03.10 Mixed Practice: Geometry & Algebra

Problem structure	Useful method	Essential check
Line through points	Slope and point-slope form; handle vertical lines separately.	Both points satisfy the equation.
Parallel or perpendicular lines	Compare slopes or direction vectors.	Check vertical and horizontal special cases.
Circle in expanded form	Complete the square.	The radius squared must be positive for a real nondegenerate circle.
Line-circle intersection	Substitution or center-to-line distance.	Count distinct real intersection points.
Ellipse or hyperbola	Normalize and identify the standard form.	Use the correct relationship between a, b, and c.
Parabola	Rewrite using the 4p convention.	Check vertex, focus, directrix, and opening direction.
Vector length or angle	Components, magnitude, and dot product.	Angle formulas require nonzero vectors.
Complex-number calculation	Expand using i² = −1; use a conjugate for division.	Write the result in the form a + bi.
Spatial distance	Use all three coordinate differences.	Do not omit the z-coordinate.
Simple solid	Choose the appropriate area or volume formula.	Distinguish radius, diameter, perpendicular height, and slant height.

3. Essential relationships

This is a revision reference. During independent practice, try to select the relationship before consulting it.

3.1 Lines and distances

For two points:

\[ A=(x_1,y_1),\qquad B=(x_2,y_2), \]

the distance is:

\[ AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. \]

If x₂ ≠ x₁, the slope is:

\[ m=\frac{y_2-y_1}{x_2-x_1}. \]

A line through (x₁, y₁) with slope m is:

\[ y-y_1=m(x-x_1). \]

For two nonvertical lines:

• Parallel lines have equal slopes.
• Perpendicular lines satisfy m₁m₂ = −1.

A vertical line is perpendicular to a horizontal line.

The distance from (x₀, y₀) to:

\[ Ax+By+C=0 \]

is:

\[ d=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}, \]

where A and B are not both zero.

3.2 Circle

\[ (x-h)^2+(y-k)^2=r^2. \]

Center: (h, k). Radius: r > 0.

For a line at distance d from the center:

• d < r: two intersections.
• d = r: tangent.
• d > r: no intersections.

3.3 Ellipse

Horizontal major axis:

\[ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1, \qquad a>b>0. \]\[ c^2=a^2-b^2,\qquad e=\frac ca. \]

For a noncircular ellipse: 0 < e < 1.

The major axis length and constant focal-distance sum both equal 2a.

3.4 Hyperbola

Horizontal opening:

\[ \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1. \]\[ c^2=a^2+b^2,\qquad e=\frac ca>1. \]

Asymptotes:

\[ y-k=\pm\frac ba(x-h). \]

For a vertical opening, the y-term is positive and the asymptote slopes are ±a/b.

3.5 Parabola

Using the 4p convention:

\[ (y-k)^2=4p(x-h) \]

has focus (h + p, k) and directrix x = h − p.

\[ (x-h)^2=4p(y-k) \]

has focus (h, k + p) and directrix y = k − p.

3.6 Vectors

For:

\[ \mathbf{u}=(u_1,u_2),\qquad \mathbf{v}=(v_1,v_2), \]\[ |\mathbf{u}|=\sqrt{u_1^2+u_2^2}, \]\[ \mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2. \]

For nonzero vectors:

\[ \cos\theta= \frac{\mathbf{u}\cdot\mathbf{v}} {|\mathbf{u}||\mathbf{v}|}. \]

Nonzero vectors are perpendicular when their dot product is zero.

3.7 Complex numbers

\[ i^2=-1. \]

For z = a + bi:

\[ \overline z=a-bi, \]\[ |z|=\sqrt{a^2+b^2}, \]\[ z\overline z=a^2+b^2. \]

3.8 Spatial coordinates and solids

For A = (x₁, y₁, z₁) and B = (x₂, y₂, z₂):

\[ AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}. \]

Useful volume formulas:

\[ V_{\text{prism}}=Bh, \]\[ V_{\text{pyramid}}=\frac13Bh, \]\[ V_{\text{cylinder}}=\pi r^2h, \]\[ V_{\text{cone}}=\frac13\pi r^2h, \]\[ V_{\text{sphere}}=\frac43\pi r^3. \]

Here B is the base area and h is the perpendicular height.

Surface areas:

\[ S_{\text{sphere}}=4\pi r^2, \]\[ S_{\text{closed cylinder}}=2\pi r^2+2\pi rh. \]

For a right circular cone with slant height l:

\[ S_{\text{total}}=\pi r^2+\pi rl. \]

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4. Worked example 1: Combining a line and a circle

A circle has equation:

\[ x^2+y^2-4x+2y-4=0. \]

Determine whether the line:

\[ 3x+4y-17=0 \]

is tangent to it.

