IB Mathematics: Analysis and Approaches Quiz: Linear Functions
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Linear FunctionsQuestion 1 of 20

Line L1L_1 has equation y=mx+2y = mx + 2. Line L2L_2 has equation y=2x+5y = 2x + 5. The lines intersect at a point with an x-coordinate of -3. Find the equation of the line L3L_3 which is perpendicular to L1L_1 and passes through the point of intersection.

y=−x−4y = -x - 4
y=−12x−52y = -\frac{1}{2}x - \frac{5}{2}
y=x−2y = x - 2
y=2x+5y = 2x + 5
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Linear Functions

Practice Linear Functions in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Linear Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Line L1L_1 has equation y=mx+2y = mx + 2. Line L2L_2 has equation y=2x+5y = 2x + 5. The lines intersect at a point with an x-coordinate of -3. Find the equation of the line L3L_3 which is perpendicular to L1L_1 and passes through the point of intersection.

  1. y=−x−4y = -x - 4 (correct answer)
  2. y=−12x−52y = -\frac{1}{2}x - \frac{5}{2}
  3. y=x−2y = x - 2
  4. y=2x+5y = 2x + 5
Explanation:
  1. Find the point of intersection. We are given x=−3x = -3. Substitute this into the equation for L2L_2 to find the y-coordinate: y=2(−3)+5=−6+5=−1y = 2(-3) + 5 = -6 + 5 = -1. The intersection point is (−3,−1)(-3, -1).\n2. Find the gradient mm of L1L_1 by substituting the intersection point into its equation: −1=m(−3)+2⇒−3=−3m⇒m=1-1 = m(-3) + 2 \Rightarrow -3 = -3m \Rightarrow m = 1.\n3. The gradient of L1L_1 is 1. The gradient of line L3L_3, which is perpendicular to L1L_1, is the negative reciprocal: m3=−11=−1m_3 = -\frac{1}{1} = -1.\n4. Use the point-gradient form to find the equation of L3L_3, which passes through (−3,−1)(-3, -1) with a gradient of -1: y−(−1)=−1(x−(−3))⇒y+1=−(x+3)⇒y+1=−x−3⇒y=−x−4y - (-1) = -1(x - (-3)) \Rightarrow y + 1 = -(x + 3) \Rightarrow y + 1 = -x - 3 \Rightarrow y = -x - 4.

Question 2

The points A(−1,2)A(-1, 2), B(3,k)B(3, k), and C(5,1)C(5, 1) are collinear. Find the value of kk.

  1. −22-22
  2. 43\frac{4}{3} (correct answer)
  3. 53\frac{5}{3}
  4. 83\frac{8}{3}
Explanation: For three points to be collinear, the gradient between any two pairs of points must be the same. Let's equate the gradient of AB and the gradient of AC.\n1. Gradient of AB: mAB=k−23−(−1)=k−24m_{AB} = \frac{k - 2}{3 - (-1)} = \frac{k-2}{4}.\n2. Gradient of AC: mAC=1−25−(−1)=−16m_{AC} = \frac{1 - 2}{5 - (-1)} = \frac{-1}{6}.\n3. Set the gradients equal: k−24=−16\frac{k-2}{4} = -\frac{1}{6}.\n4. Solve for kk: 6(k−2)=−4⇒6k−12=−4⇒6k=8⇒k=86=436(k-2) = -4 \Rightarrow 6k - 12 = -4 \Rightarrow 6k = 8 \Rightarrow k = \frac{8}{6} = \frac{4}{3}.

Question 3

A line LL passes through the point A(2,5)A(2, 5) and has a gradient of −34-\frac{3}{4}. The line intersects the x-axis at P and the y-axis at Q. Find the length of the line segment PQ.

