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Grade 12 trigonometry notes

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This is a simplied yet detailed summary for grade 10- 12 trigonometry notes. This will help you tackle basic to tricky qiestions

Institution
12th Grade

Content preview

CHAPTER 7




Analytical geometry




7.1 Revision 264
7.2 Equation of a circle 275
7.3 Equation of a tangent to a circle 294
7.4 Summary 305

, 7 Analytical geometry




7.1 Revision EMCHN

Straight line equations EMCHP
y B(x2 ; y2 )
b




b

M (x; y)
x
0
b


A(x1 ; y1 ) C(x2 ; y1 )

Theorem of Pythagoras: AB 2 = AC 2 + BC 2
p
Distance formula: AB = (x2 − x1 )2 + (y2 − y1 )2
y2 −y1
Gradient: mAB = x2 −x1 or mAB = xy11 −y2

x1 +x2 y1 +y2
 −x2
Mid-point of a line segment: M (x; y) = 2 ; 2
Points on a straight line: mAB = mAM = mM B
y
b
(x2 ; y2 )


y − y1 y2 − y1
Two-point form: =
x − x1 x2 − x1 0
x
b


(x1 ; y1 )


y

(x1 ; y1 )
b




Gradient-point form: y − y1 = m(x − x1 )

x
0



y
b
c



Gradient-intercept
y = mx + c
form: x
0

(x1 ; y1 ) b




264 7.1. Revision

, y

b
k


Horizontal lines: y=k
x
0


y




k
Vertical lines: x=k b
x
0




Worked example 1: Revision


QUESTION

Given quadrilateral P QRS with vertices P (0; 3), Q(4; 3), R(5; −1) and S(1; −1).

1. Determine the equation of the lines P S and QR.
2. Show that P S k QR.
3. Calculate the lengths of P S and QR.
4. Determine the equation of the diagonal QS.
5. What type of quadrilateral is P QRS?

SOLUTION

Step 1: Draw a sketch

y

4
P Q
3 b b




2

1

x
−1 0 1 2 3 4 5
−1 b b


S R




Chapter 7. Analytical geometry 265

, Step 2: Use the given information to determine the equation of lines P S and QR
y2 − y1
Gradient: m =
x2 − x1

y − y1 y2 − y1
Two-point form: =
x − x1 x2 − x1

Gradient-intercept form: y = mx + c

Determine the equation of the line P S using the two point form of the straight line
equation:
x1 = 0; y1 = 3; x2 = 1; y2 = −1
y − y1 y2 − y1
=
x − x1 x2 − x1
y−3 −1 − 3
=
x−0 1−0
y−3
= −4
x
y − 3 = −4x
∴ y = −4x + 3


Determine the equation of the line QR using the gradient-intercept form of the straight
line equation:
y2 − y1
mQR =
x2 − x1
−1 − 3
=
5−4
−4
=
1
= −4
y = mx + c
y = −4x + c
Substitute (4; 3) 3 = −4(4) + c
∴ c = 19
y = −4x + 19


There is often more than one method for determining the equation of a line. The
different forms of the straight line equation are used, depending on the information
provided in the problem.

Step 3: Show that line P S and line QR have equal gradients

y = −4x + 3
∴ mP S = −4
And y = −4x + 19
∴ mQR = −4
∴ mP S = mQR
∴ P S k QR



266 7.1. Revision

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Uploaded on
March 24, 2026
Number of pages
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Written in
2025/2026
Type
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Professor(s)
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Contains
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R199,33
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