12 multiple-choice questions, progressively harder.
Which equation represents the line through (−4,2)(-4, 2)(−4,2) and (2,−1)(2, -1)(2,−1)?
Solution
Correct answer: D
Find the slope, then use point-slope form with the point (2,−1)(2, -1)(2,−1).
m=−1−22−(−4)=−36=−12m = \frac{-1 - 2}{2 - (-4)} = \frac{-3}{6} = -\tfrac{1}{2}m=2−(−4)−1−2=6−3=−21
With slope −12-\tfrac{1}{2}−21 and (2,−1)(2, -1)(2,−1): y+1=−12(x−2)y + 1 = -\tfrac{1}{2}(x - 2)y+1=−21(x−2). Each distractor changes a sign and describes a different line.
Write the line through (3,−2)(3, -2)(3,−2) with slope 25\tfrac{2}{5}52 in standard form Ax+By=CAx + By = CAx+By=C with A≥0A \ge 0A≥0.
Correct answer: B
Start from point-slope form and clear the fraction.
y+2=25(x−3),5(y+2)=2(x−3),5y+10=2x−6y + 2 = \tfrac{2}{5}(x - 3), \qquad 5(y + 2) = 2(x - 3), \qquad 5y + 10 = 2x - 6y+2=52(x−3),5(y+2)=2(x−3),5y+10=2x−6
Collecting terms gives 2x−5y=162x - 5y = 162x−5y=16. Check (3,−2)(3, -2)(3,−2): 2(3)−5(−2)=6+10=162(3) - 5(-2) = 6 + 10 = 162(3)−5(−2)=6+10=16.
Write the line through (−2,5)(-2, 5)(−2,5) with slope −34-\tfrac{3}{4}−43 in standard form Ax+By=CAx + By = CAx+By=C with A≥0A \ge 0A≥0.
y−5=−34(x+2),4(y−5)=−3(x+2),4y−20=−3x−6y - 5 = -\tfrac{3}{4}(x + 2), \qquad 4(y - 5) = -3(x + 2), \qquad 4y - 20 = -3x - 6y−5=−43(x+2),4(y−5)=−3(x+2),4y−20=−3x−6
Collecting terms gives 3x+4y=143x + 4y = 143x+4y=14. Check (−2,5)(-2, 5)(−2,5): 3(−2)+4(5)=−6+20=143(-2) + 4(5) = -6 + 20 = 143(−2)+4(5)=−6+20=14.
Which equation represents the line through (4,−1)(4, -1)(4,−1) and (4,6)(4, 6)(4,6)?
Correct answer: A
The two points share the same xxx-coordinate 444, so the run is 000 and the slope is undefined: the line is vertical.
x=4x = 4x=4
Point-slope form cannot be used here, because there is no slope to substitute.
The line shown passes through the two marked points. Which equation represents it?
Correct answer: C
The line passes through (0,−1)(0, -1)(0,−1) on the yyy-axis and (2,0)(2, 0)(2,0) on the xxx-axis. Its slope is
m=0−(−1)2−0=12,y+1=12(x−0),m = \frac{0 - (-1)}{2 - 0} = \tfrac{1}{2}, \qquad y + 1 = \tfrac{1}{2}(x - 0),m=2−00−(−1)=21,y+1=21(x−0),
and clearing the fraction gives x−2y=2x - 2y = 2x−2y=2. Check both marked points: 0−2(−1)=20 - 2(-1) = 20−2(−1)=2 and 2−2(0)=22 - 2(0) = 22−2(0)=2.
Which of these is an equation of the line through (2,1)(2, 1)(2,1) and (5,7)(5, 7)(5,7)?
Find the slope, then apply point-slope form with the point (5,7)(5, 7)(5,7).
m=7−15−2=63=2,y−7=2(x−5)m = \frac{7 - 1}{5 - 2} = \frac{6}{3} = 2, \qquad y - 7 = 2(x - 5)m=5−27−1=36=2,y−7=2(x−5)
The distractors use the wrong slope or the wrong sign, so each describes a different line.
The point (k,−3)(k, -3)(k,−3) lies on the line 2x−3y=192x - 3y = 192x−3y=19. What is kkk?
Substitute y=−3y = -3y=−3 and solve for x=kx = kx=k.
2k−3(−3)=19,2k+9=19,2k=10,k=52k - 3(-3) = 19, \qquad 2k + 9 = 19, \qquad 2k = 10, \qquad k = 52k−3(−3)=19,2k+9=19,2k=10,k=5
So k=5k = 5k=5.
A line passes through (2,7)(2, 7)(2,7) and (5,7)(5, 7)(5,7). What is its equation?
The two points share the same height 777, so the line is horizontal.
m=7−75−2=0 ⇒ y=7m = \frac{7 - 7}{5 - 2} = 0 \;\Rightarrow\; y = 7m=5−27−7=0⇒y=7
A slope of 000 gives the equation y=7y = 7y=7.
Write the line through (1,4)(1, 4)(1,4) and (3,1)(3, 1)(3,1) in standard form Ax+By=CAx + By = CAx+By=C with A≥0A \ge 0A≥0.
Find the slope, write point-slope form, and clear the fraction.
m=1−43−1=−32,2(y−4)=−3(x−1),2y−8=−3x+3m = \frac{1 - 4}{3 - 1} = -\tfrac{3}{2}, \qquad 2(y - 4) = -3(x - 1), \qquad 2y - 8 = -3x + 3m=3−11−4=−23,2(y−4)=−3(x−1),2y−8=−3x+3
Collecting terms gives 3x+2y=113x + 2y = 113x+2y=11. Check (3,1)(3, 1)(3,1): 3(3)+2(1)=113(3) + 2(1) = 113(3)+2(1)=11.
Which line has slope 333 and passes through (2,−1)(2, -1)(2,−1)?
Point-slope form with slope 333 and the point (2,−1)(2, -1)(2,−1) gives, since y1=−1y_1 = -1y1=−1,
y−(−1)=3(x−2) ⇒ y+1=3(x−2).y - (-1) = 3(x - 2) \;\Rightarrow\; y + 1 = 3(x - 2).y−(−1)=3(x−2)⇒y+1=3(x−2).
The distractors change the sign of a coordinate or the slope, so each is a different line.
Write the line through (−3,6)(-3, 6)(−3,6) and (3,−2)(3, -2)(3,−2) 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,−2)(3, -2)(3,−2), and clear the fraction.
m=−2−63−(−3)=−86=−43,3(y+2)=−4(x−3)m = \frac{-2 - 6}{3 - (-3)} = \frac{-8}{6} = -\tfrac{4}{3}, \qquad 3(y + 2) = -4(x - 3)m=3−(−3)−2−6=6−8=−34,3(y+2)=−4(x−3)
Collecting terms gives 4x+3y=64x + 3y = 64x+3y=6. Check (−3,6)(-3, 6)(−3,6): 4(−3)+3(6)=−12+18=64(-3) + 3(6) = -12 + 18 = 64(−3)+3(6)=−12+18=6.
Which equation represents the line through (0,0)(0, 0)(0,0) with slope −52-\tfrac{5}{2}−25?
Point-slope form from (0,0)(0, 0)(0,0) gives y=−52xy = -\tfrac{5}{2}xy=−25x. Clear the fraction and collect terms.
2y=−5x ⇒ 5x+2y=02y = -5x \;\Rightarrow\; 5x + 2y = 02y=−5x⇒5x+2y=0
The line passes through the origin, so C=0C = 0C=0; solving for yyy returns the slope −52-\tfrac{5}{2}−25.
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