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?
To find the equation of a line perpendicular to a given line, we first need to determine the slope of the given line. In this case, the given line has the form y = mx + b, where m is the slope and b is the y-intercept. From the equation y = -8x + 8, we can see that the slope is -8.
Since we want to find a line perpendicular to the given line, we need to take the negative reciprocal of the slope. The negative reciprocal of -8 is 1/8. Therefore, the slope of the desired line is 1/8.
Now, let's find the y-intercept of the desired line. We are told that the y-intercept of the desired line is 2 units below the y-intercept of the given line. The y-intercept of the given line is 8. So, 2 units below that would be 8 - 2 = 6. Therefore, the y-intercept of the desired line is 6.
Putting this information together, 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 equations cross each other, we set the two equations equal to each other and solve for x and y:
-8x + 8 = (1/8)x + 6
To solve for x, we can add 8x and subtract 6 from both sides of the equation:
-8x - (1/8)x = 6 - 8
-64/8x - 1/8x = -2
-65/8x = -2
Now, multiply both sides by -8/65 to isolate x:
x = (-2)(-8/65)
x = 16/65
To find the value of y, we substitute this value of x back into either of the original equations. Let's use y = -8x + 8:
y = -8(16/65) + 8
y = -128/65 + 8
To simplify this expression, we find a common denominator for the two fractions:
y = (-128 + 520)/65
y = 392/65
Therefore, the two equations intersect at the point (16/65, 392/65).