Coordinate Geometry - IB Mathematics Analysis and Approaches

1. Three points in space are P(1, 2, 3), Q(3, 5, 7), and R(-2, k, -3). The points P, Q, and R are collinear. Find the value of k.
   Answer: -5/2 For three points to be collinear, the vectors between them must be parallel. Let's find the vectors $\vec{PQ}$ and $\vec{PR}$. $$\vec{PQ} = Q - P = (3-1, 5-2, 7-3) = (2, 3, 4)$$ $$\vec{PR} = R - P = (-2-1, k-2, -3-3) = (-3, k-2, -6)$$ If the vectors are parallel, then $\vec{PR} = c \cdot \vec{PQ}$ for some scalar c. We can find c by comparing the components. From the x-components: $-3 = c \cdot 2 \Rightarrow c = -3/2$. From the z-components: $-6 = c \cdot 4 \Rightarrow c = -6/4 = -3/2$. The values of c are consistent. Now use this value of c for the y-components: $$k-2 = c \cdot 3$$ $$k-2 = (-3/2) \cdot 3 = -9/2$$ $$k = 2 - 9/2 = 4/2 - 9/2 = -5/2$$
2. The distance between the midpoints of the segments AB and CD is $\sqrt{17}$. The coordinates are A(1, 2), B(3, 8), C(-2, 5), and D(k, 3). Find a possible value for k.
   Answer: -2 First, find the midpoint of segment AB: $$M_{AB} = \left(\frac{1+3}{2}, \frac{2+8}{2}\right) = (2, 5)$$ Next, find the midpoint of segment CD in terms of k: $$M_{CD} = \left(\frac{-2+k}{2}, \frac{5+3}{2}\right) = \left(\frac{k-2}{2}, 4\right)$$ The distance between these midpoints is $\sqrt{17}$, so: $$\left( \frac{k-2}{2} - 2 \right)^2 + (4-5)^2 = 17$$ $$\left( \frac{k-6}{2} \right)^2 + 1 = 17$$ $$\frac{(k-6)^2}{4} = 16$$ $$(k-6)^2 = 64$$ $$k-6 = \pm 8$$ This gives k = 14 or k = -2. Among the given choices, k = -2 is a possible value.
3. Find the perimeter of a triangle with vertices A(0, 0, 3), B(0, 4, 0), and C(5, 0, 0).
   Answer: 5 + $\sqrt{34}$ + $\sqrt{41}$ The perimeter is the sum of the lengths of the three sides AB, BC, and AC. We use the 3D distance formula to find the length of each side. $$\text{Length of AB} = \sqrt{(0-0)^2 + (4-0)^2 + (0-3)^2} = \sqrt{0^2+4^2+(-3)^2} = \sqrt{16+9} = \sqrt{25} = 5$$ $$\text{Length of BC} = \sqrt{(5-0)^2 + (0-4)^2 + (0-0)^2} = \sqrt{5^2+(-4)^2+0^2} = \sqrt{25+16} = \sqrt{41}$$ $$\text{Length of AC} = \sqrt{(5-0)^2 + (0-0)^2 + (0-3)^2} = \sqrt{5^2+0^2+(-3)^2} = \sqrt{25+9} = \sqrt{34}$$ The perimeter is the sum of these lengths: Perimeter = 5 + $\sqrt{41}$ + $\sqrt{34}$.
4. The points A(1, 2), B(5, 4), and C(3, 8) are three vertices of a parallelogram ABCD, in order. Find the sum of the coordinates of the fourth vertex D.
   Answer: 5 In a parallelogram ABCD, the diagonals AC and BD bisect each other. This means they share the same midpoint. Let the coordinates of D be (x, y). First, find the midpoint of diagonal AC: $$M_{AC} = \left(\frac{1+3}{2}, \frac{2+8}{2}\right) = (2, 5)$$ This must also be the midpoint of diagonal BD: $$M_{BD} = \left(\frac{5+x}{2}, \frac{4+y}{2}\right)$$ Setting the coordinates equal: $$\frac{5+x}{2} = 2 \Rightarrow 5+x=4 \Rightarrow x=-1$$ $$\frac{4+y}{2} = 5 \Rightarrow 4+y=10 \Rightarrow y=6$$ So, the coordinates of D are (-1, 6). The sum of the coordinates is -1 + 6 = 5.