  1. 132\frac{13}{2}
  2. 263\frac{26}{3}
  3. 656\frac{65}{6} (correct answer)
  4. 1313
Explanation:
  1. Find the equation of the line LL using the point-gradient form: y−5=−34(x−2)y - 5 = -\frac{3}{4}(x - 2). 4(y−5)=−3(x−2)⇒4y−20=−3x+6⇒3x+4y=264(y-5) = -3(x-2) \Rightarrow 4y - 20 = -3x + 6 \Rightarrow 3x + 4y = 26.\n2. Find the x-intercept (P) by setting y=0y=0: 3x=26⇒x=2633x = 26 \Rightarrow x = \frac{26}{3}. So P=(263,0)P = (\frac{26}{3}, 0).\n3. Find the y-intercept (Q) by setting x=0x=0: 4y=26⇒y=264=1324y = 26 \Rightarrow y = \frac{26}{4} = \frac{13}{2}. So Q=(0,132)Q = (0, \frac{13}{2}).\n4. Find the distance between P and Q using the distance formula: d=(263−0)2+(0−132)2=(263)2+(−132)2d = \sqrt{(\frac{26}{3}-0)^2 + (0-\frac{13}{2})^2} = \sqrt{(\frac{26}{3})^2 + (-\frac{13}{2})^2}.d=6769+1694=2704+152136=422536=656d = \sqrt{\frac{676}{9} + \frac{169}{4}} = \sqrt{\frac{2704 + 1521}{36}} = \sqrt{\frac{4225}{36}} = \frac{65}{6}.

Question 4

Line L1L_1 has equation y=mx+2y = mx + 2. Line L2L_2 has equation y=2x+5y = 2x + 5. The lines intersect at a point with an x-coordinate of -3. Find the equation of the line L3L_3 which is perpendicular to L1L_1 and passes through the point of intersection.

  1. y=−x−4y = -x - 4 (correct answer)
  2. y=−12x−52y = -\frac{1}{2}x - \frac{5}{2}
  3. y=x−2y = x - 2
  4. y=2x+5y = 2x + 5
Explanation:
  1. Find the point of intersection. We are given x=−3x = -3. Substitute this into the equation for L2L_2 to find the y-coordinate: y=2(−3)+5=−6+5=−1y = 2(-3) + 5 = -6 + 5 = -1. The intersection point is (−3,−1)(-3, -1).\n2. Find the gradient mm of L1L_1 by substituting the intersection point into its equation: −1=m(−3)+2⇒−3=−3m⇒m=1-1 = m(-3) + 2 \Rightarrow -3 = -3m \Rightarrow m = 1.\n3. The gradient of L1L_1 is 1. The gradient of line L3L_3, which is perpendicular to L1L_1, is the negative reciprocal: m3=−11=−1m_3 = -\frac{1}{1} = -1.\n4. Use the point-gradient form to find the equation of L3L_3, which passes through (−3,−1)(-3, -1) with a gradient of -1: y−(−1)=−1(x−(−3))⇒y+1=−(x+3)⇒y+1=−x−3⇒y=−x−4y - (-1) = -1(x - (-3)) \Rightarrow y + 1 = -(x + 3) \Rightarrow y + 1 = -x - 3 \Rightarrow y = -x - 4.

Question 5

The lines L1:y=(k2−3)x+2L_1: y = (k^2-3)x + 2 and L2:y=k2x−1L_2: y = \frac{k}{2}x - 1 are perpendicular. What is a possible value of kk?

  1. −1-1
  2. 1 (correct answer)
  3. 2
  4. 3
Explanation: For two lines to be perpendicular, the product of their gradients must be -1.\n1. The gradient of L1L_1 is m1=k2−3m_1 = k^2-3.\n2. The gradient of L2L_2 is m2=k2m_2 = \frac{k}{2}.\n3. Set the product of the gradients to -1: m1m2=(k2−3)(k2)=−1m_1 m_2 = (k^2-3)(\frac{k}{2}) = -1.\n4. Simplify the equation: k(k2−3)=−2⇒k3−3k=−2⇒k3−3k+2=0k(k^2-3) = -2 \Rightarrow k^3 - 3k = -2 \Rightarrow k^3 - 3k + 2 = 0.\n5. We can test the given options by substituting them into the equation.\nFor k=1k=1: (1)3−3(1)+2=1−3+2=0(1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0. This is a valid solution.\n(For completeness, the other solutions can be found by factoring (k−1)(k-1) from the polynomial, yielding (k−1)(k2+k−2)=0(k-1)(k^2+k-2)=0, which further factors to (k−1)(k+2)(k−1)=0(k-1)(k+2)(k-1)=0. The solutions are k=1k=1 (repeated root) and k=−2k=-2.)

