Chapter 1, Part 2
1. Find the equation of a line that passes through (-4, 2) and (-6,6)
m = (6-2)/(-6 + 4) = -4/2 = -2; y = -2x + b
2 = -2(-4) + b, b = -6; y = -2x - 6
Check: 2 = -2(-4) – 6 = 8 – 6 OK; 6 = -2(-6) – 6 = 12 – 6 OK
2. The x intercept of a line is -4 and the y intercept 8; find the line.
y = mx + b; if x = 0, y = 8, so b = 8
If y = 0, x = -4 then m(-4) +b = -4m + 8 = 0, m = 2
y = 2x + 8
3. Find a line with slope ¼ the passes through the point (-2, -3)
y = mx + b = ¼ x + b; put in point, -3 = ¼ (-2) + b = -1/2 + b
b = -3 + ½ = -5/2 or y = ¼ x – 5/2
Check: -3 = ¼ (-2) – 5/2 = -1/2 – 5/2 = -6/2 = -3 OK
4. Find a line with slope -10 and y intercept 0
y = mx + b; m = -10, b = 0
y = -10x
5. Find a line perpendicular to the line x + y + 1 = 0 that passes through (1, 2)
y = -x – 1, slope is -1. Negative reciprocal is 1 for our new line
y = x + b, 2 = 1 + b, b = 1
y=x+1
6. Find a line that passes through the origin and the midpoint of the line segment joining
(-2, -3) and (6, -5)
Midpoint is (x1 + x2)/2 , (y1 + y2)/2 or (-2+6)/2, (-3 – 5)/2 which is (2, -4)
7. Find the line that passes through (2,4) and whose y intercept is twice its x intercept.
y = mx + b; know 4 = m(2) + b, where b is the y intercept. The x intercept is b/2 and
occurs when y = 0. So, we have 0 = m(b/2) + b. Divide by b, and we have
m/2 + 1 = 0, m = -2. 4 = (-2)(2) + b, b = 8, y = -2x + 8
Check: 4 = -2(2) + 8; OK. x intercept is 8/2 = 4; if y = 0, 0 = -2(4)+ 8 OK
8. Find the perimeter of the triangle with vertices (3,1), (7, 4), and (-2, 13)
Need distance formulas. Between 1st two points d= √16 + 9 = 5. Between 2nd two points,
d = √81 + 81 = 9√2, between the 1st and 3rd √25 + 144 = 13. Perimeter is 5 + 13 + 9√2
= 18 + 9√2
2𝑦−5
𝑦−1
9. Find the solution to the following equation: 4𝑦+1 = 2𝑦+5.
Multiply by denominator: 4y2 – 25 = 4y2 -3y - 1; 3y = 24, y = 8
8 causes no problem in the denominator, not extraneous
𝑥
−2
10. Solve: 5−𝑥 = 11−𝑥
Multiply by denominator: -x2 + 11x = -10 + 2x, x2 -9x – 10 = 0
Factor (x-10)(x+1)= 0, x = -1, x = 10; neither is extraneous
11. Find the solution to |x- ½ | < 1
x-1/2 < 1 and x – ½ > -1; the first gives x < 3/2 and the second x > -1/2
-1/2 < x < 3/2
12. What symmetries does y = x3 – 5x + 1 have?
Subst –x get y = -x3 + 5x + 1
subst –y, -y = x3 – 5x + 1
None
13. What symmetries does y = 2x – 2 –x have?
subst –x for x we have 2-x – 2x = -y
symmetry about origin