5. A triangle PQR has vertices P(-3, 2), Q(5, 2), and R(k, 7). Given that the triangle is isosceles with base PQ, find the value of k.
   Answer: 1 If PQ is the base of the isosceles triangle, then the lengths of the other two sides, PR and QR, must be equal. The vertex R must lie on the perpendicular bisector of the base PQ. The segment PQ is horizontal, so its perpendicular bisector is a vertical line. The x-coordinate of this line is the midpoint of the x-coordinates of P and Q. $$x = \frac{-3+5}{2} = \frac{2}{2} = 1$$ Since R(k, 7) must lie on this line, its x-coordinate k must be 1. Alternatively, set the distances PR and QR equal: PR² = QR². $$(k - (-3))^2 + (7-2)^2 = (k-5)^2 + (7-2)^2$$ $$(k+3)^2 + 5^2 = (k-5)^2 + 5^2$$ $$(k+3)^2 = (k-5)^2$$ $$k^2 + 6k + 9 = k^2 - 10k + 25$$ $$16k = 16$$ $$k=1$$
6. Point P lies on the y-axis and is equidistant from A(-3, 4) and B(5, -2). Find the coordinates of P.
   Answer: (0, -1/3) When you encounter a problem about a point being equidistant from two given points, you're dealing with the perpendicular bisector concept. Since point P lies on the y-axis, it has coordinates (0, y) for some value y that you need to find. To find P, use the fact that it's equidistant from A(-3, 4) and B(5, -2). This means the distance from P to A equals the distance from P to B. Using the distance formula: $\sqrt{(0-(-3))^2 + (y-4)^2} = \sqrt{(0-5)^2 + (y-(-2))^2}$ Simplifying: $\sqrt{9 + (y-4)^2} = \sqrt{25 + (y+2)^2}$ Square both sides to eliminate the square roots: $9 + (y-4)^2 = 25 + (y+2)^2$ Expand the squared terms: $9 + y^2 - 8y + 16 = 25 + y^2 + 4y + 4$ $25 - 8y = 29 + 4y$ $-4 = 12y$ $y = -\frac{1}{3}$ Therefore, P is at (0, -1/3), which is answer D. Answer A (0, -1) results from making calculation errors in the algebra. Answer B (0, 1) comes from a sign error when solving for y. Answer C (0, 1/3) occurs if you get the correct fraction but miss the negative sign. Remember: equidistant problems often involve setting up equal distance expressions and squaring both sides to eliminate square roots. Always double-check your algebra, especially with fractions and negative signs.
7. The midpoint of the line segment joining A(2, -3, 1) and B is M(4, 1, -2). Find the distance from the origin to point B.
   Answer: $\sqrt{86}$ First, find the coordinates of point B(x, y, z). The midpoint M is given by $M = \left(\frac{x_A+x_B}{2}, \frac{y_A+y_B}{2}, \frac{z_A+z_B}{2}\right)$. We have: $$\left(\frac{2+x}{2}, \frac{-3+y}{2}, \frac{1+z}{2}\right) = (4, 1, -2)$$ This gives three equations: $$\frac{2+x}{2} = 4 \Rightarrow 2+x=8 \Rightarrow x=6$$ $$\frac{-3+y}{2} = 1 \Rightarrow -3+y=2 \Rightarrow y=5$$ $$\frac{1+z}{2} = -2 \Rightarrow 1+z=-4 \Rightarrow z=-5$$ So, point B is (6, 5, -5). The distance from the origin O(0, 0, 0) to B is: $$d = \sqrt{(6-0)^2 + (5-0)^2 + (-5-0)^2} = \sqrt{36+25+25} = \sqrt{86}$$
8. The midpoint of the line segment joining point P to the origin O(0,0,0) is M(1, -2, k). The distance OP is $\sqrt{60}$. Find the positive value of k.