Question 6

Three points are given as A(1,5)A(1, 5), B(−2,−1)B(-2, -1), and C(x,3)C(x, 3). Find the value of xx such that the line segment BCBC is perpendicular to the line segment ABAB.

  1. −10-10 (correct answer)
  2. −4-4
  3. 0
  4. 6
Explanation:
  1. Find the gradient of AB: mAB=5−(−1)1−(−2)=63=2m_{AB} = \frac{5 - (-1)}{1 - (-2)} = \frac{6}{3} = 2.\n2. For BC to be perpendicular to AB, the gradient of BC must be the negative reciprocal of the gradient of AB. So, mBC=−12m_{BC} = -\frac{1}{2}.\n3. Calculate the gradient of BC using the coordinates of B and C: mBC=3−(−1)x−(−2)=4x+2m_{BC} = \frac{3 - (-1)}{x - (-2)} = \frac{4}{x+2}.\n4. Set the gradients equal and solve for xx: 4x+2=−12⇒8=−(x+2)⇒8=−x−2⇒x=−10\frac{4}{x+2} = -\frac{1}{2} \Rightarrow 8 = -(x+2) \Rightarrow 8 = -x - 2 \Rightarrow x = -10.

Question 7

The vertices of a quadrilateral ABCD are A(1,6)A(1, 6), B(2,9)B(2, 9), C(5,8)C(5, 8), and D(4,5)D(4, 5). Which of the following statements most accurately describes the quadrilateral?

  1. A parallelogram that is not a rectangle or rhombus.
  2. A rhombus that is not a square.
  3. A rectangle that is not a square.
  4. A square. (correct answer)
Explanation: To classify the quadrilateral, we need to examine the gradients and lengths of its sides.\n1. Gradients:\n mAB=9−62−1=3m_{AB} = \frac{9-6}{2-1} = 3\n mBC=8−95−2=−13m_{BC} = \frac{8-9}{5-2} = -\frac{1}{3}\n mCD=5−84−5=−3−1=3m_{CD} = \frac{5-8}{4-5} = \frac{-3}{-1} = 3\n mDA=6−51−4=−13m_{DA} = \frac{6-5}{1-4} = -\frac{1}{3}\nSince mAB=mCDm_{AB} = m_{CD} and mBC=mDAm_{BC} = m_{DA}, opposite sides are parallel, so it is a parallelogram. Since mAB×mBC=3×(−13)=−1m_{AB} \times m_{BC} = 3 \times (-\frac{1}{3}) = -1, adjacent sides are perpendicular, so it is a rectangle.\n2. Lengths:\n dAB2=(2−1)2+(9−6)2=12+32=10d_{AB}^2 = (2-1)^2 + (9-6)^2 = 1^2 + 3^2 = 10\n dBC2=(5−2)2+(8−9)2=32+(−1)2=10d_{BC}^2 = (5-2)^2 + (8-9)^2 = 3^2 + (-1)^2 = 10\nSince adjacent sides AB and BC have equal length, the rectangle is a square. All sides have length 10\sqrt{10}.

Question 8

The line LL has equation (k−2)x+(3−k)y+5=0(k-2)x + (3-k)y + 5 = 0. For what value of kk is the line LL parallel to the line with equation y=2x+1y = 2x + 1?

