12 multiple-choice questions, progressively harder.
The line shown passes through the two marked points. Which equation represents it?
Solution
Correct answer: A
The line passes through (0,2)(0, 2)(0,2) on the yyy-axis and (3,0)(3, 0)(3,0) on the xxx-axis. Its slope is
m=0−23−0=−23,y−2=−23(x−0),m = \frac{0 - 2}{3 - 0} = -\tfrac{2}{3}, \qquad y - 2 = -\tfrac{2}{3}(x - 0),m=3−00−2=−32,y−2=−32(x−0),
and clearing the fraction gives 2x+3y=62x + 3y = 62x+3y=6. Check both marked points: 2(0)+3(2)=62(0) + 3(2) = 62(0)+3(2)=6 and 2(3)+3(0)=62(3) + 3(0) = 62(3)+3(0)=6.
What is the slope of the line y+8=−37(x−2)y + 8 = -\tfrac{3}{7}(x - 2)y+8=−73(x−2)?
Correct answer: D
The slope is the number multiplying the parenthesis.
y+8=−37(x−2) ⇒ m=−37y + 8 = -\tfrac{3}{7}(x - 2) \;\Rightarrow\; m = -\tfrac{3}{7}y+8=−73(x−2)⇒m=−73
So the slope is −37-\tfrac{3}{7}−73.
The line through (2,3)(2, 3)(2,3) with slope 444 also passes through (4,k)(4, k)(4,k). What is kkk?
Correct answer: C
Write point-slope form, then substitute x=4x = 4x=4.
y−3=4(x−2),k−3=4(4−2)=8,k=11y - 3 = 4(x - 2), \qquad k - 3 = 4(4 - 2) = 8, \qquad k = 11y−3=4(x−2),k−3=4(4−2)=8,k=11
So k=11k = 11k=11.
Which equation represents the line through (0,−6)(0, -6)(0,−6) with slope −3-3−3?
Correct answer: B
Substitute the point (0,−6)(0, -6)(0,−6) and slope −3-3−3, with x1=0x_1 = 0x1=0 and y1=−6y_1 = -6y1=−6.
y−(−6)=−3(x−0) ⇒ y+6=−3xy - (-6) = -3(x - 0) \;\Rightarrow\; y + 6 = -3xy−(−6)=−3(x−0)⇒y+6=−3x
So the equation is y+6=−3xy + 6 = -3xy+6=−3x.
Write the line through (1,−2)(1, -2)(1,−2) and (4,4)(4, 4)(4,4) in standard form Ax+By=CAx + By = CAx+By=C with A≥0A \ge 0A≥0.
Find the slope, write point-slope form, then convert.
m=4−(−2)4−1=63=2,y+2=2(x−1)m = \frac{4 - (-2)}{4 - 1} = \frac{6}{3} = 2, \qquad y + 2 = 2(x - 1)m=4−14−(−2)=36=2,y+2=2(x−1)
Distributing and collecting gives 2x−y=42x - y = 42x−y=4. Check (4,4)(4, 4)(4,4): 2(4)−4=42(4) - 4 = 42(4)−4=4.
Write the line through (−3,−1)(-3, -1)(−3,−1) and (1,7)(1, 7)(1,7) in standard form Ax+By=CAx + By = CAx+By=C with A≥0A \ge 0A≥0.
m=7−(−1)1−(−3)=84=2,y−7=2(x−1)m = \frac{7 - (-1)}{1 - (-3)} = \frac{8}{4} = 2, \qquad y - 7 = 2(x - 1)m=1−(−3)7−(−1)=48=2,y−7=2(x−1)
Collecting terms gives 2x−y=−52x - y = -52x−y=−5. Check (−3,−1)(-3, -1)(−3,−1): 2(−3)−(−1)=−52(-3) - (-1) = -52(−3)−(−1)=−5.
Which point does NOT lie on the line y−1=3(x−2)y - 1 = 3(x - 2)y−1=3(x−2)?
A point lies on the line when it satisfies the equation. Test (0,1)(0, 1)(0,1).
