Question
Question: Represent the following pair of equations graphically and write the coordinates of points where the ...
Represent the following pair of equations graphically and write the coordinates of points where the line intersects the y-axis.
x+3y=62x−3y=12
Solution
Hint: Here, first we have to find the two points to draw the graph of the system of equations x+3y=6 and 2x−3y=12. While choosing the points, first take x=0 and then find the value of y. Similarly, for y=0, find the value of x. Using the two points of the equations, plot the graph. From the graph identify the points that intersect x+3y=6 and 2x−3y=12 with the y-axis which are the required coordinates.
Complete step-by-step answer:
Here, we are given with the equations:
x+3y=62x−3y=12
Now, we have to represent the above equations graphically and have to find the coordinates of points where the line intersects the y-axis.
First consider the equation:
x+3y=6
Now, we have to draw the line corresponding to the given equation. For that consider two points.
When x=0, then the equation,
⇒0+3y=6⇒3y=6
⇒y=36⇒y=2
Now, consider the equation x+3y=6 when y=0,
⇒x+3×0=6⇒x+0=6⇒x=6
So, we got two points to plot the graph of the equation x+3y=6. Now we will get the table as follows:
Next, consider the equation:
2x−3y=12
Now, we have to find two points to draw the graph of the equation.
For that take x=0, our equation,
⇒2×0−3y=12⇒0−3y=12⇒−3y=12
⇒y=−312⇒y=−4
Now, for y=0 we will obtain:
2x−3×0=12⇒2x−0=12⇒2x=12
⇒x=212⇒x=6
Now, we can write the table for the above data. It is as follows:
Now, we can plot the graph with the points (0, 2), (6, 0) and (0, -4), (6, 0).
We can see that the lines represented by the equations x+3y=6 and 2x−3y=12 meet the y-axis at B (0, 2) and C (0, -4) respectively.
Hence, the required coordinates are (0, 2) and (0, -4).
Note: Here, we can also check the coordinates by substituting it into any one of the equations. Since every point on the line is a solution of the equation, the point satisfies the equation if it lies on the line. Also, every solution of the equation is a point on the line. Otherwise, we can also find the points using the elimination method.