Step 1: Find the center and radius

Complete the square:

\[ (x-2)^2+(y+1)^2=9. \]

Therefore:

Center: (2, −1).
Radius: 3.

Step 2: Find the center-to-line distance

\[ d=\frac{|3(2)+4(-1)-17|}{\sqrt{3^2+4^2}}. \]\[ d=\frac{15}{5}=3. \]

Step 3: Interpret

Since d = r, the line is tangent.

Answer: The line is tangent to the circle.

Method-selection lesson: When only the number of intersections is requested, comparing distance with radius can be faster than solving simultaneously.

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5. Worked example 2: Distinguishing two conics

Find the foci of each curve:

\[ \text{Curve A: }9x^2+25y^2=225, \]\[ \text{Curve B: }9x^2-25y^2=225. \]

Curve A

Normalize:

\[ \frac{x^2}{25}+\frac{y^2}{9}=1. \]

This is an ellipse with a = 5 and b = 3.

\[ c^2=25-9=16. \]

Foci: (−4, 0), (4, 0).

Curve B

Normalize:

\[ \frac{x^2}{25}-\frac{y^2}{9}=1. \]

This is a horizontal hyperbola.

\[ c^2=25+9=34. \]

Foci: (−√34, 0), (√34, 0).

Method-selection lesson: One sign changes the curve type and the focal relationship.

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6. Worked example 3: A quadratic equation as a parabola

Find the vertex, focus, and directrix of:

\[ x^2-4x-8y+12=0. \]

Step 1: Complete the square

\[ x^2-4x=8y-12. \]\[ (x-2)^2=8y-8. \]\[ (x-2)^2=8(y-1). \]

Step 2: Compare with standard form

\[ (x-h)^2=4p(y-k). \]

Thus:

\[ h=2,\qquad k=1,\qquad p=2. \]

Vertex: (2, 1).
Focus: (2, 3).
Directrix: y = −1.

Verification: The vertex is 2 units from both the focus and directrix.

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7. Worked example 4: Vectors and triangle geometry

Let:

• A = (1, 1).
• B = (4, 2).
• C = (0, 4).

Show that the angle at A is a right angle, then find the triangle’s area.

Step 1: Form vectors from the same vertex

\[ \overrightarrow{AB}=(4-1,2-1)=(3,1). \]\[ \overrightarrow{AC}=(0-1,4-1)=(-1,3). \]

Step 2: Calculate the dot product

\[ \overrightarrow{AB}\cdot\overrightarrow{AC} =3(-1)+1(3)=0. \]

Both vectors are nonzero, so they are perpendicular.

Therefore, angle BAC is 90°.

Step 3: Find the side lengths

\[ AB=\sqrt{10},\qquad AC=\sqrt{10}. \]

Step 4: Find the area

Since these sides are perpendicular:

\[ \text{Area}=\frac12(\sqrt{10})(\sqrt{10})=5. \]

Answer: The triangle is right-angled at A and has area 5 square units.

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8. Worked example 5: Complex-number division

Simplify and find the modulus of:

\[ z=\frac{3+4i}{1-2i}. \]

Step 1: Multiply by the denominator’s conjugate

\[ z=\frac{(3+4i)(1+2i)}{(1-2i)(1+2i)}. \]

Step 2: Expand

Numerator:

\[ 3+6i+4i+8i^2=-5+10i. \]

Denominator:

\[ 1-(2i)^2=5. \]

Therefore:

\[ z=-1+2i. \]

Step 3: Find the modulus

\[ |z|=\sqrt{(-1)^2+2^2}=\sqrt5. \]

Verification: Multiply −1 + 2i by 1 − 2i. The result is 3 + 4i.

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9. Worked example 6: Spatial coordinates and a solid

A rectangular cuboid has opposite vertices:

\[ A=(0,0,0),\qquad G=(3,4,12). \]

Its edges are parallel to the coordinate axes. Find its volume and space diagonal.

Step 1: Identify the edge lengths

The edge lengths are 3, 4, and 12.

The condition that the edges are parallel to the coordinate axes is essential.

Step 2: Find the volume

\[ V=3(4)(12)=144. \]

Step 3: Find the space diagonal

\[ AG=\sqrt{3^2+4^2+12^2} =\sqrt{169}=13. \]

Volume: 144 cubic units.
Space diagonal: 13 units.

