Linear Functions - IB Mathematics Analysis and Approaches

1. The line segment joining points $A(-2, 5)$ and $B(4, -1)$ is a chord of a circle. The perpendicular bisector of this chord is a diameter of the circle. Find the equation of this diameter.
   Answer: $y = x + 1$ The perpendicular bisector passes through the midpoint of AB and has a gradient that is the negative reciprocal of the gradient of AB. 1. Find the midpoint M of AB: $M = \left(\frac{-2+4}{2}, \frac{5+(-1)}{2}\right) = (1, 2)$. 2. Find the gradient of AB: $m_{AB} = \frac{-1 - 5}{4 - (-2)} = \frac{-6}{6} = -1$. 3. Find the perpendicular gradient: $m_{\perp} = -\frac{1}{-1} = 1$. 4. Use the point-gradient form $y - y_1 = m(x - x_1)$ with point M(1, 2) and gradient 1: $y - 2 = 1(x - 1) \Rightarrow y = x - 1 + 2 \Rightarrow y = x + 1$.
2. The line $L$ is defined by the equation $x - 2y + 6 = 0$. The line $M$ is the reflection of $L$ in the y-axis. Find the x-intercept of line $M$.
   Answer: 6 1. To find the equation of the reflection of a line in the y-axis, we replace $x$ with $-x$ in the original equation. 2. The equation of line $L$ is $x - 2y + 6 = 0$. 3. Replacing $x$ with $-x$ gives the equation for line $M$: $(-x) - 2y + 6 = 0$, which simplifies to $-x - 2y + 6 = 0$ or $x + 2y - 6 = 0$. 4. To find the x-intercept of line $M$, we set $y=0$ in its equation. 5. $x + 2(0) - 6 = 0 \Rightarrow x - 6 = 0 \Rightarrow x = 6$.
3. The line segment PQ has midpoint $M(3, -1)$. Point P has coordinates $(-2, 5)$. Find the equation of the line that is perpendicular to PQ and passes through point Q.
   Answer: $5x - 6y - 82 = 0$ 1. Find the coordinates of point Q. Let $Q=(x,y)$. Using the midpoint formula: $\frac{-2+x}{2} = 3 \Rightarrow -2+x=6 \Rightarrow x=8$. And $\frac{5+y}{2} = -1 \Rightarrow 5+y=-2 \Rightarrow y=-7$. So, $Q=(8, -7)$. 2. Find the gradient of PQ: $m_{PQ} = \frac{5 - (-7)}{-2 - 8} = \frac{12}{-10} = -\frac{6}{5}$. 3. The gradient of the line perpendicular to PQ is $m_{\perp} = -\frac{1}{-6/5} = \frac{5}{6}$. 4. Find the equation of the line passing through $Q(8, -7)$ with gradient $\frac{5}{6}$: $y - (-7) = \frac{5}{6}(x - 8) \Rightarrow y + 7 = \frac{5}{6}(x - 8) \Rightarrow 6(y+7) = 5(x-8) \Rightarrow 6y + 42 = 5x - 40 \Rightarrow 5x - 6y - 82 = 0$.
4. The vertices of a quadrilateral ABCD are $A(1, 6)$, $B(2, 9)$, $C(5, 8)$, and $D(4, 5)$. Which of the following statements most accurately describes the quadrilateral?
   Answer: A square. To classify the quadrilateral, we need to examine the gradients and lengths of its sides. 1. Gradients: $m_{AB} = \frac{9-6}{2-1} = 3$ $m_{BC} = \frac{8-9}{5-2} = -\frac{1}{3}$ $m_{CD} = \frac{5-8}{4-5} = \frac{-3}{-1} = 3$ $m_{DA} = \frac{6-5}{1-4} = -\frac{1}{3}$ Since $m_{AB} = m_{CD}$ and $m_{BC} = m_{DA}$, opposite sides are parallel, so it is a parallelogram. Since $m_{AB} \times m_{BC} = 3 \times (-\frac{1}{3}) = -1$, adjacent sides are perpendicular, so it is a rectangle. 2. Lengths: $d_{AB}^2 = (2-1)^2 + (9-6)^2 = 1^2 + 3^2 = 10$ $d_{BC}^2 = (5-2)^2 + (8-9)^2 = 3^2 + (-1)^2 = 10$ Since adjacent sides AB and BC have equal length, the rectangle is a square. All sides have length $\sqrt{10}$.
5. The line $L_1$ has the equation $3x + 2y - 12 = 0$. The line $L_2$ is parallel to $L_1$ and passes through the point $P(k, 3)$ and the point $Q(2, -3)$. Find the value of $k$.
