Q.Find a point on the x-axis, which is equidistant from the points $(7, 6)$ and $(3, 4)$.

Showing the 12 most recent of 13 on this concept.

1. COMEDK 2026Set 2026-M1 markMCQ
   Q.

   [!FORMULA] $\begin{aligned} &\text { Find the co-ordinates of the orthocentre of the triangle formed by the lines }\\ &\begin{aligned} & L_1: y-x=2 \\ & L_2: y+2 x=8 \\ & L_3: 3 y-x=18 \end{aligned} \end{aligned}$

   (A) (2, 4) (B) $\left(\frac{22}{7}, \frac{43}{7}\right)$ (C) $\left(\frac{10}{7}, \frac{40}{7}\right)$ (D) (6, 8)

   ›Reveal solutionSolution

   The orthocentre is the intersection of two altitudes. Finding the vertices of the triangle and then the equations of two altitudes and solving them gives the orthocentre. The correct coordinates are $\left(\frac{10}{7}, \frac{40}{7}\right)$, which corresponds to option (C).

   Concept & Intuition

   The orthocentre of a triangle is the point where all three altitudes meet. Instead of solving three altitude equations, we only need two — their intersection gives the orthocentre. The trick is to find the vertices first (intersections of the given lines), then pick a vertex and find the altitude from it to the opposite side. The altitude is perpendicular to that side, so we use the negative reciprocal slope. Then repeat for another vertex and solve the two altitude equations.

   ---

   Step-by-step solution

   1. Find the vertices of the triangle The triangle is formed by the three lines:

   $L_1: y - x = 2 \quad\Rightarrow\quad y = x + 2$

   $L_2: y + 2x = 8 \quad\Rightarrow\quad y = 8 - 2x$

   $L_3: 3y - x = 18 \quad\Rightarrow\quad y = \frac{x + 18}{3}$

   • Vertex A = intersection of $L_1$ and $L_2$:

     Set $x + 2 = 8 - 2x$ → $3x = 6$ → $x = 2$, then $y = 4$. So $A = (2, 4)$.

   • Vertex B = intersection of $L_2$ and $L_3$:

     Set $8 - 2x = \frac{x + 18}{3}$ → multiply by 3: $24 - 6x = x + 18$ → $24 - 18 = 7x$ → $x = \frac{6}{7}$, then $y = 8 - 2\cdot\frac{6}{7} = \frac{56 - 12}{7} = \frac{44}{7}$. So $B = \left(\frac{6}{7}, \frac{44}{7}\right)$.

   • Vertex C = intersection of $L_3$ and $L_1$:

     Set $\frac{x + 18}{3} = x + 2$ → multiply by 3: $x + 18 = 3x + 6$ → $12 = 2x$ → $x = 6$, then $y = 8$. So $C = (6, 8)$.

   Tip

   Notice that option (A) is vertex A and option (D) is vertex C — common distractors! The orthocentre is not a vertex unless the triangle is right-angled.

   2. Find the equation of the altitude from vertex A to side BC Side BC is between $B\left(\frac{6}{7}, \frac{44}{7}\right)$ and $C(6, 8)$. Slope of BC:

   $m_{BC} = \frac{8 - \frac{44}{7}}{6 - \frac{6}{7}} = \frac{\frac{56 - 44}{7}}{\frac{42 - 6}{7}} = \frac{12/7}{36/7} = \frac{12}{36} = \frac{1}{3}$

   The altitude from A is perpendicular to BC, so its slope is the negative reciprocal: $m_{\perp} = -3$.

   It passes through $A(2, 4)$:

   $y - 4 = -3(x - 2) \quad\Rightarrow\quad y = -3x + 10$

   This is altitude $h_A$.

   3. Find the equation of the altitude from vertex B to side AC Side AC is between $A(2, 4)$ and $C(6, 8)$. Slope of AC:

   $m_{AC} = \frac{8 - 4}{6 - 2} = \frac{4}{4} = 1$

   The altitude from B is perpendicular to AC, so its slope is $-1$.