  1. 1
  2. 73\frac{7}{3}
  3. 83\frac{8}{3}
  4. 4 (correct answer)
Explanation:
  1. Find the gradient of line LL in terms of kk. Rearrange the equation to the form y=mx+cy = mx + c:\n(3−k)y=−(k−2)x−5⇒y=−(k−2)3−kx−53−k(3-k)y = -(k-2)x - 5 \Rightarrow y = \frac{-(k-2)}{3-k}x - \frac{5}{3-k}.\n2. The gradient is m=−(k−2)3−k=2−k3−km = \frac{-(k-2)}{3-k} = \frac{2-k}{3-k}.\n3. For line LL to be parallel to y=2x+1y=2x+1, their gradients must be equal. The gradient of y=2x+1y=2x+1 is 2.\n4. Set the gradients equal and solve for kk:\n2−k3−k=2⇒2−k=2(3−k)⇒2−k=6−2k⇒k=4\frac{2-k}{3-k} = 2 \Rightarrow 2-k = 2(3-k) \Rightarrow 2-k = 6-2k \Rightarrow k = 4.

Question 9

The line L1L_1 passes through (1,1)(1, 1) and (3,5)(3, 5). The line L2L_2 has equation x+y=5x + y = 5. The lines intersect at point P. Find the sum of the coordinates of P.

  1. 5 (correct answer)
  2. 6
  3. 7
  4. 8
Explanation:
  1. First, find the equation of line L1L_1. The gradient is m=5−13−1=42=2m = \frac{5-1}{3-1} = \frac{4}{2} = 2.\n2. Using the point-gradient form with point (1,1)(1,1): y−1=2(x−1)⇒y=2x−2+1⇒y=2x−1y - 1 = 2(x - 1) \Rightarrow y = 2x - 2 + 1 \Rightarrow y = 2x - 1.\n3. To find the point of intersection P, solve the system of simultaneous equations:\n y=2x−1y = 2x - 1 (from L1L_1)\n x+y=5x + y = 5 (from L2L_2)\n4. Substitute the expression for yy from the first equation into the second: x+(2x−1)=5⇒3x−1=5⇒3x=6⇒x=2x + (2x - 1) = 5 \Rightarrow 3x - 1 = 5 \Rightarrow 3x = 6 \Rightarrow x = 2.\n5. Substitute x=2x = 2 back into the equation for L1L_1 to find yy: y=2(2)−1=4−1=3y = 2(2) - 1 = 4 - 1 = 3.\n6. The point of intersection is P(2,3)P(2, 3).\n7. The sum of the coordinates of P is 2+3=52 + 3 = 5.

Question 10

A line passes through the points A(1,5)A(1, 5) and B(3,1)B(3, 1). Find the equation of the perpendicular bisector of the line segment AB.

  1. y=12x+92y = \frac{1}{2}x + \frac{9}{2}
  2. y=−2x+7y = -2x + 7
  3. y=12x+2y = \frac{1}{2}x + 2 (correct answer)
  4. y=2x−1y = 2x - 1
Explanation: When you encounter a perpendicular bisector problem, you need to find a line that passes through the midpoint of the given segment and is perpendicular to the original line. This combines coordinate geometry concepts of midpoint, slope, and perpendicular lines. First, find the midpoint of segment AB. Using the midpoint formula with A(1, 5) and B(3, 1): M=(1+32,5+12)=(2,3)M = \left(\frac{1+3}{2}, \frac{5+1}{2}\right) = (2, 3) Next, determine the slope of line AB: mAB=1−53−1=−42=−2m_{AB} = \frac{1-5}{3-1} = \frac{-4}{2} = -2 Since perpendicular lines have slopes that are negative reciprocals, the perpendicular bisector has slope: m⊥=−1−2=12m_{\perp} = -\frac{1}{-2} = \frac{1}{2} Using point-slope form with the midpoint (2, 3) and slope 12\frac{1}{2}: y−3=12(x−2)y - 3 = \frac{1}{2}(x - 2) y=12x−1+3=12x+2y = \frac{1}{2}x - 1 + 3 = \frac{1}{2}x + 2 This confirms answer C is correct. Answer A has the correct slope but wrong y-intercept (92\frac{9}{2} instead of 2), suggesting an error in applying the point-slope formula. Answer B gives the slope of the original line (-2) rather than its perpendicular, showing confusion about perpendicular relationships. Answer D uses slope 2, which is the negative of the original slope but not the negative reciprocal. Remember: perpendicular bisectors require both finding the midpoint AND using the negative reciprocal of the original slope. Double-check your arithmetic when substituting into point-slope form.

