MAT 104
ELEMENTARY MATHEMATICS III
3 CREDITS
CO-ORDINATE GEOMETRY (CONT’D)
Recall from the last class, we considered that the
1. Distance between two points 𝐴(𝑥1 , 𝑦1 ) and 𝐵(𝑥2 , 𝑦2 ) is given by
√(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
2. Midpoint of two points
𝑥1 + 𝑥2 𝑦1 + 𝑦2
(
2
,
2
𝐴(𝑥1 , 𝑦1 )
and
𝐵(𝑥2 , 𝑦2 )
is
given
by
)
3. Coordinates of the point dividing the join of 𝐴(𝑥1 , 𝑦1 ) and 𝐵(𝑥2 , 𝑦2 ) in the
𝑐𝑥2 + 𝑥1 𝑐𝑦2 + 𝑦1
ratio 𝑐: 1 are (
𝑐+1
,
𝑐+1
) provided that 𝑐 ≠ 1
EXAMPLES
1. 𝑃 and 𝑄 divide the line 𝐴𝐵, 𝐴(−1, 2) and 𝐵(3, 4) internally and externally
respectively in the ratio 3: 2. Find the coordinates of 𝑃 and 𝑄
Solution
𝑃 divides the line 𝐴𝐵 internally, thus the ratio of division is
3
3
(3) + (−1) (4)+ (2)
2
2
3
3
+1
+1
2
2
Therefore 𝑃 = (
,
7
16
5
5
=( ,
3
2
∶1
)
)
3
𝑄 divides the line 𝐴𝐵 externally, thus the ratio of division is − ∶ 1
2
Therefore 𝑄 = (
3
− 2 (3) + (−1)
3
− 2 +1
,
3
−2 (4)+ (2)
3
− 2 +1
)
= (11, 8)
2. Find the ratios in which 𝑃(2, 3) and 𝑄(−12, −11) divide the line joining
𝐴(−2, −1) and 𝐵(3, 4)
Solution
For the point 𝑃, let the ratio of division be 𝑝: 1
Thus using the 𝑥 − 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒, we have 2 =
𝑝(3) + (−2)
𝑝+ 1
2𝑝 + 2 = 3𝑝 − 2
𝑝=4
Hence the point 𝑃 divides line 𝐴𝐵 in the ratio 4: 1
Similarly for the point 𝑄, let the ratio of division be 𝑞: 1
Thus using the 𝑦 − 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒, we have −11 =
𝑞(4) + (−1)
𝑞+ 1
−11𝑞 − 11 = 4𝑞 − 1
𝑞=−
2
3
Hence the point 𝑄 divides line 𝐴𝐵 in the ratio −2: 3
GRADIENTS
The gradient (or slope) – 𝑚 of a line is defined as the ratio
𝑦−𝑠𝑡𝑒𝑝
𝑥−𝑠𝑡𝑒𝑝
in going from
one point to another on the line. That is, the gradient (or slope) of the line joining
𝐴(𝑥1 , 𝑦1 ) and 𝐵(𝑥2 , 𝑦2 ) is
𝑦2 − 𝑦1
𝑚=
𝑥2 − 𝑥1
Similarly, if the line joining the points 𝐴(𝑥1 , 𝑦1 ) and 𝐵(𝑥2 , 𝑦2 ) makes an angle 𝜃
with the (positive) 𝑥 − 𝑎𝑥𝑖𝑠, then 𝜃 is the angle of slope of the line.
Thus 𝑚 = tan 𝜃 ⟹ tan 𝜃 =
𝑦2 − 𝑦1
𝑥2 − 𝑥1
EXAMPLE
1. Find the gradient of the line joining (3, −2) and (−5, −8) and the angle the
line make with the 𝑥 − 𝑎𝑥𝑖𝑠.