   It passes through $B\left(\frac{6}{7}, \frac{44}{7}\right)$:

   $y - \frac{44}{7} = -1\left(x - \frac{6}{7}\right) \quad\Rightarrow\quad y = -x + \frac{6}{7} + \frac{44}{7} = -x + \frac{50}{7}$ …

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2. COMEDK 2025Set 2025-A1 markMCQ
   Q.In a triangle $A B C$ the coordinate of the vertex $A$ is $(1,2)$. Equations of the median through $B$ and $C$ are respectively $x+y=5$ and $x=4$. Then the equation of side $\mathbf{A B}$ is (A) $2 x-3 y+4=0$ (B) $2 x+3 y=8$ (C) $3 x-2 y+1=0$ (D) $3 x+2 y=5$

   ›Reveal solutionSolution

   The key idea is to find the coordinates of B and C using the intersection of medians (centroid) and the given median equations, then compute the slope of AB to match one of the options. The correct equation is $2x + 3y = 8$.

   We are given triangle $ABC$ with $A = (1,2)$. The median through $B$ is the line $x + y = 5$, and the median through $C$ is the line $x = 4$. A median goes from a vertex to the midpoint of the opposite side. So the median through $B$ goes from $B$ to the midpoint of $AC$; the median through $C$ goes from $C$ to the midpoint of $AB$. The intersection of any two medians is the centroid $G$. This is our starting point.

   1. Find the centroid $G$. The centroid is the intersection of the medians. Solve the equations of the medians through $B$ and $C$:

   $x + y = 5 \quad \text{and} \quad x = 4.$

   Substituting $x = 4$ into $x + y = 5$ gives $4 + y = 5 \Rightarrow y = 1$.

   So the centroid is $G = (4, 1)$.

   2. Use the centroid formula. The centroid $G$ is the average of the vertices:

   $G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right).$

   With $A = (1,2)$ and $G = (4,1)$, we have:

   $\frac{1 + x_B + x_C}{3} = 4 \quad \Rightarrow \quad 1 + x_B + x_C = 12 \quad \Rightarrow \quad x_B + x_C = 11.$

   $\frac{2 + y_B + y_C}{3} = 1 \quad \Rightarrow \quad 2 + y_B + y_C = 3 \quad \Rightarrow \quad y_B + y_C = 1.$

   3. Interpret the median through $C$. The median through $C$ is the line $x = 4$. This median goes from $C$ to the midpoint of $AB$. Let $M_{AB}$ be the midpoint of $AB$. Since $C$ and $M_{AB}$ lie on $x = 4$, both have $x$-coordinate 4. The midpoint $M_{AB}$ is $\left( \frac{1 + x_B}{2}, \frac{2 + y_B}{2} \right)$. Its $x$-coordinate is 4, so:

   $\frac{1 + x_B}{2} = 4 \quad \Rightarrow \quad 1 + x_B = 8 \quad \Rightarrow \quad x_B = 7.$

   4. Find $x_C$ and then $y_B, y_C$. From $x_B + x_C = 11$ and $x_B = 7$, we get $x_C = 4$. From $y_B + y_C = 1$, we need another relation. Use the median through $B$: the line $x + y = 5$ goes through $B$ and the midpoint of $AC$. The midpoint of $AC$ is $M_{AC} = \left( \frac{1 + 4}{2}, \frac{2 + y_C}{2} \right) = \left( \frac{5}{2}, \frac{2 + y_C}{2} \right)$. …

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3. COMEDK 2025Set 2025-E1 markMCQ
   Q.If the line $(3 x+14 y+7)+k(5 x+7 y+6)=0$ is perpendicular to $x$-axis then the value of ' $k$ ' is (A) $-2$ (B) $2$ (C) $-\frac{3}{5}$ (D) $\frac{1}{3}$

   ›Reveal solutionSolution

   The line is perpendicular to the x‑axis, so it must be vertical. A vertical line has no y‑term, so we set the coefficient of y to zero. This gives $k = -2$, which corresponds to option (A).

   Concept & Intuition

   A line perpendicular to the x‑axis is vertical. Vertical lines have equations of the form $x = \text{constant}$ — they contain no $y$ term. Our given equation is a family of lines depending on $k$. We need the one member of that family that is vertical, so we must force the coefficient of $y$ to vanish. That single condition determines $k$.