   Answer: $-2$ 1. Find the gradient of $L_1$: $2y = -3x + 12 \Rightarrow y = -\frac{3}{2}x + 6$. The gradient is $m = -\frac{3}{2}$. 2. Since $L_2$ is parallel to $L_1$, its gradient is also $-\frac{3}{2}$. 3. Calculate the gradient of $L_2$ using points P and Q: $m = \frac{-3 - 3}{2 - k} = \frac{-6}{2 - k}$. 4. Set the gradients equal and solve for $k$: $\frac{-6}{2 - k} = -\frac{3}{2} \Rightarrow -12 = -3(2 - k) \Rightarrow -12 = -6 + 3k \Rightarrow -6 = 3k \Rightarrow k = -2$.
6. The line $L$ passes through the point $(8, -1)$ and is perpendicular to the line with equation $2x - y = 7$. Find the area of the triangle enclosed by the line $L$, the x-axis, and the y-axis.
   Answer: 9 1. The line $2x - y = 7$ can be written as $y = 2x - 7$. Its gradient is 2. 2. The gradient of the perpendicular line $L$ is $m = -\frac{1}{2}$. 3. The equation of $L$ passes through $(8, -1)$: $y - (-1) = -\frac{1}{2}(x - 8) \Rightarrow y + 1 = -\frac{1}{2}x + 4 \Rightarrow y = -\frac{1}{2}x + 3$. 4. Find the intercepts of $L$: - x-intercept (set $y=0$): $0 = -\frac{1}{2}x + 3 \Rightarrow \frac{1}{2}x = 3 \Rightarrow x = 6$. Point (6,0). - y-intercept (set $x=0$): $y = 3$. Point (0,3). 5. The area of the triangle with vertices (0,0), (6,0), and (0,3) is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 3 = 9$.
7. Line $L_1$ has equation $y = mx + 2$. Line $L_2$ has equation $y = 2x + 5$. The lines intersect at a point with an x-coordinate of -3. Find the equation of the line $L_3$ which is perpendicular to $L_1$ and passes through the point of intersection.
   Answer: $y = -x - 4$ 1. Find the point of intersection. We are given $x = -3$. Substitute this into the equation for $L_2$ to find the y-coordinate: $y = 2(-3) + 5 = -6 + 5 = -1$. The intersection point is $(-3, -1)$. 2. Find the gradient $m$ of $L_1$ by substituting the intersection point into its equation: $-1 = m(-3) + 2 \Rightarrow -3 = -3m \Rightarrow m = 1$. 3. The gradient of $L_1$ is 1. The gradient of line $L_3$, which is perpendicular to $L_1$, is the negative reciprocal: $m_3 = -\frac{1}{1} = -1$. 4. Use the point-gradient form to find the equation of $L_3$, which passes through $(-3, -1)$ with a gradient of -1: $y - (-1) = -1(x - (-3)) \Rightarrow y + 1 = -(x + 3) \Rightarrow y + 1 = -x - 3 \Rightarrow y = -x - 4$.
8. The line $L$ has equation $(k-2)x + (3-k)y + 5 = 0$. For what value of $k$ is the line $L$ parallel to the line with equation $y = 2x + 1$?
   Answer: 4 1. Find the gradient of line $L$ in terms of $k$. Rearrange the equation to the form $y = mx + c$: $(3-k)y = -(k-2)x - 5 \Rightarrow y = \frac{-(k-2)}{3-k}x - \frac{5}{3-k}$. 2. The gradient is $m = \frac{-(k-2)}{3-k} = \frac{2-k}{3-k}$. 3. For line $L$ to be parallel to $y=2x+1$, their gradients must be equal. The gradient of $y=2x+1$ is 2. 4. Set the gradients equal and solve for $k$: $\frac{2-k}{3-k} = 2 \Rightarrow 2-k = 2(3-k) \Rightarrow 2-k = 6-2k \Rightarrow k = 4$.
9. The line $L_1$ passes through $(1, 1)$ and $(3, 5)$. The line $L_2$ has equation $x + y = 5$. The lines intersect at point P. Find the sum of the coordinates of P.
   Answer: 5 1. First, find the equation of line $L_1$. The gradient is $m = \frac{5-1}{3-1} = \frac{4}{2} = 2$. 2. Using the point-gradient form with point $(1,1)$: $y - 1 = 2(x - 1) \Rightarrow y = 2x - 2 + 1 \Rightarrow y = 2x - 1$. 3. To find the point of intersection P, solve the system of simultaneous equations: $y = 2x - 1$ (from $L_1$) $x + y = 5$ (from $L_2$) 4. Substitute the expression for $y$ from the first equation into the second: $x + (2x - 1) = 5 \Rightarrow 3x - 1 = 5 \Rightarrow 3x = 6 \Rightarrow x = 2$. 5. Substitute $x = 2$ back into the equation for $L_1$ to find $y$: $y = 2(2) - 1 = 4 - 1 = 3$. 6. The point of intersection is $P(2, 3)$. 7. The sum of the coordinates of P is $2 + 3 = 5$.