Solution
Gradient 𝑚 =
Thus 𝑚 =
𝑦2 − 𝑦1
𝑥2 − 𝑥1
−8 –(−2)
−5 − 3
=
3
4
Since tan 𝜃 = 𝑚 ⟹ tan 𝜃 =
𝜃 = tan−1
𝜃 = 36.87°
3
4
3
4
COLLINEAR POINTS
Three or more points are said to be collinear, if they lie on the same straight line
which has a constant gradient.
Examples
1. The points 𝐴(2, 5), 𝐵(−1, −1) and 𝐶(−4, −7) are collinear since
Gradient 𝐴 to 𝐵 =
Gradient 𝐵 to 𝐶 =
Gradient 𝐴 to 𝐶 =
−1 −5
−1 −2
−7+ 1
−4 +1
−7 −5
−4 −2
=
=
=
−6
−2
−6
−3
−12
−6
=2
=2
=2
2. If the points (−1, 𝑡 − 1), (𝑡, 𝑡 − 3) and (𝑡 − 6, 3) are collinear, find the
possible values of 𝑡.
(𝑡−3)− (𝑡−1)
𝑡−(−1)
−2
𝑡+1
=
=
3−(𝑡−3)
(𝑡−6)− 𝑡
6−𝑡
−6
12 = (6 − 𝑡)(𝑡 + 1)
𝑡 2 − 5𝑡 + 6 = 0
𝑡 = 2 or 𝑡 = 3
EQUATION OF A STRAIGHT LINE
The equation of any line can be written in the form 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0 and this is
the general linear equation. Therefore the equation of a line can be derived using
any of the following methods.
I.
Gradient-intercept form: 𝑦 = 𝑚𝑥 + 𝑐, where 𝑚 is the gradient and 𝑐 is
the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡.
Examples
2
1. Find the equation of a line with gradient of and 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 of 5
3
Solution
Using 𝑦 = 𝑚𝑥 + 𝑐
We have 𝑦 =
2
3
𝑥 + 5 ⟹ 3𝑦 = 2𝑥 + 10
2. Find the equation of the line that cuts the 𝑦 − 𝑎𝑥𝑖𝑠 at (0, −2) and has a
1
gradient of − .
7
Solution
Using 𝑦 = 𝑚𝑥 + 𝑐
We have 𝑦 = −
II.
1
7
𝑥 + (−2) ⟹ 7𝑦 = −𝑥 − 14
Gradient and one point form: 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) where 𝑚 is the
gradient and (𝑥1 , 𝑦1 ) is any given fixed point on the line.
Example
Find the equation of the line which has gradient 3 and passes through the
point (−2, 5)
Solution
Using 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
We have 𝑦 − 5 = −2(𝑥 − [−2])
𝑦 − 5 = −2(𝑥 + 2) ⟹ 𝑦 = −2𝑥 − 4 + 5
𝑦 = 1 − 2𝑥
III.
Two point form: Given any two points 𝐴(𝑥1 , 𝑦1 ) and 𝐵(𝑥2 , 𝑦2 ), the
equation of such a line is given by 𝑦 − 𝑦1 =
𝑦2 − 𝑦1
𝑥2 − 𝑥1
(𝑥 − 𝑥1 )
Example
Find the equation of the line 𝐴𝐵 which passes through the points
𝐴(−3, −2) and 𝐵 (−2, 5)
Solution
Using 𝑦 − 𝑦1 =
𝑦2 − 𝑦1
𝑥2 − 𝑥1
We have, 𝑦 − (−2) =
𝑦 + 2 = 7(𝑥 + 3)
𝑦 = 7𝑥 + 19
(𝑥 − 𝑥1 )
5− (−2)
−2 –(−3)
(𝑥 − (−3))
ANGLE BETWEEN TWO LINES
If two lines 𝐿1 and 𝐿2 have gradients 𝑚1 and 𝑚2 and makes an angle 𝜃1 and 𝜃2
with the positive 𝑥 − 𝑎𝑥𝑖𝑠 respectively. Then, 𝑚1 = tan 𝜃1 and 𝑚2 = tan 𝜃2 .