   ---

   1. Write the given equation in standard linear form The line is

   $(3x + 14y + 7) + k(5x + 7y + 6) = 0.$

   Expand and collect like terms:

   $3x + 14y + 7 + 5kx + 7ky + 6k = 0.$

   Group the $x$ terms, the $y$ terms, and the constants:

   $(3 + 5k)x + (14 + 7k)y + (7 + 6k) = 0.$

   2. Condition for a vertical line A vertical line has no $y$‑dependence, meaning the coefficient of $y$ must be zero. So we set

   $14 + 7k = 0.$

   3. Solve for $k$

   $7k = -14 \quad\Rightarrow\quad k = -2.$

   4. Verify the result Substituting $k = -2$ into the original equation:

   $(3x + 14y + 7) - 2(5x + 7y + 6) = 0$

   becomes

   $$3x + 14y + 7 - 10x - 14y - 12 = 0, …$$

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4. COMEDK 2025Set 2025-E1 markMCQ
   Q.A straight line makes positive intercepts on the coordinate axes whose sum is 5 . If the line passes through the point $P(-3,4)$ then the equation of a line is (A) $2 x-y+10=0$ (B) $2 x+3 y=6$ (C) $3 x+2 y=6$ (D) $x+4 y=13$

   ›Reveal solutionSolution

   Intercept form $\tfrac{x}{a}+\tfrac{y}{b}=1$ with $a+b=5$ and $(-3,4)$ on it gives $a=3,\,b=2$, i.e. $2x+3y=6$.

   Let the intercepts be $a$ (on $x$-axis) and $b$ (on $y$-axis), both positive:

   $\frac{x}{a}+\frac{y}{b}=1, \qquad a+b=5 \implies b = 5-a$

   The line passes through $P(-3,4)$:

   $\frac{-3}{a}+\frac{4}{5-a}=1$

   Multiply through by $a(5-a)$:

   $-3(5-a) + 4a = a(5-a)$

   $-15 + 3a + 4a = 5a - a^{2}$ …

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5. COMEDK 2025Set 2025-M1 markMCQ
   Q.If two vertices of a triangle are $(3,-2)$ and $(-2,3)$ and its orthocentre is $(-6,1)$. Then the difference between ordinate and abscissa of the third vertex of the triangle is (A) 2 (B) 5 (C) $-$5 (D) 7

   ›Reveal solutionSolution

   Using the two altitude conditions $\vec{AH}\perp\vec{BC}$ and $\vec{BH}\perp\vec{AC}$, the third vertex is $(-1,6)$, so ordinate $-$ abscissa $=6-(-1)=7$ — option (D).

   Let $A=(3,-2)$, $B=(-2,3)$, orthocentre $H=(-6,1)$, and third vertex $C=(x,y)$.

   The altitude from $A$ passes through $H$ and is perpendicular to $BC$:

   $\vec{AH} = (-9,\,3),\quad \vec{BC} = (x+2,\;y-3),$

   $\vec{AH}\cdot\vec{BC} = -9(x+2) + 3(y-3) = 0 \;\Rightarrow\; -3x + y - 9 = 0 \;\Rightarrow\; y = 3x+9. \qquad(1)$

   The altitude from $B$ passes through $H$ and is perpendicular to $AC$:

   $\vec{BH} = (-4,\,-2),\quad \vec{AC} = (x-3,\;y+2),$ …

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6. COMEDK 2024Set 2024-A1 markMCQ
   Q.Let $\mathrm{ABC}$ be a triangle with equations of its sides $\mathrm{AB}, \mathrm{BC}$. $\mathrm{CA}$ respectively are $x-2=0, y-5=0$ and $5 x+2 y-10=0$. Then the orthocentre of triangle lies on the line (A) $3 x+y=1$ (B) $x-2 y=1$ (C) $4 x+y=13$ (D) $x-y=0$

   ›Reveal solutionSolution

   The orthocenter is the intersection of two altitudes; by finding the vertices and using the fact that altitudes are perpendicular to opposite sides, we compute the orthocenter and check which given line it satisfies — it lies on x − 2y = 1.

   We are given three side equations:

   AB: $x - 2 = 0$ (vertical line x = 2)

   BC: $y - 5 = 0$ (horizontal line y = 5)

   CA: $5x + 2y - 10 = 0$

   Since AB is vertical and BC is horizontal, triangle ABC is right-angled at B (the intersection of AB and BC). In a right triangle, the orthocenter is simply the vertex at the right angle. That’s the key insight — no need to compute altitudes.

