Coordinates and Geometry — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for coordinate geometry.

UT-1: Circle from Three Non-Collinear Points

Question:

Find the equation of the circle passing through the three points $A(1, 2)$, $B(5, 4)$, and $C(3, 8)$.

Express your answer in the form $(x-a)^2 + (y-b)^2 = r^2$ and in the expanded form $x^2 + y^2 + px + qy + r = 0$.

[Difficulty: hard. Tests the perpendicular bisector method and algebraic verification.]

Solution:

Step 1: Find the perpendicular bisector of $AB$.

Midpoint of $AB$: $M_1 = \left(\frac{1+5}{2}, \frac{2+4}{2}\right) = (3, 3)$.

Gradient of $AB$: $m_{AB} = \frac{4-2}{5-1} = \frac{1}{2}$.

Gradient of perpendicular bisector: $m_1 = -2$.

Equation: $y - 3 = -2(x - 3)$, i.e. $y = -2x + 9$.

Step 2: Find the perpendicular bisector of $BC$.

Midpoint of $BC$: $M_2 = \left(\frac{5+3}{2}, \frac{4+8}{2}\right) = (4, 6)$.

Gradient of $BC$: $m_{BC} = \frac{8-4}{3-5} = \frac{4}{-2} = -2$.

Gradient of perpendicular bisector: $m_2 = \frac{1}{2}$.

Equation: $y - 6 = \frac{1}{2}(x - 4)$, i.e. $y = \frac{1}{2}x + 4$.

Step 3: Find the centre (intersection of perpendicular bisectors).

$-2x + 9 = \frac{1}{2}x + 4$ $5 = \frac{5}{2}x$ $x = 2$

$y = -2(2) + 9 = 5$

The centre is $O = (2, 5)$.

Step 4: Find the radius.

$r = |OA| = \sqrt{(2-1)^2 + (5-2)^2} = \sqrt{1 + 9} = \sqrt{10}$

Verification with point $C$: $|OC| = \sqrt{(2-3)^2 + (5-8)^2} = \sqrt{1+9} = \sqrt{10}$. Confirmed.

Step 5: Write the equations.

Standard form: $(x - 2)^2 + (y - 5)^2 = 10$.

Expanded form: $x^2 - 4x + 4 + y^2 - 10y + 25 = 10$, giving:

$x^2 + y^2 - 4x - 10y + 19 = 0$

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UT-2: Circle-Line Tangency via Discriminant

Question:

Find the value of $k$ such that the line $y = kx + 5$ is tangent to the circle $x^2 + y^2 - 4x - 6y + 9 = 0$.

Hence find the coordinates of the point of tangency for each valid value of $k$.

[Difficulty: hard. Tests the discriminant method for tangency and the geometric interpretation.]

Solution:

Step 1: Substitute the line into the circle equation.

$x^2 + (kx+5)^2 - 4x - 6(kx+5) + 9 = 0$ $x^2 + k^2x^2 + 10kx + 25 - 4x - 6kx - 30 + 9 = 0$ $(1+k^2)x^2 + (10k - 4 - 6k)x + 4 = 0$ $(1+k^2)x^2 + (4k - 4)x + 4 = 0$

Step 2: Set the discriminant to zero for tangency.

$\Delta = (4k-4)^2 - 4(1+k^2)(4) = 0$ $16k^2 - 32k + 16 - 16 - 16k^2 = 0$ $-32k = 0$ $k = 0$

Step 3: Find the point of tangency.

With $k = 0$, the line is $y = 5$. Substituting into the circle:

$x^2 + 25 - 4x - 30 + 9 = 0$ $x^2 - 4x + 4 = 0$ $(x-2)^2 = 0$ $x = 2$

The point of tangency is $(2, 5)$.

Geometric check: The circle $(x-2)^2 + (y-3)^2 = 4$ has centre $(2, 3)$ and radius $2$. The line $y = 5$ is at distance $|5-3| = 2$ from the centre, equal to the radius. Confirmed.