Question 11

Find the equation of the line that passes through the point P(−1,6)P(-1, 6) and the point of intersection of the lines L1:x−2y=1L_1: x - 2y = 1 and L2:3x+y=10L_2: 3x + y = 10.

  1. 5x−4y−11=05x - 4y - 11 = 0
  2. 4x+5y−17=04x + 5y - 17 = 0
  3. 5x+4y−19=05x + 4y - 19 = 0 (correct answer)
  4. x−2y+13=0x - 2y + 13 = 0
Explanation:
  1. Find the point of intersection by solving the system of equations. From L2L_2, y=10−3xy = 10 - 3x.\n2. Substitute this into L1L_1: x−2(10−3x)=1⇒x−20+6x=1⇒7x=21⇒x=3x - 2(10 - 3x) = 1 \Rightarrow x - 20 + 6x = 1 \Rightarrow 7x = 21 \Rightarrow x = 3.\n3. Find yy: y=10−3(3)=1y = 10 - 3(3) = 1. The intersection point is Q(3,1)Q(3, 1).\n4. Find the gradient of the line passing through P(−1,6)P(-1, 6) and Q(3,1)Q(3, 1): m=1−63−(−1)=−54m = \frac{1 - 6}{3 - (-1)} = \frac{-5}{4}.\n5. Use the point-gradient form with Q(3, 1): y−1=−54(x−3)⇒4(y−1)=−5(x−3)⇒4y−4=−5x+15⇒5x+4y−19=0y - 1 = -\frac{5}{4}(x - 3) \Rightarrow 4(y-1) = -5(x-3) \Rightarrow 4y - 4 = -5x + 15 \Rightarrow 5x + 4y - 19 = 0.

Question 12

The lines L1:y=(k2−3)x+2L_1: y = (k^2-3)x + 2 and L2:y=k2x−1L_2: y = \frac{k}{2}x - 1 are perpendicular. What is a possible value of kk?

  1. −1-1
  2. 1 (correct answer)
  3. 2
  4. 3
Explanation: For two lines to be perpendicular, the product of their gradients must be -1.\n1. The gradient of L1L_1 is m1=k2−3m_1 = k^2-3.\n2. The gradient of L2L_2 is m2=k2m_2 = \frac{k}{2}.\n3. Set the product of the gradients to -1: m1m2=(k2−3)(k2)=−1m_1 m_2 = (k^2-3)(\frac{k}{2}) = -1.\n4. Simplify the equation: k(k2−3)=−2⇒k3−3k=−2⇒k3−3k+2=0k(k^2-3) = -2 \Rightarrow k^3 - 3k = -2 \Rightarrow k^3 - 3k + 2 = 0.\n5. We can test the given options by substituting them into the equation.\nFor k=1k=1: (1)3−3(1)+2=1−3+2=0(1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0. This is a valid solution.\n(For completeness, the other solutions can be found by factoring (k−1)(k-1) from the polynomial, yielding (k−1)(k2+k−2)=0(k-1)(k^2+k-2)=0, which further factors to (k−1)(k+2)(k−1)=0(k-1)(k+2)(k-1)=0. The solutions are k=1k=1 (repeated root) and k=−2k=-2.)

Question 13

The line LL passes through the point (8,−1)(8, -1) and is perpendicular to the line with equation 2x−y=72x - y = 7. Find the area of the triangle enclosed by the line LL, the x-axis, and the y-axis.

