Answer: y= - 4x+18
Step-by-step explanation:
Equation: y=mx+b
***remember: b is the y-intercept and m is the slope.
m=
3= x1
2= x2
6= y1
10=y2
m==
= -4
m=-4
Now we have y=-4x+b , so let's find b.
You can use either (x,y) such as (3,6) or (2,10) point you want..the answer will be the same:
(3,6). y=mx+b or 6=-4 × 3+b, or solving for b: b=6-(-4)(3). b=18.
(2,10). y=mx+b or 10=-4 × 2+b, or solving for b: b=10-(-4)(2). b=18.
Equation of the line: y=-4x+18
Answer:
and
Step-by-step explanation:
has parametric equations:
.
Let's solve these for x and y respectively.
can be solved for x by adding h on both sides:
.
can be solve for y by adding k on both sides:
.
We can verify this works by plugging these back in for x and y respectively.
Let's do that:
By a Pythagorean Identity.
which is what we had on the right hand side.
We have confirmed our parametric equations are correct.
Now here your h=0 while your k=1 and r=2.
So we are going to play with these parametric equations:
and
We want to travel clockwise so we need to put -t and instead of t.
If we were going counterclockwise it would be just the t.
and
Now cosine is even function while sine is an odd function so you could simplify this and say:
and
.
We want to find such that
when t=0.
Let's start with the first equation:
Divide both sides by 2:
We wanted to find for when
Cosine is an even function:
This happens when where n is an integer.
Let's do the second equation:
Subtract 2 on both sides:
Divide both sides by -2:
Recall we are trying to find what is when t=0:
Recall sine is an odd function:
Divide both sides by -1:
So this means we don't have to shift the cosine parametric equation at all because we can choose n=0 which means .
We also don't have to shift the sine parametric equation either since at n=0, we have .
So let's see what our equations look like now:
and
Let's verify these still work in our original equation:
It still works.
Now let's see if we are being moving around the circle once around for values of t between and
.
This first table will be the first half of the rotation.
t 0 pi/4 pi/2 3pi/4 pi
x 2 sqrt(2) 0 -sqrt(2) -2
y 1 -sqrt(2)+1 -1 -sqrt(2)+1 1
Ok this is the fist half of the rotation. Are we moving clockwise from (2,1)?
If we are moving clockwise around a circle with radius 2 and center (0,1) starting at (2,1) our x's should be decreasing and our y's should be decreasing at the beginning we should see a 4th of a circle from the point (x,y)=(2,1) and the point (x,y)=(0,-1).
Now after that 4th, the x's will still decrease until we make half a rotation but the y's will increase as you can see from point (x,y)=(0,-1) to (x,y)=(-2,1). We have now made half a rotation around the circle whose center is (0,1) and radius is 2.
Let's look at the other half of the circle:
t pi 5pi/4 3pi/2 7pi/4 2pi
x -2 -sqrt(2) 0 sqrt(2) 2
y 1 sqrt(2)+1 3 sqrt(2)+1 1
So now for the talk half going clockwise we should see the x's increase since we are moving right for them. The y's increase after the half rotation but decrease after the 3/4th rotation.
We also stopped where we ended at the point (2,1).
