Find the midpoint between the two points:
(–1, 3) and (5, 9)
(2, 6)
What is the center and radius of this circle:
(x – 3)2 + (y – 1)2 = 36
(3, 1) is the center.
6 is the radius.
Is 3 a solution of the equation:
2x + 4 = 10?
Yes! 2(3) + 4 = 6 + 4 = 10!
What is the slope between (–1, 2) and (2, 5)?
m = (y2 – y1)/(x2 – x1) = (5 – 2)/(2 – (-1)) = 3/2+1 = 3/3 = 1!
2x – 1 ≤ 4x + 3
–4 ≤ 2x, so x ≥ –2
Find the distance between the two points:
(–2, –3) and (1, 1)
5
Find the center and radius of the circle:
(x – 2)2 + (y + 6)2 = 25
(2, –6) is the center.
5 is the radius.
Is x = –4 a solution of the equation?
–3(x – 10) = x + 14
Nope!
–3(x – 10) = x + 14 becomes –3x + 30 = x + 14
which becomes 16 = 4x, which becomes x = 4,
not –4.
Write a slope-intercept equation with a point of
(–2, 5) and a slope of 3.
y = 3x + b so 5 = 3(–2) + b so 5 = –6 + b so b is 11, there fore the equation is y = 3x + 11.
4(1 – x) + 5(1 + x) > 3x – 1
4 – 4x + 5 + 5x > 3x – 1
x + 9 > 3x – 1
10 > 2x
5 > x
Explain how the Pythagorean theorem and the Distance formula are related.
a2 + b2 = c2 if a and b are lines that span two points (x1, y1) and (x2, y2), then it can be written as
(x2 – x1)2 + (y2 – y1)2 = c2, so then to solve for c, the distance between a and b, you take the sq. root:
c = √[(x2 – x1)2 + (y2 – y1)2], which is the distance formula.
Write a circle equation with a center of (0, 0) and a radius of 3.
x2 + y2 = 9
Is x = 0 a solution of this equation?
√(1 – x2) + 2 = 3
Yes! 02 is 0, so 1-0 is 1 and √1 is 1. Therefore, 1 + 2 = 3 is true!
Write a point–slope equation with the points (4, 2) and (–3, 1).
the slope is 1–2/–3–4 = –1/–7 = 1/7. Then, I can plug in either point. so both of these are correct answers:
y – 2 = (1/7)(x – 4)
y – 1 = (1/7)(x + 3)
0 ≤ –2x + 6 < 8
0 ≤ –2x + 6 < 8
–6 ≤ –2x < 2
3 ≥ x > –1
Do these points make a parallelogram?
(–7, –1) (–2, 4) (3, –1) and (–2, –6)
Yes, in fact it is a square.
The midpoint between (–2, 4) and (–2, –6) is
(–2, –1) and
the midpoint between (–7, –1) and (3, –1) is
(–2, –1) also, so it must be a parallelogram.
Write a circle equation with a center of (1, 4) and a radius of √5?
(x – 1)2 + (y – 4)2 = 5
Solve the equation: 2x – 4 = (4x – 5)/3. Is x = 3 a solution?
Nope. 2(3) – 4 = (4(3) – 5)/3 becomes
6 – 4 = (12 – 5)/3 so, 2 = 7/3 is not true.
Write the slope-intercept form of the equation with the points (0, 0) and (3, –6).
Once you do, convert that into standard form.
–6–0/3–0 = –6/3 = –2 and (0, 0) is the y-intercept, so y = –2x + 0 or y = –2x
Now, to bring all of the variables to one side, I will add 2x to both sides and I get y + 2x = 0.
1 > (3y – 1)/4 > –1
4 > 3y – 1 > –4
5 > 3y > –3
5/3 > y > –1
What kind of triangle do these points make?
(1, 3), (4, 7), (8, 4)
Isosceles
Write a circle equation with a center of (–3, –2) and a radius of √121.
(x + 3)2 + (y + 2)2 = 11
Plug in the coordinate point (2, 4). Is it a solution point?
4(2x + 2) + 3y = 36
Yes! 4(2x + 2) + 3y = 36 becomes
8x + 16 + 3y = 36 and 8(2) + 16 + 3(4) = 36 becomes 16 + 16 + 12 = 36 is true!
Find the value of y so that the given points become the given slope.(–3, –5) and (4, y) where the slope is m = 3
(y – (–5))/(4 – (–3)) = (y + 5)/7
21/7 is 3, so y + 5 = 21, therefore y has to be 16.
(x – 5)/4 + (3 – 2x)/3 < –2
(x – 5)/4 + (3 – 2x)/3 < –2
(3x – 15)/12 + (12 – 8x)/12 < –2
(–5x – 3)/12 < –2
–5x – 3 < –24 becomes –5x < –21 becomes x >21/5 or x > 4 and 1/5