10. The lines $L_1: y = (k^2-3)x + 2$ and $L_2: y = \frac{k}{2}x - 1$ are perpendicular. What is a possible value of $k$?
    Answer: 1 For two lines to be perpendicular, the product of their gradients must be -1. 1. The gradient of $L_1$ is $m_1 = k^2-3$. 2. The gradient of $L_2$ is $m_2 = \frac{k}{2}$. 3. Set the product of the gradients to -1: $m_1 m_2 = (k^2-3)(\frac{k}{2}) = -1$. 4. Simplify the equation: $k(k^2-3) = -2 \Rightarrow k^3 - 3k = -2 \Rightarrow k^3 - 3k + 2 = 0$. 5. We can test the given options by substituting them into the equation. For $k=1$: $(1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0$. This is a valid solution. (For completeness, the other solutions can be found by factoring $(k-1)$ from the polynomial, yielding $(k-1)(k^2+k-2)=0$, which further factors to $(k-1)(k+2)(k-1)=0$. The solutions are $k=1$ (repeated root) and $k=-2$.)
11. The points $A(-1, 2)$, $B(3, k)$, and $C(5, 1)$ are collinear. Find the value of $k$.
    Answer: $\frac{4}{3}$ 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. 1. Gradient of AB: $m_{AB} = \frac{k - 2}{3 - (-1)} = \frac{k-2}{4}$. 2. Gradient of AC: $m_{AC} = \frac{1 - 2}{5 - (-1)} = \frac{-1}{6}$. 3. Set the gradients equal: $\frac{k-2}{4} = -\frac{1}{6}$. 4. Solve for $k$: $6(k-2) = -4 \Rightarrow 6k - 12 = -4 \Rightarrow 6k = 8 \Rightarrow k = \frac{8}{6} = \frac{4}{3}$.
12. The line $L_1$ has equation $y=3x-2$. The line $L_2$ is perpendicular to $L_1$ and has the same y-intercept as the line $2x+3y=6$. Find the equation of $L_2$.
    Answer: $y = -\frac{1}{3}x + 2$ When you encounter questions about perpendicular lines and y-intercepts, you need to master two key relationships: perpendicular lines have slopes that are negative reciprocals of each other, and you can find y-intercepts by converting equations to slope-intercept form. Let's find the slope and y-intercept we need. Line $L_1$ has equation $y = 3x - 2$, so its slope is 3. Since $L_2$ is perpendicular to $L_1$, the slope of $L_2$ must be $-\frac{1}{3}$ (the negative reciprocal of 3). Next, we need the y-intercept of line $2x + 3y = 6$. Converting to slope-intercept form: $3y = -2x + 6$, so $y = -\frac{2}{3}x + 2$. The y-intercept is 2. Therefore, $L_2$ has slope $-\frac{1}{3}$ and y-intercept 2, giving us $y = -\frac{1}{3}x + 2$, which is answer C. Looking at the wrong answers: A) $y = 3x + 2$ uses the same slope as $L_1$ instead of the perpendicular slope, making these lines parallel rather than perpendicular. B) $y = -\frac{1}{3}x + 3$ has the correct perpendicular slope but uses 3 as the y-intercept, which isn't the y-intercept of the given line. D) $y = -\frac{2}{3}x + 2$ uses the slope from $2x + 3y = 6$ instead of its negative reciprocal. Remember: perpendicular slopes multiply to give -1, and always convert linear equations to $y = mx + b$ form to easily identify the y-intercept.
13. A line passes through the points $A(1, 5)$ and $B(3, 1)$. Find the equation of the perpendicular bisector of the line segment AB.
    Answer: $y = \frac{1}{2}x + 2$ 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): $M = \left(\frac{1+3}{2}, \frac{5+1}{2}\right) = (2, 3)$ Next, determine the slope of line AB: $m_{AB} = \frac{1-5}{3-1} = \frac{-4}{2} = -2$ Since perpendicular lines have slopes that are negative reciprocals, the perpendicular bisector has slope: $m_{\perp} = -\frac{1}{-2} = \frac{1}{2}$ Using point-slope form with the midpoint (2, 3) and slope $\frac{1}{2}$: $y - 3 = \frac{1}{2}(x - 2)$ $y = \frac{1}{2}x - 1 + 3 = \frac{1}{2}x + 2$ This confirms answer C is correct. Answer A has the correct slope but wrong y-intercept ($\frac{9}{2}$ 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.