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UT-3: Composite Transformation Matrices

Question:

The matrix $R$ represents a rotation anticlockwise by $90°$ about the origin, and the matrix $S$ represents a reflection in the line $y = x$.

$R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad S = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

(a) Find the matrix $RS$ and describe the single transformation it represents.

(b) Find the matrix $SR$ and describe the single transformation it represents.

(c) The point $P(3, 1)$ is first reflected in the line $y = x$ and then rotated $90°$ anticlockwise about the origin. Find the coordinates of the image of $P$.

(d) The triangle with vertices $(0, 0)$, $(4, 0)$, $(0, 3)$ is transformed by $RS$. Find the area of the image and explain why it is preserved.

[Difficulty: hard. Tests understanding that matrix multiplication is not commutative for transformations, and the relationship between determinant and area.]

Solution:

(a)

$RS = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0\cdot0+(-1)\cdot1 & 0\cdot1+(-1)\cdot0 \\ 1\cdot0+0\cdot1 & 1\cdot1+0\cdot0 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$

This is a reflection in the $y$-axis.

(b)

$SR = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0\cdot0+1\cdot1 & 0\cdot(-1)+1\cdot0 \\ 1\cdot0+0\cdot1 & 1\cdot(-1)+0\cdot0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

This is a reflection in the $x$-axis.

Note: $RS \neq SR$, confirming that transformations do not generally commute.

(c) Applying $S$ first, then $R$: the image is $R \cdot S \cdot \begin{pmatrix} 3 \\ 1 \end{pmatrix}$.

$S\begin{pmatrix}3\\1\end{pmatrix} = \begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}3\\1\end{pmatrix} = \begin{pmatrix}1\\3\end{pmatrix}$

$R\begin{pmatrix}1\\3\end{pmatrix} = \begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}1\\3\end{pmatrix} = \begin{pmatrix}-3\\1\end{pmatrix}$

The image of $P$ is $(-3, 1)$.

Alternatively, using $RS = \begin{pmatrix}-1&0\\0&1\end{pmatrix}$:

$RS\begin{pmatrix}3\\1\end{pmatrix} = \begin{pmatrix}-3\\1\end{pmatrix}$

Confirmed.

(d) The original triangle has area $\frac{1}{2} \times 4 \times 3 = 6$ square units.

The determinant of $RS$ is $\det\begin{pmatrix}-1&0\\0&1\end{pmatrix} = -1$.

The absolute value of the determinant gives the area scale factor: $|\det(RS)| = 1$. Therefore the area of the image is $6$ square units.

Both $R$ (rotation) and $S$ (reflection) are isometries (distance-preserving transformations). Their composite is also an isometry, so area is preserved.

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Integration Tests

Tests synthesis of coordinate geometry with other topics. Requires combining concepts from multiple units.

IT-1: Closest Point on a Curve Using Calculus (with Differentiation)

Question:

Find the point on the curve $y = x^2 - 4x + 7$ that is closest to the point $(1, 8)$.

[Difficulty: hard. Combines distance formula with differentiation to find the minimum distance.]

Solution:

Step 1: Set up the distance squared function. (Using distance squared avoids the square root and gives the same minimiser.)

A general point on the curve is $(x, x^2-4x+7)$.

$D^2 = (x - 1)^2 + (x^2 - 4x + 7 - 8)^2 = (x-1)^2 + (x^2-4x-1)^2$

Step 2: Differentiate with respect to $x$.

$\frac{d(D^2)}{dx} = 2(x-1) + 2(x^2-4x-1)(2x-4)$

Set this equal to zero:

$(x-1) + (x^2-4x-1)(2x-4) = 0$

$(x-1) + 2(x-2)(x^2-4x-1) = 0$

Expand $(x-2)(x^2-4x-1) = x^3 - 4x^2 - x - 2x^2 + 8x + 2 = x^3 - 6x^2 + 7x + 2$.

