It is a bit tedious to write 6 equations, but it is a straightforward process to substitute the given point values into the form provided.
For segment ab. (x1, y1) = (1, 1); (x2, y2) = (3, 4).
... x = 1 + t(3-1)
... y = 1 + t(4-1)
ab = {x=1+2t, y=1+3t}
For segment bc. (x1, y1) = (3, 4); (x2, y2) = (1, 7).
... x = 3 + t(1-3)
... y = 4 + t(7-4)
bc = {x=3-2t, y=4+3t}
For segment ca. (x1, y1) = (1, 7); (x2, y2) = (1, 1).
... x = 1 + t(1-1)
... y = 7 + t(1-7)
ca = {x=1, y=7-6t}
Answer:
x=12/23
Step-by-step explanation:
Isolate the variable by dividing each side by factors that don't contain the variable.
Answer:
its green so you got it right
and if thats not the case you are correct
Step-by-step explanation:
Answer:
2+2=4
Step-by-step explanation:
X² + 8x +y² - 2y -64 =0
We see that the equation has x² and y² . Also we see that coefficients in front of x² and y² are equal. So this is an equation of the circle.
(x² + 8x) +(y² - 2y) -64 =0
(x² + 8x) +(y² - 2y) = 64
We need to complete square for x and y groups, that means it should be written in form (a+b)² or (a-b)².
Expressions in parenthesis we will write as a²+/-2ab+b², to write it after as (a+/-b)², because a²+/-2ab+b² = (a+/-b)²
(x² + 2*4x) +(y² - 2*1y) = 64
(x² + 2*4x+4²) +(y² - 2*1y+1²) = 64+4²+1²
(x+4)² + (y-1)²= 81 Sometimes this is called a standard form of the circle.
(x+4)² + (y-1)²= 9² Sometimes it is required to write like this.
And if you are studying circles,ellipses and hyperbolas, the standard form should look like