   Let’s verify step by step.

   1. Find the vertices

      • B = intersection of AB (x = 2) and BC (y = 5) → B = (2, 5).
      • A = intersection of AB (x = 2) and CA (5x + 2y − 10 = 0). Substitute x = 2: $5(2) + 2y - 10 = 0 \Rightarrow 10 + 2y - 10 = 0 \Rightarrow 2y = 0 \Rightarrow y = 0$. So A = (2, 0).
      • C = intersection of BC (y = 5) and CA (5x + 2y − 10 = 0). Substitute y = 5: $5x + 2(5) - 10 = 0 \Rightarrow 5x + 10 - 10 = 0 \Rightarrow 5x = 0 \Rightarrow x = 0$. So C = (0, 5).

   2. Identify the right angle

      AB is vertical (x = 2), BC is horizontal (y = 5). They meet at B (2, 5) at a right angle. So ∠B = 90°.

   3. Orthocenter of a right triangle

      In any triangle, the orthocenter is the intersection of the altitudes. In a right triangle, the two legs are altitudes to each other, so the orthocenter is the vertex of the right angle.

      Therefore, orthocenter = B = (2, 5).

   4. Check which line passes through (2, 5)

      • (A) $3x + y = 1$: $3(2) + 5 = 6 + 5 = 11 \neq 1$ → no. …

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7. COMEDK 2024Set 2024-E1 markMCQ
   Q.The line joining two points $A(2,0) B(3,1)$ is rotated about $A$ in anticlockwise direction through an angle of $15^{\circ}$. If $B$ goes to $C$ in the new position, then the coordinates of $C$ is (A) $\left(2+\frac{1}{\sqrt{3}}, \sqrt{\frac{3}{2}}\right)$ (B) $\left(2, \sqrt{\frac{3}{2}}\right)$ (C) $\left(2+\frac{1}{\sqrt{3}}, 1\right)$ (D) $\left(2+\frac{1}{\sqrt{2}}, \sqrt{\frac{3}{2}}\right)$

   ›Reveal solutionSolution

   The problem rotates point B about point A by 15° anticlockwise. Using the rotation matrix and the known coordinates, we find C = (2 + 1/√2, √(3/2)), which matches option (D).

   We start with two points: A(2,0) and B(3,1). The line AB is rotated about A by 15° anticlockwise, so B moves to C. We need C's coordinates.

   Concept & Intuition

   Rotation about a point is easiest if we shift coordinates so that the center of rotation becomes the origin. Here, A is the pivot. So we first find the vector from A to B, rotate that vector by 15° anticlockwise, then add the result back to A's coordinates. The rotation of a vector (x, y) by angle θ anticlockwise is given by the standard rotation matrix:

   $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.$

   Step-by-step solution

   1. Find the vector AB B = (3,1), A = (2,0). So

   $\overrightarrow{AB} = (3-2,\; 1-0) = (1, 1).$

   2. Determine the rotation angle The rotation is 15° anticlockwise, so θ = 15°. We'll need cos 15° and sin 15°. Using known exact values:

   $\cos 15^\circ = \cos(45^\circ - 30^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4},$

   $\sin 15^\circ = \sin(45^\circ - 30^\circ) = \frac{\sqrt{6} - \sqrt{2}}{4}.$

   3. Apply the rotation matrix to (1,1)

   $\begin{pmatrix} \cos 15^\circ & -\sin 15^\circ \\ \sin 15^\circ & \cos 15^\circ \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} \cos 15^\circ - \sin 15^\circ \\ \sin 15^\circ + \cos 15^\circ \end{pmatrix}.$

   Compute each component:

   • x-component:

   $\cos 15^\circ - \sin 15^\circ = \frac{\sqrt{6}+\sqrt{2}}{4} - \frac{\sqrt{6}-\sqrt{2}}{4} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2}.$

   • y-component:

   $\sin 15^\circ + \cos 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4} + \frac{\sqrt{6}+\sqrt{2}}{4} = \frac{2\sqrt{6}}{4} = \frac{\sqrt{6}}{2}.$