$(x-1) + 2(x^3 - 6x^2 + 7x + 2) = 0$ $x - 1 + 2x^3 - 12x^2 + 14x + 4 = 0$ $2x^3 - 12x^2 + 15x + 3 = 0$

Step 3: Solve the cubic. Testing $x = 1$: $2 - 12 + 15 + 3 = 8 \neq 0$. Testing $x = -1$: $-2 - 12 - 15 + 3 = -26$. Testing $x = 3$: $54 - 108 + 45 + 3 = -6$. Testing $x = \frac{1}{2}$: $\frac{1}{4} - 3 + \frac{15}{2} + 3 = 7.25$. Testing $x = 0$: $3$.

By the intermediate value theorem, there is a root between $x = -1$ and $x = 0$, and possibly others.

Testing $x = -\frac{1}{2}$: $-\frac{1}{4} - 3 - \frac{15}{2} + 3 = -\frac{1}{4} - \frac{15}{2} = -\frac{31}{4}$. Between $x = 0$ ($D' = 3$) and $x = 1$ ($D' = 8$), no root. Between $x = 3$ ($D' = -6$) and $x = 4$ ($D' = 128-192+60+3 = -1$), and $x = 5$ ($D' = 250-300+75+3=28$), so a root between $x = 4$ and $x = 5$.

Actually, let me check $x = \frac{3}{2}$: $\frac{27}{4} - 27 + \frac{45}{2} + 3 = 6.75 - 27 + 22.5 + 3 = 5.25$. Between $x = 1$ ($8$) and $x = 3$ ($-6$), a root near $x = 2$: $16 - 48 + 30 + 3 = 1$. At $x = 2.1$: $18.522 - 52.92 + 31.5 + 3 = 0.102$. At $x = 2.12$: $19.08... - 53.95... + 31.8 + 3 = -0.06...$.

Root near $x \approx 2.11$.

The exact solution requires the cubic formula. For the closest point, we accept the numerical approximation. At $x \approx 2.11$: $y \approx 4.45 - 8.44 + 7 = 3.01$.

The closest point is approximately $(2.11, 3.01)$.

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IT-2: Vector Method for a Geometry Problem (with Vectors)

Question:

In triangle $ABC$, the point $D$ lies on $BC$ such that $BD : DC = 2 : 1$. The point $E$ is the midpoint of $AC$. The lines $AD$ and $BE$ intersect at point $F$.

Using position vectors with origin at $A$, and taking $\overrightarrow{AB} = \mathbf{b}$ and $\overrightarrow{AC} = \mathbf{c}$:

(a) Find the position vector of $F$ in terms of $\mathbf{b}$ and $\mathbf{c}$.

(b) Find the ratio $AF : FD$.

(c) If $|\mathbf{b}| = 5$, $|\mathbf{c}| = 7$, and $\mathbf{b} \cdot \mathbf{c} = 15$, find $|\overrightarrow{AF}|$.

[Difficulty: hard. Tests vector methods for concurrency problems and application of the dot product.]

Solution:

(a) Express all points in terms of $\mathbf{b}$ and $\mathbf{c}$:

• $\overrightarrow{OB} = \mathbf{b}$, $\overrightarrow{OC} = \mathbf{c}$ (since origin is at $A$)
• $\overrightarrow{OD} = \overrightarrow{OB} + \frac{2}{3}\overrightarrow{BC} = \mathbf{b} + \frac{2}{3}(\mathbf{c} - \mathbf{b}) = \frac{1}{3}\mathbf{b} + \frac{2}{3}\mathbf{c}$
• $\overrightarrow{OE} = \frac{1}{2}\mathbf{c}$

Point $F$ lies on $AD$: $\overrightarrow{OF} = \overrightarrow{OA} + s\overrightarrow{AD} = s\left(\frac{1}{3}\mathbf{b} + \frac{2}{3}\mathbf{c}\right)$ for some $0 \leq s \leq 1$.

Point $F$ also lies on $BE$: $\overrightarrow{OF} = \overrightarrow{OB} + t\overrightarrow{BE} = \mathbf{b} + t\left(\frac{1}{2}\mathbf{c} - \mathbf{b}\right) = (1-t)\mathbf{b} + \frac{t}{2}\mathbf{c}$ for some $0 \leq t \leq 1$.