Method-selection lesson: Volume multiplies three perpendicular edge lengths; diagonal length uses the three-dimensional distance formula.

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10. Worked example 7: Hidden information in a cone

A right circular cone has radius 3 cm and slant height 5 cm. Find its volume.

Step 1: Find the perpendicular height

The radius, perpendicular height, and slant height form a right triangle:

\[ h^2+3^2=5^2. \]\[ h=4. \]

Step 2: Calculate the volume

\[ V=\frac13\pi(3^2)(4)=12\pi. \]

Volume: 12π cm³.

Key lesson: The volume formula uses perpendicular height, not slant height.

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11. Independent mixed practice

Suggested time: 20–25 minutes.

Show your working. Questions intentionally omit topic labels.

Practice 1

Find the equation of the line through (2, −1) and (2, 5).

Practice 2

Find the line through (1, 2) perpendicular to:

\[ 2x-y+3=0. \]

Practice 3

Find the center and radius of:

\[ x^2+y^2+6x-4y-12=0. \]

Practice 4

Find all intersections of:

\[ x+y=4 \]

and:

\[ x^2+y^2=10. \]

Practice 5

For:

\[ \frac{x^2}{36}+\frac{y^2}{20}=1, \]

find the foci and eccentricity.

Practice 6

For:

\[ \frac{y^2}{9}-\frac{x^2}{16}=1, \]

find the vertices, foci, and asymptotes.

Practice 7

Find the vertex, focus, and directrix of:

\[ (y+1)^2=-8(x-3). \]

Practice 8

Given:

\[ \mathbf{u}=(2,-1),\qquad \mathbf{v}=(1,3), \]

find 2u − v and u · v.

Practice 9

Find the angle between:

\[ \mathbf{u}=(1,1),\qquad \mathbf{v}=(1,0). \]

Practice 10

Simplify:

\[ (2-i)(3+2i). \]

Then find the modulus of the result.

Practice 11

Find the midpoint and distance between:

\[ A=(1,-2,3),\qquad B=(5,1,15). \]

Practice 12

A closed cylinder has radius 2 cm and height 5 cm.

Find its volume and total surface area.

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12. Independent practice solutions

Solution 1

Both points have x-coordinate 2.

Equation: x = 2.

The line is vertical; its slope is undefined.

Solution 2

Rewrite the given line:

\[ y=2x+3. \]

Its slope is 2, so the perpendicular slope is −1/2.

Through (1, 2):

\[ y-2=-\frac12(x-1). \]

Equation: x + 2y − 5 = 0.

Solution 3

Complete the square:

\[ (x+3)^2+(y-2)^2=25. \]

Center: (−3, 2).
Radius: 5.

Solution 4

Substitute y = 4 − x:

\[ x^2+(4-x)^2=10. \]\[ 2x^2-8x+6=0. \]\[ (x-1)(x-3)=0. \]

Intersection points: (1, 3), (3, 1).

Solution 5

\[ a=6,\qquad c=\sqrt{36-20}=4. \]

Foci: (−4, 0), (4, 0).

\[ e=\frac46=\frac23. \]

Solution 6

The opening is vertical.

\[ a=3,\qquad b=4,\qquad c=5. \]

Vertices: (0, −3), (0, 3).
Foci: (0, −5), (0, 5).

Asymptotes:

\[ y=\pm\frac34x. \]

Solution 7

\[ h=3,\qquad k=-1,\qquad p=-2. \]

Vertex: (3, −1).
Focus: (1, −1).
Directrix: x = 5.

Solution 8

\[ 2\mathbf{u}-\mathbf{v} =(4,-2)-(1,3)=(3,-5). \]\[ \mathbf{u}\cdot\mathbf{v}=2(1)+(-1)(3)=-1. \]

Solution 9

\[ \cos\theta=\frac{1}{\sqrt2}. \]

Angle: 45°, or π/4 radians.

Solution 10

\[ (2-i)(3+2i)=6+4i-3i-2i^2=8+i. \]\[ |8+i|=\sqrt{65}. \]

Solution 11

Midpoint:

\[ M=\left(\frac{1+5}{2},\frac{-2+1}{2},\frac{3+15}{2}\right). \]

Midpoint: (3, −1/2, 9).

Distance:

\[ AB=\sqrt{4^2+3^2+12^2}=13. \]

Solution 12

Volume:

\[ V=\pi(2^2)(5)=20\pi. \]

Total surface area:

\[ S=2\pi(2^2)+2\pi(2)(5)=28\pi. \]

Volume: 20π cm³.
Total surface area: 28π cm².

