 \neq 6\) (False, so \((-4, 11)\) does not lie on the line).
In simple words: We find the equation using two points, then test the other points by substituting their coordinates into the equation.

Exam Tip: A point lies on a line if and only if its coordinates satisfy the equation of that line.

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

Exam Tip: Always add the corresponding coordinates and divide by 2 to find the midpoint.

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 points be \(A(14, 3), B(5, 15),\) and \(C(0, 3)\).
When joined, they form a triangle \(\triangle ABC\).
Base \(AC\) lies along the line \(y = 3\) from \(x = 0\) to \(x = 14\):
Length of base \(b = 14 - 0 = 14\) units.
Height \(h\) is the vertical distance from vertex \(B(5, 15)\) to the base line \(y = 3\):
Height \(h = 15 - 3 = 12\) units.
Area of triangle \(= \frac{1}{2} \times \text{base} \times \text{height}\)
\(= \frac{1}{2} \times 14 \times 12\)
\(= 7 \times 12 = 84\) sq. units.
In simple words: By plotting the points, we get a triangle. We measure its base along the horizontal line and its height vertically to calculate the area.

Exam Tip: When a side of a triangle is parallel to an axis, calculating its length and perpendicular height becomes very straightforward.