Then the angle 𝜙 between the two lines is thus given by
tan 𝜙 = tan(𝜃2 − 𝜃1 )
=
=
tan 𝜃2 − tan 𝜃1
1+ tan 𝜃2 tan 𝜃1
𝑚2 − 𝑚 1
1+ 𝑚2 𝑚1
Hence, the angle between two lines of gradient is given by tan 𝜙 = |
where 𝑚2 𝑚1 ≠ −1
Example
Find the angle between the lines 2𝑦 = 3𝑥 − 8 and 5𝑦 = 𝑥 + 7
Solution
2𝑦 = 3𝑥 − 8
3
3
2
2
𝑦 = 𝑥 − 4 ⟹ 𝑚1 =
Also 5𝑦 = 𝑥 + 7
1
7
5
5
𝑦= 𝑥−
Thus, tan 𝜙 = |
tan 𝜙 = |
−13
10
13
10
tan 𝜙 = |−1|
𝜙 = tan−1 1
𝜙 = 45°
1
⟹ 𝑚2 =
|
1 3
−
5 2
1 3
1+(5)(2)
5
|
𝑚2 − 𝑚1
1+ 𝑚2 𝑚1
|
Parallel and Perpendicular Lines
Parallel lines have the same gradient, as they make equal angles with the positive
𝑥 − 𝑎𝑥𝑖𝑠 i. e. angle between two parallel lines is zero, which means tan 0 = 0,
thus two parallel lines 𝐿1 and 𝐿2 having gradients 𝑚1 and 𝑚2 ,
𝑚1 = 𝑚2
For perpendicular lines angle between then is 90°, since tan 90° = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑, it
implies that 1 + 𝑚2 𝑚1 = 0. So for two perpendicular lines 𝐿1 and 𝐿2 have
gradients 𝑚1 and 𝑚2
𝑚1 𝑚2 = 1 ⟹ 𝑚1 = −
1
𝑚2
Examples
1. Find the equation of the line passing through the point (−1, 3) which is
parallel to 𝑦 = 3𝑥 + 1
Solution
From 𝑦 = 3𝑥 + 1 and comparing with 𝑦 = 𝑚𝑥 + 𝑐, gradient 𝑚 = 3
Thus using the one point form [𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )] given that the point is
parallel to the equation since 𝑚 = 3 and the given point is (−1, 3)
𝑦 − 3 = 3(𝑥 + 1)
𝑦 = 3𝑥 + 6
2. Find the equation of the line passing through the point (−1, 3) which is
perpendicular to 𝑦 = 3𝑥 − 1
Solution
From 𝑦 = 3𝑥 − 1 and comparing with 𝑦 = 𝑚𝑥 + 𝑐, gradient 𝑚 = 3
Thus using the one point form [𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )] given that the point is
perpendicular to the equation and since 𝑚1 = 3, then the gradient of the
new equation would be 𝑚2 = −
1
𝑦 − 3 = − (𝑥 + 1)
3
3𝑦 = 8 − 𝑥
1
3
and the given point is (−1, 3)
Intersection of lines
If two lines 𝐿1 and 𝐿2 intersect, the coordinates of the point it intersection is
obtained by solving the equations of the two lines simultaneously.
Example
Find the coordinates of the point of intersection of the lines 2𝑦 + 3𝑥 = 16 and
7𝑦 = 2𝑥 + 6
Solution
2𝑦 + 3𝑥 = 16 ⟹ 4𝑦 + 6𝑥 = 32
7𝑦 − 2𝑥 = 6 ⟹ 21𝑦 − 6𝑥 = 18
Solving by elimination 25𝑦 = 50 ⟹ 𝑦 = 2
Substituting for 𝑦 we have 2(2) + 3𝑥 = 16 ⟹ 4 + 3𝑥 = 16, 3𝑥 = 12 ⟹ 𝑥 = 4
Hence the coordinates point if intersection of the lines is (4, 2)