What is the simplified form of i^15 ?
1 answer:
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Answer:
-8
Step-by-step explanation:
Let's first establish the reference line (the one that the second line will be perpendicular to). We are told that this line passes through two points:
(1,1) and (-3,-1).
We'll find a line equation using the point slope form format to start:
(y - y1) = m * (x - x1), where m is the slope and the x and y are from two points.
(x,y) = (1,1)
(x1,y1) = (-3,-1)
Rearrange the equation:
(y - y1) = m * (x - x1)
m = (y - y1)/ (x - x1)
m = (1-(-1))/(1-(-3))
m = 2/4, or 1/2: The slope is 1/2. [<u>This is "m."]</u>
We can use the slope-intercept form for this line (y=mx + b) and then calculate b, the y-intercept:
y = (1/2)x + b
Use either of the two given points. I'll use (1,1) since I have memorized the "1" math tables.
y = (1/2)x + b
1 = (1/2)(1) + b for (1,1)
b = 1/2
This makes the reference line: y = (1/2)x+(1/2)
===
The line perpendicular must have a slope that is the negative inverse of (1/2). This would be -(2/1), or -2.
We can then write y = -2x + b
To find be, enter the one given point for this line: (-3,-2)
y = -2x + b
-2 = -2(-3) + b
-2 = 6 + b
b = -8
The perpendicular line is thus:
y = -2x - 8
It has a y-intercept of -8