14. A line $L$ passes through the point $A(2, 5)$ and has a gradient of $-\frac{3}{4}$. The line intersects the x-axis at P and the y-axis at Q. Find the length of the line segment PQ.
    Answer: $\frac{65}{6}$ 1. Find the equation of the line $L$ using the point-gradient form: $y - 5 = -\frac{3}{4}(x - 2)$. $4(y-5) = -3(x-2) \Rightarrow 4y - 20 = -3x + 6 \Rightarrow 3x + 4y = 26$. 2. Find the x-intercept (P) by setting $y=0$: $3x = 26 \Rightarrow x = \frac{26}{3}$. So $P = (\frac{26}{3}, 0)$. 3. Find the y-intercept (Q) by setting $x=0$: $4y = 26 \Rightarrow y = \frac{26}{4} = \frac{13}{2}$. So $Q = (0, \frac{13}{2})$. 4. Find the distance between P and Q using the distance formula: $d = \sqrt{(\frac{26}{3}-0)^2 + (0-\frac{13}{2})^2} = \sqrt{(\frac{26}{3})^2 + (-\frac{13}{2})^2}$.$d = \sqrt{\frac{676}{9} + \frac{169}{4}} = \sqrt{\frac{2704 + 1521}{36}} = \sqrt{\frac{4225}{36}} = \frac{65}{6}$.
15. Three points are given as $A(1, 5)$, $B(-2, -1)$, and $C(x, 3)$. Find the value of $x$ such that the line segment $BC$ is perpendicular to the line segment $AB$.
    Answer: $-10$ 1. Find the gradient of AB: $m_{AB} = \frac{5 - (-1)}{1 - (-2)} = \frac{6}{3} = 2$. 2. For BC to be perpendicular to AB, the gradient of BC must be the negative reciprocal of the gradient of AB. So, $m_{BC} = -\frac{1}{2}$. 3. Calculate the gradient of BC using the coordinates of B and C: $m_{BC} = \frac{3 - (-1)}{x - (-2)} = \frac{4}{x+2}$. 4. Set the gradients equal and solve for $x$: $\frac{4}{x+2} = -\frac{1}{2} \Rightarrow 8 = -(x+2) \Rightarrow 8 = -x - 2 \Rightarrow x = -10$.
16. The line $L$ is given by the equation $ax + 4y - 8 = 0$. The line $L$ has a gradient of 3 and passes through the point $(b, 4)$. Find the value of $a+b$.
    Answer: $-\frac{34}{3}$ 1. Find $a$ using the gradient. Rearrange the equation to $y = mx + c$ form: $4y = -ax + 8 \Rightarrow y = -\frac{a}{4}x + 2$. The gradient is $-\frac{a}{4}$. 2. We are given the gradient is 3, so $-\frac{a}{4} = 3 \Rightarrow a = -12$. 3. The equation of the line is $-12x + 4y - 8 = 0$. Substitute the point $(b, 4)$ into the equation to find $b$: $-12(b) + 4(4) - 8 = 0 \Rightarrow -12b + 16 - 8 = 0 \Rightarrow -12b + 8 = 0 \Rightarrow 12b = 8 \Rightarrow b = \frac{8}{12} = \frac{2}{3}$. 4. Calculate $a+b$: $a+b = -12 + \frac{2}{3} = -\frac{36}{3} + \frac{2}{3} = -\frac{34}{3}$.
17. Find the equation of the line that passes through the point $P(-1, 6)$ and the point of intersection of the lines $L_1: x - 2y = 1$ and $L_2: 3x + y = 10$.
    Answer: $5x + 4y - 19 = 0$ 1. Find the point of intersection by solving the system of equations. From $L_2$, $y = 10 - 3x$. 2. Substitute this into $L_1$: $x - 2(10 - 3x) = 1 \Rightarrow x - 20 + 6x = 1 \Rightarrow 7x = 21 \Rightarrow x = 3$. 3. Find $y$: $y = 10 - 3(3) = 1$. The intersection point is $Q(3, 1)$. 4. Find the gradient of the line passing through $P(-1, 6)$ and $Q(3, 1)$: $m = \frac{1 - 6}{3 - (-1)} = \frac{-5}{4}$. 5. Use the point-gradient form with Q(3, 1): $y - 1 = -\frac{5}{4}(x - 3) \Rightarrow 4(y-1) = -5(x-3) \Rightarrow 4y - 4 = -5x + 15 \Rightarrow 5x + 4y - 19 = 0$.

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