  1. 4.5
  2. 9 (correct answer)
  3. 18
  4. 25
Explanation:
  1. The line 2x−y=72x - y = 7 can be written as y=2x−7y = 2x - 7. Its gradient is 2.\n2. The gradient of the perpendicular line LL is m=−12m = -\frac{1}{2}.\n3. The equation of LL passes through (8,−1)(8, -1): y−(−1)=−12(x−8)⇒y+1=−12x+4⇒y=−12x+3y - (-1) = -\frac{1}{2}(x - 8) \Rightarrow y + 1 = -\frac{1}{2}x + 4 \Rightarrow y = -\frac{1}{2}x + 3.\n4. Find the intercepts of LL:\n - x-intercept (set y=0y=0): 0=−12x+3⇒12x=3⇒x=60 = -\frac{1}{2}x + 3 \Rightarrow \frac{1}{2}x = 3 \Rightarrow x = 6. Point (6,0).\n - y-intercept (set x=0x=0): y=3y = 3. Point (0,3).\n5. The area of the triangle with vertices (0,0), (6,0), and (0,3) is 12×base×height=12×6×3=9\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 3 = 9.

Question 14

The points A(−1,2)A(-1, 2), B(3,k)B(3, k), and C(5,1)C(5, 1) are collinear. Find the value of kk.

  1. −22-22
  2. 43\frac{4}{3} (correct answer)
  3. 53\frac{5}{3}
  4. 83\frac{8}{3}
Explanation: For three points to be collinear, the gradient between any two pairs of points must be the same. Let's equate the gradient of AB and the gradient of AC.\n1. Gradient of AB: mAB=k−23−(−1)=k−24m_{AB} = \frac{k - 2}{3 - (-1)} = \frac{k-2}{4}.\n2. Gradient of AC: mAC=1−25−(−1)=−16m_{AC} = \frac{1 - 2}{5 - (-1)} = \frac{-1}{6}.\n3. Set the gradients equal: k−24=−16\frac{k-2}{4} = -\frac{1}{6}.\n4. Solve for kk: 6(k−2)=−4⇒6k−12=−4⇒6k=8⇒k=86=436(k-2) = -4 \Rightarrow 6k - 12 = -4 \Rightarrow 6k = 8 \Rightarrow k = \frac{8}{6} = \frac{4}{3}.

Question 15

The line segment joining points A(−2,5)A(-2, 5) and B(4,−1)B(4, -1) is a chord of a circle. The perpendicular bisector of this chord is a diameter of the circle. Find the equation of this diameter.

  1. y=−x+3y = -x + 3
  2. y=x+1y = x + 1 (correct answer)
  3. y=x+6y = x + 6
  4. y=x+7y = x + 7
Explanation: The perpendicular bisector passes through the midpoint of AB and has a gradient that is the negative reciprocal of the gradient of AB.\n1. Find the midpoint M of AB: M=(−2+42,5+(−1)2)=(1,2)M = \left(\frac{-2+4}{2}, \frac{5+(-1)}{2}\right) = (1, 2).\n2. Find the gradient of AB: mAB=−1−54−(−2)=−66=−1m_{AB} = \frac{-1 - 5}{4 - (-2)} = \frac{-6}{6} = -1.\n3. Find the perpendicular gradient: m⊥=−1−1=1m_{\perp} = -\frac{1}{-1} = 1.\n4. Use the point-gradient form y−y1=m(x−x1)y - y_1 = m(x - x_1) with point M(1, 2) and gradient 1: y−2=1(x−1)⇒y=x−1+2⇒y=x+1y - 2 = 1(x - 1) \Rightarrow y = x - 1 + 2 \Rightarrow y = x + 1.

Question 16

The line LL has equation (k−2)x+(3−k)y+5=0(k-2)x + (3-k)y + 5 = 0. For what value of kk is the line LL parallel to the line with equation y=2x+1y = 2x + 1?

