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
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
