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Sagot :
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
[tex]y = mx+b[/tex]
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
[tex]y = mx+b[/tex]
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.
![View image absor201](https://us-static.z-dn.net/files/d9a/09d9f9ac91c2ca51f9ddce8a1b2dcdc1.png)
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