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13. Integrated assessment — Geometry & Algebra Mixed Quiz

Suggested time: 20 minutes
Questions: 15 single-answer multiple-choice questions
Marks: 1 per question
Suggested course mastery target: 12/15

Complete the quiz without consulting the reference section.

Question 1

Find the slope of the line through (−1, 2) and (3, −6).

A. 2
B. −2
C. −1/2
D. 1/2

Question 2

Find the distance from (1, 2) to:

\[ 3x+4y-1=0. \]

A. 1
B. 5
C. 10
D. 2

Question 3

Find the center and radius of:

\[ x^2+y^2-2x+6y+6=0. \]

A. Center (1, −3), radius 2
B. Center (−1, 3), radius 2
C. Center (1, −3), radius 4
D. Center (−1, −3), radius 2

Question 4

How many intersection points do the line y = 5 and circle x² + y² = 16 have?

A. One
B. Two
C. None
D. Infinitely many

Question 5

Find the foci of:

\[ \frac{x^2}{49}+\frac{y^2}{24}=1. \]

A. (−7, 0), (7, 0)
B. (−5, 0), (5, 0)
C. (0, −5), (0, 5)
D. (−√73, 0), (√73, 0)

Question 6

Find the eccentricity of:

\[ \frac{x^2}{16}-\frac{y^2}{9}=1. \]

A. 3/4
B. √7/4
C. 4/5
D. 5/4

Question 7

Find the asymptotes of:

\[ \frac{y^2}{25}-\frac{x^2}{9}=1. \]

A. \(y=\pm\frac53x\)

B. \(y=\pm\frac35x\)

C. \(y=\pm\frac{25}{9}x\)

D. \(y=\pm\frac95x\)

Question 8

Find the focus of:

\[ (x-1)^2=-12(y+2). \]

A. (1, 1)
B. (−2, −2)
C. (1, −5)
D. (1, −3)

Question 9

For:

\[ \mathbf{u}=(k,2),\qquad \mathbf{v}=(4,-6), \]

find k if the vectors are perpendicular.

A. −3
B. 3
C. 2
D. 12

Question 10

Find the angle between:

\[ \mathbf{u}=(1,0),\qquad \mathbf{v}=(1,\sqrt3). \]

A. 30°
B. 45°
C. 90°
D. 60°

Question 11

Simplify:

\[ \frac{1+i}{1-i}. \]

A. i
B. −i
C. 1
D. −1

Question 12

If z = 3 − 4i, find:

\[ z\overline z. \]

A. 5
B. −7
C. 25
D. 9 − 16i

Question 13

Find the distance between:

\[ A=(1,2,3),\qquad B=(3,-1,9). \]

A. √35
B. 7
C. 11
D. 49

Question 14

A right circular cone has radius 3 cm and slant height 5 cm.

Find its volume.

A. 15π cm³
B. 45π cm³
C. 9π cm³
D. 12π cm³

Question 15

A sphere has surface area 100π cm².

Find its volume.

A. 500π/3 cm³
B. 100π/3 cm³
C. 125π cm³
D. 500π cm³

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14. Instructor answer key

Tutor LMS placement: Put the assessment in a separate Quiz item. Keep the key and explanations separate from student-facing content until submission.

Question	Answer	Skill assessed
1	B	Line slope
2	D	Point-to-line distance
3	A	Circle standard form
4	C	Line-circle position
5	B	Ellipse foci
6	D	Hyperbola eccentricity
7	A	Hyperbola asymptotes
8	C	Parabola focus
9	B	Perpendicular vectors
10	D	Vector angle
11	A	Complex division
12	C	Complex conjugate product
13	B	Spatial distance
14	D	Cone height and volume
15	A	Sphere area and volume

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15. Assessment solutions

Answer 1: B

\[ m=\frac{-6-2}{3-(-1)}=\frac{-8}{4}=-2. \]

Answer 2: D

\[ d=\frac{|3(1)+4(2)-1|}{5} =\frac{10}{5}=2. \]

Answer 3: A

Complete the square:

\[ (x-1)^2+(y+3)^2=4. \]

Center: (1, −3).
Radius: 2.

Answer 4: C

The circle has center (0, 0) and radius 4.

The distance from the center to y = 5 is 5.

Since 5 > 4, there are no intersections.

Answer 5: B

\[ c^2=49-24=25. \]

The major axis is horizontal.

Foci: (−5, 0), (5, 0).