   So the rotated vector is

   $\left( \frac{\sqrt{2}}{2},\; \frac{\sqrt{6}}{2} \right).$ …

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8. COMEDK 2024Set 2024-M1 markMCQ
   Q.Find the direction in which a straight line must be drawn through the point $(1,2)$ so that its point of intersection with the line $x+y=4$ may be at a distance of $\sqrt{\frac{2}{3}}$ from this point. (A) $60^{\circ} \text { or } 120^{\circ}$ (B) $50^{\circ} \text { or } 100^{\circ}$ (C) $15^{\circ} \text { or } 75^{\circ}$ (D) $30^{\circ} \text { or } 150^{\circ}$

   ›Reveal solutionSolution

   We find the slope of the line through (1,2) that meets x+y=4 at a point exactly √(2/3) away. Using the distance formula and solving for the angle gives two directions: 15° and 75°. The correct option is (C).

   ---

   Concept & Intuition

   We have a fixed point $P(1,2)$ and a fixed line $L: x+y=4$. Any line through $P$ will intersect $L$ at some point $Q$. The distance $PQ$ depends on the direction (angle) of the line. We want that distance to be exactly $\sqrt{2/3}$.

   Instead of solving for the intersection point directly, we can use the parametric form of a line through $P$:

   $(x,y) = (1 + t\cos\theta,\; 2 + t\sin\theta)$

   where $t$ is the signed distance from $P$ along the direction $\theta$. Then we find $t$ such that the point lies on $x+y=4$, and set $|t| = \sqrt{2/3}$. This gives an equation for $\theta$.

   ---

   Step-by-step solution

   1. Parametrize the line through (1,2) Let the line make an angle $\theta$ with the positive x-axis. Then any point on it is

   $(x,y) = (1 + t\cos\theta,\; 2 + t\sin\theta)$

   where $t$ is the directed distance from $(1,2)$.

   2. Find the intersection with $x+y=4$ Substitute into the line equation:

   $(1 + t\cos\theta) + (2 + t\sin\theta) = 4$

   $3 + t(\cos\theta + \sin\theta) = 4$

   $t(\cos\theta + \sin\theta) = 1$

   So the distance from $P$ to the intersection point is

   $|t| = \frac{1}{|\cos\theta + \sin\theta|}.$

   3. Set this equal to the required distance We need $|t| = \sqrt{\frac{2}{3}}$. Hence

   $\frac{1}{|\cos\theta + \sin\theta|} = \sqrt{\frac{2}{3}}$

   $|\cos\theta + \sin\theta| = \sqrt{\frac{3}{2}}.$

   4. Simplify using a trigonometric identity Recall: $\cos\theta + \sin\theta = \sqrt{2}\sin(\theta + 45^\circ)$. So the equation becomes

   $|\sqrt{2}\sin(\theta + 45^\circ)| = \sqrt{\frac{3}{2}}$

   $\sqrt{2}\,|\sin(\theta + 45^\circ)| = \sqrt{\frac{3}{2}}$

   Multiply both sides by $\sqrt{2}$:

   $2\,|\sin(\theta + 45^\circ)| = \sqrt{3}$

   $|\sin(\theta + 45^\circ)| = \frac{\sqrt{3}}{2}.$

   5. Solve for $\theta$ …

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9. KCET 2023Set A-21 markMCQ
   Q.The point of intersection of the line $x+1 = \frac{y+3}{3} = \frac{-z+2}{2}$ with the plane $3x + 4y + 5z = 10$ is (A) (2, -6, -4) (B) (2, 6, -4) (C) (2, 6, 4) (D) (-2, 6, -4)

   ›Reveal solutionSolution

   Put the line in parametric form, substitute into the plane equation, solve for the parameter and read off the point.

   Step 1 — Read the line correctly.

   $x+1=\frac{y+3}{3}=\frac{-z+2}{2}\quad\Longrightarrow\quad \frac{x+1}{1}=\frac{y+3}{3}=\frac{2-z}{2}=t$

   Note the third fraction has $2-z$ on top, not $z-2$ — this sign is what fixes the $z$-coordinate.

   Step 2 — Parametric coordinates.

