Name: Solving a Linear Problem by Using Table and Graph
Uploaded: 2019-07-19
Duration: 27 min 40 s
Description: How can you solve a problem using a linear relationship if you are not given an initial value? Let's watch this interesting video to learn about this!

Question

Draw a graph for the following:

Distance(in meters)	$5$	$10$	$15$	$20$	$25$	$30$
Time(in seconds)	$1$	$2$	$3$	$4$	$5$	$6$

Is it a linear graph?

Accepted Answer

Steps to plot the points:

Draw the $x$ and $5$ axes with a scale $x$ -axis; $1 division = 1 second$ and $y$ -axis $1 division = 5 meters$ .

We need to plot the point $(1, 5)$ . Here, the $x$ -coordinate is $1$ , which is positive and the $y$ -coordinate is $5$ , which is positive as well. Hence, the point $(1, 5)$ lies in the first quadrant. Start from the origin, move $1$ unit to the right along the $x$ -axis and then $1$ unit up, mark the point.

We need to plot the point $(2, 10)$ . Here, the $x$ -coordinate is $2$ , which is positive and the $y$ -coordinate is $10$ , which is positive as well. Hence, the point $(2, 10)$ lies in the first quadrant. Start from the origin, move $2$ units to the right along the $x$ -axis and then $2$ units up, mark the point.

We need to plot the point $(3, 15)$ . Here, the $x$ -coordinate is $3$ , which is positive and the $y$ -coordinate is $15$ , which is positive as well. Hence, the point $(3, 15)$ lies in the first quadrant. Start from the origin, move $3$ units to the right along the $x$ -axis and then $3$ units up, mark the point.

We need to plot the point $(4, 20)$ . Here, the $x$ -coordinate is $4$ , which is positive and the $y$ -coordinate is $30$ , which is positive as well. Hence, the point $(4, 20)$ lies in the first quadrant. Start from the origin, move $2$ units to the right along the $x$ -axis and then $4$ units up, mark the point.

We need to plot the point $5$ . Here, the $x$ -coordinate is $5$ , which is positive and the $y$ -coordinate is $25$ , which is positive as well. Hence, the point $(5, 25)$ lies in the first quadrant. Start from the origin, move $5$ units to the right along the $x$ -axis and then $5$ units up, mark the point.

We need to plot the point $(6, 30)$ . Here, the $x$ -coordinate is $6$ , which is positive and the $y$ -coordinate is $30$ , which is positive as well. Hence, the point $(6, 30)$ lies in the first quadrant. Start from the origin, move $6$ units to the right along the $x$ -axis and then $6$ units up, mark the point.

Now, join the points.

After following the given steps, we can obtain the graph as shown below.

Hence, it is a linear graph.

Months	June	July	Aug	Sep	Oct	Nov	Dec
Output of tin mine(in tonne)	$2$	$10$	$6$	$9$	$5$	$5$	$8$

Months	June	July	Aug	Sep	Oct	Nov	Dec
Output of tin mine(in tonne)	$2$	$10$	$6$	$9$	$5$	$5$	$8$

Height(in $cm$ )	$75$	$90$	$110$	$120$	$130$
Age(in year)	$2$	$4$	$6$	$8$	$10$

Height(in cm)	$75$	$90$	$110$	$120$	$130$
Age(in yr)	$2$	$4$	$6$	$8$	$10$

Draw a graph for the following:

Distance(in meters) $5$ $10$ $15$ $20$ $25$ $30$

Time(in seconds) $1$ $2$ $3$ $4$ $5$ $6$

Is it a linear graph?

Important Questions on Introduction to Graphs

Learn and Prepare for any exam you want !

Draw a graph for the following: Distance(in meters) 5 10 15 20 25 30 Time(in seconds) 1 2 3 4 5 6 Is it a linear graph?

Important Questions on Introduction to Graphs

Learn and Prepare for any exam you want !

Draw a graph for the following:

Distance(in meters) $5$ $10$ $15$ $20$ $25$ $30$

Time(in seconds) $1$ $2$ $3$ $4$ $5$ $6$

Is it a linear graph?