Give the equation of a line perpendicular to the line y = -8x + 8 having the y-intercept 2 units below the y-intercept of the given equation. At what point the two equations cross each other?
-6x + 8?
I have no clue, that's just a guess.
Hope that helps.
To find the equation of a line perpendicular to the given line y = -8x + 8, we need to find its slope first. The slope of the given line is -8 because it is in the form y = mx + b, where m represents the slope.
Since we are looking for a line perpendicular to this, the slope of the new line will be the negative reciprocal of -8. To find the negative reciprocal, we flip the fraction and change its sign. Therefore, the slope of the new line will be 1/8.
Now, let's find the y-intercept of the new line. We know that it will be 2 units below the y-intercept of the given line. The y-intercept of the given line is 8, so the y-intercept of the new line will be 8 - 2 = 6.
Therefore, the equation of the line perpendicular to y = -8x + 8, with a y-intercept 2 units below, is y = (1/8)x + 6.
To find the point where the two lines intersect, we equate the equations of the two lines and solve for x and y.
-8x + 8 = (1/8)x + 6
First, let's isolate the x-term:
-8x - (1/8)x = 6 - 8
Simplifying:
-64/8x - 1/8x = -2
-65/8x = -2
To get rid of the fraction, we can multiply both sides by -8:
(-65/8x) * -8 = -2 * -8
65x = 16
Now, divide both sides by 65 to solve for x:
x = 16/65
Substituting this value of x back into either of the original equations, we can solve for y:
y = -8 * (16/65) + 8
Simplifying:
y = -128/65 + 8
Now, let's convert the fraction to a decimal:
y = -1.969 + 8
y = 6.031
Therefore, the two lines intersect at the point (16/65, 6.031).