Answer 6: D

\[ c=\sqrt{16+9}=5. \]\[ e=\frac54. \]

Answer 7: A

The hyperbola is vertical, with a = 5 and b = 3.

\[ y=\pm\frac abx=\pm\frac53x. \]

Answer 8: C

The vertex is (1, −2).

\[ 4p=-12,\qquad p=-3. \]

The focus lies 3 units below the vertex.

Focus: (1, −5).

Answer 9: B

Perpendicular vectors have zero dot product:

\[ 4k+2(-6)=0. \]\[ 4k=12. \]

Answer: k = 3.

Answer 10: D

\[ |\mathbf{u}|=1,\qquad |\mathbf{v}|=2, \qquad \mathbf{u}\cdot\mathbf{v}=1. \]\[ \cos\theta=\frac12. \]

Angle: 60°.

Answer 11: A

\[ \frac{1+i}{1-i} =\frac{(1+i)^2}{(1-i)(1+i)} =\frac{2i}{2}=i. \]

Answer 12: C

\[ z\overline z=3^2+(-4)^2=25. \]

This is the modulus squared, not the modulus.

Answer 13: B

\[ AB=\sqrt{(3-1)^2+(-1-2)^2+(9-3)^2}. \]\[ AB=\sqrt{4+9+36}=7. \]

Answer 14: D

Find the perpendicular height:

\[ h=\sqrt{5^2-3^2}=4. \]

Then:

\[ V=\frac13\pi(3^2)(4)=12\pi. \]

Volume: 12π cm³.

Answer 15: A

From surface area:

\[ 4\pi r^2=100\pi. \]\[ r^2=25,\qquad r=5. \]

Then:

\[ V=\frac43\pi(5^3)=\frac{500\pi}{3}. \]

Volume: 500π/3 cm³.

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16. Error analysis: Identify and repair the mistake

Error A: A vertical line assigned slope zero

A student says that the line through (2, 1) and (2, 5) has slope 0.

Correction: Its slope is undefined because the change in x is zero. Its equation is x = 2.

A horizontal line has slope 0.

Error B: Reading radius squared as radius

A student reads:

\[ (x-1)^2+(y+2)^2=25 \]

and gives radius 25.

Correction: The radius is √25 = 5.

Error C: Using the wrong conic relationship

A student finds the foci of:

\[ \frac{x^2}{25}-\frac{y^2}{9}=1 \]

using c² = 25 − 9.

Correction: This is a hyperbola:

\[ c^2=25+9=34. \]

Error D: Forgetting the factor 4

A student reads x² = 12y and gives focus (0, 12).

Correction:

\[ 4p=12,\qquad p=3. \]

The focus is (0, 3).

Error E: Treating dot product as vector multiplication by components

A student writes:

\[ (2,3)\cdot(4,-1)=(8,-3). \]

Correction: The dot product is a scalar:

\[ 2(4)+3(-1)=5. \]

Error F: Using i² = 1

A student expands a complex product using i² = 1.

Correction: Always use:

\[ i^2=-1. \]

Error G: Using slant height in a cone’s volume

A student substitutes the slant height directly into the cone-volume formula.

Correction: Find the perpendicular height first. For a right circular cone:

\[ l^2=r^2+h^2. \]

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17. Verifying answers efficiently

Use substitution

A claimed point should satisfy the original equation.

A claimed line should pass through the required points.

Check geometric position

• An ellipse’s foci lie inside its major-axis endpoints.
• A hyperbola’s foci lie beyond its vertices.
• A parabola’s focus lies on its opening side.
• A circle’s radius must be positive.

Check numerical limits

For nonzero vectors:

\[ -1\leq\cos\theta\leq1. \]

If a calculation produces a value outside this interval, the arithmetic or formula application is wrong.

Check units

• Distance: units.
• Area: square units.
• Volume: cubic units.

Check scale

If every linear dimension of a solid is multiplied by k:

• Surface area is multiplied by k².
• Volume is multiplied by k³.

For example, doubling a sphere’s radius multiplies its volume by 8, not 2.

Check exactness

Keep square roots and π in exact form unless a decimal approximation is requested.

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18. Review your performance

Assessment questions	Revision focus
1–2	Lines, slopes, and point-to-line distance.
3–4	Circle equations and line-circle position.
5–8	Ellipses, hyperbolas, and parabolas.
9–10	Dot products, perpendicularity, and vector angles.
11–12	Complex arithmetic, conjugates, and modulus.
13–15	Spatial coordinates, solid dimensions, areas, and volumes.