Give three points that lie on a line with aslope of -2/5
1 answer:
Answer:
The three points that lie on a line y = -2/5x with a slope -2/5 will be:
- (0, 0)
- (5, -2)
- (10, -4)
Please check the attached graph also.
Step-by-step explanation:
We know that the slope-intercept form of the line equation is
where m is the slope and b is the y-intercept
Given
slope = m = -2/5
We suppose the line passes through the origin.
so b = 0
substituting m = -2/5 and b = 0 in the slope-intercept form
y = -2/5x + 0
y = -2/5x
Thus, the equation of line with the slope m = -2/5 and passes through the origin (0, 0) will be:
y = -2/5x
As the equation of line passes through (0, 0), thus the point (0, 0) lies on the line.
Putting x = 0 in the equation y = -2/5x
y = -2/5 × (0)
y = 0
Thus, (0, 0) is the point which also passes through the line
Putting x = 5 in the equation y = -2/5x
y = -2/5 × 5
y = -2
Thus, (5, -2) is the point which also passes through the line
Now, putting x = 10 in the equation y = -2/5x
y = -2/5 × 10
y = -4
Thus, (10, -4) is the point which also passes through the line.
Thus, the three points that lie on a line y = -2/5x with a slope -2/5 will be:
- (0, 0)
- (5, -2)
- (10, -4)
Please check the attached graph also.
You might be interested in
Answer:
Step-by-step explanation:
From the graph,
The initial red incline is a straight line passing through the origin.
So, a straight line passing through the origin is of the form:
Where, 'm' is the slope.
Now, the slope of a given as the change in y value to change in x value.
Consider the two end points of the red line (0, 0) and (4, 6).
The slope with two points on the line is given as:
Plug in . Therefore, slope is:
Hence, the equation of the initial red incline is: