QUESTION 7
iv
Easy
Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.
3y = 6x − 11
SOLUTION
1
Find slope and y-intercept from the given equation
\(3y = 6x − 11\)
⇒ \(y = 2x − \frac{11}{3}\)
Here, slope \(= 2\) and \(y-\)intercept \(= \frac{−11}{3}\)
The line cuts \(y-\)axis at (\(0,\frac{−11}{3}\)).
⇒ \(y = 2x − \frac{11}{3}\)
Here, slope \(= 2\) and \(y-\)intercept \(= \frac{−11}{3}\)
The line cuts \(y-\)axis at (\(0,\frac{−11}{3}\)).
2
Find two points on the line to draw the graph
When \(x = 1, y = 2 − \frac{11}{3} = \frac{−5}{3}\)
So the point is (\(1, \frac{−5}{3})\).
When \(x = 2, y = 4 − \frac{11}{3} = \frac{1}{3}\)
So the point is (\(2, \frac{1}{3})\).
So the point is (\(1, \frac{−5}{3})\).
When \(x = 2, y = 4 − \frac{11}{3} = \frac{1}{3}\)
So the point is (\(2, \frac{1}{3})\).
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Final Answer : slope \(= 2\), \(y-\)intercept \(= \frac{−11}{3}\), and the line cuts the \(y-\)axis at (\(0,\frac{−11}{3}\)).
Concept Note
The equation of a straight line is of the form
\(y = mx + c\), where
\(m = \)slope
\(c = \)y-intercept
The values \(m\) and \(c\) completely describe the line's direction, steepness and position.