  1. 1
  2. 73\frac{7}{3}
  3. 83\frac{8}{3}
  4. 4 (correct answer)
Explanation:
  1. Find the gradient of line LL in terms of kk. Rearrange the equation to the form y=mx+cy = mx + c:\n(3−k)y=−(k−2)x−5⇒y=−(k−2)3−kx−53−k(3-k)y = -(k-2)x - 5 \Rightarrow y = \frac{-(k-2)}{3-k}x - \frac{5}{3-k}.\n2. The gradient is m=−(k−2)3−k=2−k3−km = \frac{-(k-2)}{3-k} = \frac{2-k}{3-k}.\n3. For line LL to be parallel to y=2x+1y=2x+1, their gradients must be equal. The gradient of y=2x+1y=2x+1 is 2.\n4. Set the gradients equal and solve for kk:\n2−k3−k=2⇒2−k=2(3−k)⇒2−k=6−2k⇒k=4\frac{2-k}{3-k} = 2 \Rightarrow 2-k = 2(3-k) \Rightarrow 2-k = 6-2k \Rightarrow k = 4.

Question 17

The line segment PQ has midpoint M(3,−1)M(3, -1). Point P has coordinates (−2,5)(-2, 5). Find the equation of the line that is perpendicular to PQ and passes through point Q.

  1. 5x−6y−82=05x - 6y - 82 = 0 (correct answer)
  2. 5x−6y+2=05x - 6y + 2 = 0
  3. 6x+5y−13=06x + 5y - 13 = 0
  4. 6x+5y−5=06x + 5y - 5 = 0
Explanation:
  1. Find the coordinates of point Q. Let Q=(x,y)Q=(x,y). Using the midpoint formula: −2+x2=3⇒−2+x=6⇒x=8\frac{-2+x}{2} = 3 \Rightarrow -2+x=6 \Rightarrow x=8. And 5+y2=−1⇒5+y=−2⇒y=−7\frac{5+y}{2} = -1 \Rightarrow 5+y=-2 \Rightarrow y=-7. So, Q=(8,−7)Q=(8, -7).\n2. Find the gradient of PQ: mPQ=5−(−7)−2−8=12−10=−65m_{PQ} = \frac{5 - (-7)}{-2 - 8} = \frac{12}{-10} = -\frac{6}{5}.\n3. The gradient of the line perpendicular to PQ is m⊥=−1−6/5=56m_{\perp} = -\frac{1}{-6/5} = \frac{5}{6}.\n4. Find the equation of the line passing through Q(8,−7)Q(8, -7) with gradient 56\frac{5}{6}: y−(−7)=56(x−8)⇒y+7=56(x−8)⇒6(y+7)=5(x−8)⇒6y+42=5x−40⇒5x−6y−82=0y - (-7) = \frac{5}{6}(x - 8) \Rightarrow y + 7 = \frac{5}{6}(x - 8) \Rightarrow 6(y+7) = 5(x-8) \Rightarrow 6y + 42 = 5x - 40 \Rightarrow 5x - 6y - 82 = 0.

Question 18

The line LL is defined by the equation x−2y+6=0x - 2y + 6 = 0. The line MM is the reflection of LL in the y-axis. Find the x-intercept of line MM.

  1. −6-6
  2. −3-3
  3. 3
  4. 6 (correct answer)
Explanation:
  1. To find the equation of the reflection of a line in the y-axis, we replace xx with −x-x in the original equation.\n2. The equation of line LL is x−2y+6=0x - 2y + 6 = 0.\n3. Replacing xx with −x-x gives the equation for line MM: (−x)−2y+6=0(-x) - 2y + 6 = 0, which simplifies to −x−2y+6=0-x - 2y + 6 = 0 or x+2y−6=0x + 2y - 6 = 0.\n4. To find the x-intercept of line MM, we set y=0y=0 in its equation.\n5. x+2(0)−6=0⇒x−6=0⇒x=6x + 2(0) - 6 = 0 \Rightarrow x - 6 = 0 \Rightarrow x = 6.