   $x=t-1,\qquad y=3t-3,\qquad 2-z=2t\ \Rightarrow\ z=2-2t$

   Step 3 — Impose the plane condition.

   The point of intersection lies on both, so its coordinates must satisfy $3x+4y+5z=10$: …

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10. COMEDK 2023Set 2023-E1 markMCQ
    Q.In a $\triangle A B C$, if coordinates of point $A$ is $(1,2)$ and equation of the medians through $B$ and $C$ are $x+y=5$ and $x=4$ respectively, then the coordinates of $B$ is (A) (4, 1) (B) $(7,-2)$ (C) $(1,4)$ (D) $(-2,7)$

    ›Reveal solutionSolution

    Condition 2 (consistency check) - the median from B passes through the midpoint of AC, which must satisfy x + y = 5: midpoint of AC = ((1 + 4)/2, (2 + c)/2) = (2.5, (2 + c)/2). 2.5 + (2 + c)/2 = 5 => (2 + c)/2 = 2.5 => c = 3, so C = (4, 3) - a consistent point on x = 4.

    Concept: a median through a vertex passes through the midpoint of the opposite side; every vertex lies on the median drawn from it.

    A = (1, 2). Median through B is x + y = 5; median through C is x = 4.

    B lies on its own median: B = (b, 5 - b).

    C lies on its own median: C = (4, c).

    Condition 1 - the median from C passes through the midpoint of AB, and that median is the line x = 4:

    midpoint of AB = ((1 + b)/2, (2 + 5 - b)/2). Its x-coordinate must be 4:

    (1 + b)/2 = 4 => b = 7.

    So B = (7, -2). …

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11. COMEDK 2022Set 20221 markMCQ
    Q.Let the equation of the pair of lines $y=p x$ and $y=q x$ can be written as $(y-p x)(y-q x)=0$. Then the equation of the pair of the angle bisectors of the line $x^2-4 x y-5 y^2=0$ is (A) $x^2-3 x y+y^2=0$ (B) $x^2+4 x y-y^2=0$ (C) $x^2+3 x y-y^2=0$ (D) $x^2-3 x y-y^2=0$

    ›Reveal solutionSolution

    (Sanity check: the given pair factorises as (x − 5y)(x + y) = 0, i.e. lines of slope 1/5 and −1. Their bisectors have slopes satisfying x² + 3xy − y² = 0 → y/x = (3 ± √13)/2, giving m ≈ 3.30 and −0.30, which are indeed perpendicular to each other and bisect the angles between slopes 0.2 and −1. ✓)

    Concept: For the pair ax² + 2hxy + by² = 0, the pair of angle bisectors is (x² − y²)/(a − b) = xy/h.

    Here x² − 4xy − 5y² = 0 → a = 1, 2h = −4 → h = −2, b = −5.

    (x² − y²)/(1 − (−5)) = xy/(−2)

    (x² − y²)/6 = −xy/2

    −2(x² − y²) = 6xy

    −2x² + 2y² − 6xy = 0

    Divide by −2: x² − y² + 3xy = 0 → x² + 3xy − y² = 0. …

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12. KCET 2021Set A-11 markMCQ
    Q.The mid points of the sides of a triangle are (1, 5, $-1$) (0, 4, $-2$) and (2, 3, 4) then centroid of the triangle (A) (1, 4, 3) (B) (1, 4, $\dfrac{1}{3}$) (C) ($-1$, 4, 3) (D) ($\dfrac{1}{3}$, 2, 4)

    ›Reveal solutionSolution

    The triangle formed by the midpoints shares the centroid of the original triangle, so simply average the three given midpoints.

    Step 1 — The key fact. If $A, B, C$ are the vertices and $D, E, F$ the midpoints of the sides, then

    $D = \frac{B+C}{2},\quad E = \frac{C+A}{2},\quad F = \frac{A+B}{2}.$

    Adding,

    $D + E + F = \frac{2(A+B+C)}{2} = A + B + C.$

    Therefore

    $\frac{D+E+F}{3} = \frac{A+B+C}{3},$

    i.e. the centroid of the medial triangle equals the centroid of the original triangle. So we may average the midpoints directly — we never need the vertices.

    Step 2 — Average the three midpoints $(1, 5, -1)$, $(0, 4, -2)$, $(2, 3, 4)$: …

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