Find an equation of the line. Write the equation using function notation.
1 answer:
Answer:
f(x) = -5x +48
Step-by-step explanation:
A fairly easy way to write the equation of the perpendicular line is to swap the x- and y-coefficients, negating one of them. Then choose the constant in the equation to make it true at the given point.
<h3>Swapped coefficients</h3>In this instance, we find it convenient to leave the coefficient of y as positive.
5y = x -10 . . . . . . original equation
y = -5x +c . . . . . . with coefficients swapped, x-coefficient negated
<h3>New constant</h3>At the given point, the equation becomes ...
8 = -5(8) +c
48 = c . . . . . . . add 40
<h3>Equation of the perpendicular line</h3>The equation of the line is then ...
y = -5x +48
f(x) = -5x +48 . . . . . . in functional notation
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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.