Question 19

A line LL passes through the point A(2,5)A(2, 5) and has a gradient of −34-\frac{3}{4}. The line intersects the x-axis at P and the y-axis at Q. Find the length of the line segment PQ.

  1. 132\frac{13}{2}
  2. 263\frac{26}{3}
  3. 656\frac{65}{6} (correct answer)
  4. 1313
Explanation:
  1. Find the equation of the line LL using the point-gradient form: y−5=−34(x−2)y - 5 = -\frac{3}{4}(x - 2). 4(y−5)=−3(x−2)⇒4y−20=−3x+6⇒3x+4y=264(y-5) = -3(x-2) \Rightarrow 4y - 20 = -3x + 6 \Rightarrow 3x + 4y = 26.\n2. Find the x-intercept (P) by setting y=0y=0: 3x=26⇒x=2633x = 26 \Rightarrow x = \frac{26}{3}. So P=(263,0)P = (\frac{26}{3}, 0).\n3. Find the y-intercept (Q) by setting x=0x=0: 4y=26⇒y=264=1324y = 26 \Rightarrow y = \frac{26}{4} = \frac{13}{2}. So Q=(0,132)Q = (0, \frac{13}{2}).\n4. Find the distance between P and Q using the distance formula: d=(263−0)2+(0−132)2=(263)2+(−132)2d = \sqrt{(\frac{26}{3}-0)^2 + (0-\frac{13}{2})^2} = \sqrt{(\frac{26}{3})^2 + (-\frac{13}{2})^2}.d=6769+1694=2704+152136=422536=656d = \sqrt{\frac{676}{9} + \frac{169}{4}} = \sqrt{\frac{2704 + 1521}{36}} = \sqrt{\frac{4225}{36}} = \frac{65}{6}.

Question 20

A line passes through the points A(1,5)A(1, 5) and B(3,1)B(3, 1). Find the equation of the perpendicular bisector of the line segment AB.

  1. y=12x+92y = \frac{1}{2}x + \frac{9}{2}
  2. y=−2x+7y = -2x + 7
  3. y=12x+2y = \frac{1}{2}x + 2 (correct answer)
  4. y=2x−1y = 2x - 1
Explanation: When you encounter a perpendicular bisector problem, you need to find a line that passes through the midpoint of the given segment and is perpendicular to the original line. This combines coordinate geometry concepts of midpoint, slope, and perpendicular lines. First, find the midpoint of segment AB. Using the midpoint formula with A(1, 5) and B(3, 1): M=(1+32,5+12)=(2,3)M = \left(\frac{1+3}{2}, \frac{5+1}{2}\right) = (2, 3) Next, determine the slope of line AB: mAB=1−53−1=−42=−2m_{AB} = \frac{1-5}{3-1} = \frac{-4}{2} = -2 Since perpendicular lines have slopes that are negative reciprocals, the perpendicular bisector has slope: m⊥=−1−2=12m_{\perp} = -\frac{1}{-2} = \frac{1}{2} Using point-slope form with the midpoint (2, 3) and slope 12\frac{1}{2}: y−3=12(x−2)y - 3 = \frac{1}{2}(x - 2) y=12x−1+3=12x+2y = \frac{1}{2}x - 1 + 3 = \frac{1}{2}x + 2 This confirms answer C is correct. Answer A has the correct slope but wrong y-intercept (92\frac{9}{2} instead of 2), suggesting an error in applying the point-slope formula. Answer B gives the slope of the original line (-2) rather than its perpendicular, showing confusion about perpendicular relationships. Answer D uses slope 2, which is the negative of the original slope but not the negative reciprocal. Remember: perpendicular bisectors require both finding the midpoint AND using the negative reciprocal of the original slope. Double-check your arithmetic when substituting into point-slope form.