   Answer: $\sqrt{10}$ This question tests your understanding of midpoints and distance formulas in 3D coordinate geometry. When you see problems involving midpoints and distances, think about how these concepts connect to find unknown coordinates. Since M(1, -2, k) is the midpoint of segment PO where O is the origin, you can use the midpoint formula to find point P's coordinates. The midpoint formula gives us: $M = \frac{P + O}{2}$, so $(1, -2, k) = \frac{(x_P, y_P, z_P) + (0, 0, 0)}{2}$. This means $P = (2, -4, 2k)$. Now use the distance formula with the given information that $OP = \sqrt{60}$. The distance from O(0,0,0) to P(2, -4, 2k) is: $\sqrt{(2-0)^2 + (-4-0)^2 + (2k-0)^2} = \sqrt{4 + 16 + 4k^2} = \sqrt{20 + 4k^2}$ Setting this equal to the given distance: $\sqrt{20 + 4k^2} = \sqrt{60}$ Squaring both sides: $20 + 4k^2 = 60$, so $4k^2 = 40$, giving $k^2 = 10$. Since we want the positive value, $k = \sqrt{10}$. Choice A ($\sqrt{5}$) comes from incorrectly solving $k^2 = 5$ instead of $k^2 = 10$. Choice B (10) results from forgetting to take the square root of $k^2 = 10$. Choice C (5) appears if you make both errors: solving $k^2 = 5$ and then forgetting the square root. Remember: in midpoint problems, always double the midpoint coordinates to find the other endpoint when one point is the origin, then apply the distance formula systematically.
9. The distance between point A(t, 2) and point B(1, 5) is 5. Find the sum of the possible values of t.
   Answer: 2 The distance formula is $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. We are given d=5. Squaring both sides, d²=25. $$(t-1)^2 + (2-5)^2 = 5^2$$ $$(t-1)^2 + (-3)^2 = 25$$ $$(t-1)^2 + 9 = 25$$ $$(t-1)^2 = 16$$ Taking the square root of both sides gives two possibilities: $$t-1 = 4 \quad \text{or} \quad t-1 = -4$$ $$t = 5 \quad \text{or} \quad t = -3$$ The possible values of t are 5 and -3. The sum of these values is 5 + (-3) = 2. Alternatively, expanding $(t-1)^2=16$ gives $t^2-2t+1=16$, which simplifies to the quadratic equation $t^2-2t-15=0$. The sum of the roots of a quadratic equation $ax^2+bx+c=0$ is $-b/a$. Here, the sum is $-(-2)/1 = 2$.
10. A sphere has a diameter with endpoints P(-2, 4, 1) and Q(4, -2, 7). Find the square of the radius of the sphere.
    Answer: 27 The length of the diameter is the distance between points P and Q. The radius is half the length of the diameter. First, find the square of the diameter's length (d²). $$d^2 = (4 - (-2))^2 + (-2 - 4)^2 + (7 - 1)^2$$ $$d^2 = (6)^2 + (-6)^2 + (6)^2$$ $$d^2 = 36 + 36 + 36 = 108$$ The radius r is half the diameter, so $r = d/2$. The square of the radius is $r^2 = (d/2)^2 = d^2/4$. $$r^2 = \frac{108}{4} = 27$$ Alternatively, one could find the center of the sphere (the midpoint of PQ) and then find the distance from the center to one of the endpoints. Center C = ((-2+4)/2, (4-2)/2, (1+7)/2) = (1, 1, 4). Then r² = distance from C to P squared = $(1 - (-2))^2 + (1 - 4)^2 + (4 - 1)^2 = 3^2 + (-3)^2 + 3^2 = 9 + 9 + 9 = 27$.
11. A line segment has one endpoint at A(-3, 5) and its midpoint at M(2, 1). What are the coordinates of the other endpoint, B?
    Answer: (7, -3) Let the coordinates of endpoint B be (x, y). The midpoint formula states that $M = \left(\frac{x_A+x_B}{2}, \frac{y_A+y_B}{2}\right)$. We are given A(-3, 5) and M(2, 1). We can set up two equations: $$\frac{-3+x}{2} = 2 \quad \text{and} \quad \frac{5+y}{2} = 1$$ Solving for x: $$-3+x = 4 \Rightarrow x = 7$$ Solving for y: $$5+y = 2 \Rightarrow y = -3$$ Therefore, the coordinates of endpoint B are (7, -3).
12. Point M is the midpoint of AB and point N is the midpoint of AM. If A is (-5, 12) and B is (7, -4), find the coordinates of N.
    Answer: (-2, 8) This is a two-step problem. First, find the coordinates of M, the midpoint of AB. $$M = \left(\frac{-5+7}{2}, \frac{12+(-4)}{2}\right) = \left(\frac{2}{2}, \frac{8}{2}\right) = (1, 4)$$ Next, find the coordinates of N, the midpoint of AM. A is (-5, 12) and M is (1, 4). $$N = \left(\frac{-5+1}{2}, \frac{12+4}{2}\right) = \left(\frac{-4}{2}, \frac{16}{2}\right) = (-2, 8)$$
13. Point A is (2, 3) and point B is (10, k). The midpoint M of the segment AB lies on the line y = 2x - 7. Find the value of k.