Equating coefficients of $\mathbf{b}$ and $\mathbf{c}$:

$\frac{s}{3} = 1 - t \quad \text{and} \quad \frac{2s}{3} = \frac{t}{2}$

From the second equation: $t = \frac{4s}{3}$.

Substituting into the first: $\frac{s}{3} = 1 - \frac{4s}{3}$, giving $\frac{5s}{3} = 1$, so $s = \frac{3}{5}$.

$\overrightarrow{OF} = \frac{3}{5}\left(\frac{1}{3}\mathbf{b} + \frac{2}{3}\mathbf{c}\right) = \frac{1}{5}\mathbf{b} + \frac{2}{5}\mathbf{c}$

(b) $AF : FD$. Since $s = 3/5$, point $F$ divides $AD$ in the ratio $s : (1-s) = 3/5 : 2/5 = 3 : 2$.

So $AF : FD = 3 : 2$.

(c)

$\overrightarrow{AF} = \frac{1}{5}\mathbf{b} + \frac{2}{5}\mathbf{c}$

$|\overrightarrow{AF}|^2 = \left(\frac{1}{5}\mathbf{b} + \frac{2}{5}\mathbf{c}\right) \cdot \left(\frac{1}{5}\mathbf{b} + \frac{2}{5}\mathbf{c}\right)$

$= \frac{1}{25}|\mathbf{b}|^2 + \frac{4}{25}\mathbf{b}\cdot\mathbf{c} + \frac{4}{25}|\mathbf{c}|^2$

$= \frac{1}{25}(25) + \frac{4}{25}(15) + \frac{4}{25}(49)$

$= 1 + \frac{60}{25} + \frac{196}{25}$

$= 1 + \frac{256}{25} = \frac{281}{25}$

$|\overrightarrow{AF}| = \frac◆LB◆\sqrt{281}◆RB◆◆LB◆5◆RB◆$

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IT-3: Parametric Curve via Function Composition (with Functions)

Question:

A curve is defined parametrically by:

$x = t^2 - 2t, \quad y = (t^2 - 2t)^2 - 2(t^2 - 2t)$

for $t \in \mathbb{R}$.

(a) Show that $y = x^2 - 2x$ and hence identify the curve as a standard form.

(b) Find the coordinates of the turning points of the curve and determine their nature.

(c) The curve is translated so that its minimum point moves to the origin. Find the equation of the translated curve in the form $y = ax^2 + bx + c$.

[Difficulty: hard. Tests recognition of function composition in parametric equations and completing the square.]

Solution:

(a) Let $f(t) = t^2 - 2t = (t-1)^2 - 1$.

Then $x = f(t)$ and $y = f(f(t)) = f(x)$.

Therefore $y = x^2 - 2x = (x-1)^2 - 1$.

This is a parabola with vertex at $(1, -1)$, opening upward.

(b) $y = x^2 - 2x$. Since $x = (t-1)^2 - 1 \geq -1$ for all $t$, the domain of the curve is $x \geq -1$.

$\frac{dy}{dx} = 2x - 2$. Setting this to zero: $x = 1$.

At $x = 1$: $y = 1 - 2 = -1$. This is the vertex.

$\frac{d^2y}{dx^2} = 2 > 0$, confirming a minimum at $(1, -1)$.

The turning point is $(1, -1)$, which is a minimum.

Note: there are no other turning points. The curve is a standard parabola restricted to $x \geq -1$.

At $x = -1$ (when $t = 1$): $y = 1 + 2 = 3$. This is an endpoint of the parametric curve, not a turning point of the parabola itself.

(c) The translation moves $(1, -1)$ to $(0, 0)$. The translation vector is $(-1, 1)$.

Let $X = x - 1$ and $Y = y + 1$. Then $y = (x-1)^2 - 1$ becomes $Y - 1 = X^2 - 1$, i.e.:

$Y = X^2$

In terms of the original variables (after translation, using $x$ and $y$ for the new coordinates):

$y = x^2$

So $a = 1$, $b = 0$, $c = 0$.