1−1=3(0−2) ⇒ 0=−6 (false)1 - 1 = 3(0 - 2) \;\Rightarrow\; 0 = -6 \;\text{(false)}1−1=3(0−2)⇒0=−6(false)
So (0,1)(0, 1)(0,1) is not on the line. The other three points all satisfy the equation.
Which of these is an equation of the line through (1,−1)(1, -1)(1,−1) and (4,5)(4, 5)(4,5)?
Find the slope, then apply point-slope form with the point (4,5)(4, 5)(4,5).
m=5−(−1)4−1=63=2,y−5=2(x−4)m = \frac{5 - (-1)}{4 - 1} = \frac{6}{3} = 2, \qquad y - 5 = 2(x - 4)m=4−15−(−1)=36=2,y−5=2(x−4)
The distractors use the wrong sign or the wrong point, so each is a different line.
Write the line through (4,1)(4, 1)(4,1) with slope −32-\tfrac{3}{2}−23 in standard form Ax+By=CAx + By = CAx+By=C with A≥0A \ge 0A≥0.
Start from point-slope form and clear the fraction.
y−1=−32(x−4),2(y−1)=−3(x−4),2y−2=−3x+12y - 1 = -\tfrac{3}{2}(x - 4), \qquad 2(y - 1) = -3(x - 4), \qquad 2y - 2 = -3x + 12y−1=−23(x−4),2(y−1)=−3(x−4),2y−2=−3x+12
Collecting terms gives 3x+2y=143x + 2y = 143x+2y=14. Check (4,1)(4, 1)(4,1): 3(4)+2(1)=143(4) + 2(1) = 143(4)+2(1)=14.
Write the line through (−1,−2)(-1, -2)(−1,−2) and (3,6)(3, 6)(3,6) in standard form Ax+By=CAx + By = CAx+By=C with A≥0A \ge 0A≥0.
Find the slope, write point-slope form with (3,6)(3, 6)(3,6), then convert.
m=6−(−2)3−(−1)=84=2,y−6=2(x−3)m = \frac{6 - (-2)}{3 - (-1)} = \frac{8}{4} = 2, \qquad y - 6 = 2(x - 3)m=3−(−1)6−(−2)=48=2,y−6=2(x−3)
Collecting terms gives 2x−y=02x - y = 02x−y=0. Both points check: 2(−1)−(−2)=02(-1) - (-2) = 02(−1)−(−2)=0 and 2(3)−6=02(3) - 6 = 02(3)−6=0.
The equation y+2=−13(x−6)y + 2 = -\tfrac{1}{3}(x - 6)y+2=−31(x−6) is in point-slope form. Which point and slope does it use?
Match to y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1=m(x−x1). The term y+2y + 2y+2 is y−(−2)y - (-2)y−(−2), so y1=−2y_1 = -2y1=−2; the term x−6x - 6x−6 gives x1=6x_1 = 6x1=6; and the multiplier is m=−13m = -\tfrac{1}{3}m=−31.
y+2=−13(x−6) ⇒ (x1,y1)=(6,−2), m=−13y + 2 = -\tfrac{1}{3}(x - 6) \;\Rightarrow\; (x_1, y_1) = (6, -2), \; m = -\tfrac{1}{3}y+2=−31(x−6)⇒(x1,y1)=(6,−2),m=−31
So the point is (6,−2)(6, -2)(6,−2) and the slope is −13-\tfrac{1}{3}−31.
Which equation has BOTH (−1,−4)(-1, -4)(−1,−4) and (2,5)(2, 5)(2,5) as solutions?
Find the slope through the two points, then convert to standard form.
m=5−(−4)2−(−1)=93=3,y−5=3(x−2)m = \frac{5 - (-4)}{2 - (-1)} = \frac{9}{3} = 3, \qquad y - 5 = 3(x - 2)m=2−(−1)5−(−4)=39=3,y−5=3(x−2)
Collecting terms gives 3x−y=13x - y = 13x−y=1. Both points check: 3(−1)−(−4)=13(-1) - (-4) = 13(−1)−(−4)=1 and 3(2)−5=13(2) - 5 = 13(2)−5=1.
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