    Answer: 7 First, find the coordinates of the midpoint M in terms of k. $$M = \left(\frac{2+10}{2}, \frac{3+k}{2}\right) = \left(6, \frac{3+k}{2}\right)$$ Since M lies on the line y = 2x - 7, substitute the coordinates of M into the equation: $$\frac{3+k}{2} = 2(6) - 7$$ $$\frac{3+k}{2} = 12 - 7 = 5$$ $$3+k = 10$$ $$k = 7$$
14. Find the length of the median from vertex A for a triangle with vertices A(5, -1), B(0, 4), and C(-2, -2).
    Answer: $2\sqrt{10}$ When you encounter a question about finding the length of a median, remember that a median connects a vertex to the midpoint of the opposite side. Your approach should be: find the midpoint, then calculate the distance. First, find the midpoint of side BC (opposite vertex A). Using the midpoint formula $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ with B(0, 4) and C(-2, -2): Midpoint M = $\left(\frac{0+(-2)}{2}, \frac{4+(-2)}{2}\right) = (-1, 1)$ Now find the distance from A(5, -1) to M(-1, 1) using the distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ $d = \sqrt{(-1-5)^2 + (1-(-1))^2} = \sqrt{(-6)^2 + (2)^2} = \sqrt{36 + 4} = \sqrt{40}$ Simplifying: $\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$ This confirms answer D is correct. The wrong answers likely come from calculation errors: A) $2\sqrt{5}$ equals $\sqrt{20}$, suggesting someone found the wrong midpoint or made an arithmetic mistake. B) $5\sqrt{2}$ equals $\sqrt{50}$, possibly from incorrectly calculating the distance components. C) $\sqrt{58}$ might result from using the wrong vertices or forgetting to find the midpoint first. Study tip: Always break median problems into two clear steps: find the midpoint of the opposite side, then calculate the distance. Double-check your midpoint calculation before moving to the distance formula, as errors compound quickly in coordinate geometry.
15. A line segment has endpoints A(a, -2, 5) and B(3, b, 1). Its midpoint is M(-1, 4, c). Find the value of a + b + c.
    Answer: 8 The coordinates of the midpoint M are the average of the coordinates of the endpoints A and B. We set up equations for each coordinate: $$\frac{a+3}{2} = -1 \Rightarrow a+3 = -2 \Rightarrow a = -5$$ $$\frac{-2+b}{2} = 4 \Rightarrow -2+b = 8 \Rightarrow b = 10$$ $$\frac{5+1}{2} = c \Rightarrow \frac{6}{2} = c \Rightarrow c = 3$$ Now, we find the sum a + b + c: $$a + b + c = -5 + 10 + 3 = 8$$
16. Point M(3, 1) is the midpoint of the line segment AB. Point A has coordinates (-1, 4). Find the length of the segment MB.
    Answer: 5 The length of the segment MB is the distance from M to B. However, since M is the midpoint of AB, the length of MB is equal to the length of AM. We can find the length of AM directly from the given coordinates of A and M. Using the distance formula: $$\text{Length of AM} = \sqrt{(3 - (-1))^2 + (1 - 4)^2}$$ $$= \sqrt{(4)^2 + (-3)^2}$$ $$= \sqrt{16 + 9} = \sqrt{25} = 5$$ Since AM = MB, the length of MB is 5. An alternative, longer method is to first find the coordinates of B and then calculate the distance MB.
17. A cuboid has opposite vertices at A(1, 1, 1) and G(5, 7, 4). Find the square of the length of the space diagonal AG.
    Answer: 61 The length of the space diagonal AG can be found using the distance formula in three dimensions. Let d be the length of AG. Then the square of the length, d², is given by: $$d^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$$ Substituting the coordinates of A(1, 1, 1) and G(5, 7, 4): $$d^2 = (5-1)^2 + (7-1)^2 + (4-1)^2$$ $$d^2 = 4^2 + 6^2 + 3^2$$ $$d^2 = 16 + 36 + 9$$ $$d^2 = 61$$

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