1. Higher Unit 1 Distance Formula The Midpoint Formula Gradients
Collinearity
Gradients of Perpendicular Lines
The Equation of a Straight Line
Median, Altitude & Perpendicular Bisector
Exam Type Questions

2. Starter Questions 1. Calculate the length of the length AC. A 6 m B C
Calculate the coordinates
that are halfway between.
8 m
( 1, 2) and ( 5, 10) (b) ( -4, -10) and ( -2,-6)

3. Distance Formula Length of a straight linex y O
B(x2,y2)
This is just
Pythagoras’ Theorem
y2 – y1
A(x1,y1) C
x2 – x1

4. Distance Formula
The length (distance ) of ANY line can be given by the formula :
Just Pythagoras Theorem in disguise

5. Finding Mid-Point of a line
The mid-point between
2 points is given by x y O
B(x2,y2) y2
Simply add both x coordinates together and divide by 2.
Then do the same with the y coordinates. M
A(x1,y1) y1 x1 x2

6. Another version of the straight line general formula is:
Straight line Facts
y = mx + c
Y – axis Intercept
SLOPE
Another version of the straight line general formula is:
ax + by + c = 0

7. Gradient Facts SLOPE Sloping left to right up has +ve SLOPE
m > 0
Sloping left to right down has -ve SLOPE
m < 0
Horizontal line has zero SLOPE
m = 0
y = c
Vertical line has undefined SLOPE.
x = a
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8. Gradient Facts m = tan θ Lines with the same gradient
means lines are Parallel
m > 0
The gradient of a line is ALWAYS
equal to the tangent of the angle
made with the line and the positive x-axis θ
m = tan θ
21-Apr-17

9. x Collinearity y C A Points are said to be collinear
if they lie on the same straight. A C x y O x1 x2 B
The coordinates A,B C are collinear since they lie on the same straight line.
D,E,F are not collinear they do not lie on the same straight line. D E F

10. SLOPE of Perpendicular Lines
If 2 lines with SLOPES m1 and m2 are perpendicular then m1 × m2 = -1
When rotated through 90º about the origin A (a, b) → B (-b, a) y
B(-b,a) -a
A(a,b) -b O a x -b
Conversely:
If m1 × m2 = -1 then the two lines with slopes m1 and m2 are perpendicular.

11. The Equation of the Straight Line y – b = m (x - a)
The equation of any line can be found if we know
the slope and one point on the line. O y x
P (x, y) m
A (a, b)
y - b y m
m = =
y - b
(x – a)
y - b b
x - a
x – a
Slope, m a x
Point (a, b)
y – b = m ( x – a )
Point on the line ( a, b )

12. Terminology A B C D A B C D Median means a line from vertex
to midpoint of the base. B C D A B C D
Altitude means a perpendicular line
from a vertex to the base.
21-Apr-17

13. Terminology
Perpendicular bisector - a line that cuts another line
into two equal parts at an angle of 90o A B C D
21-Apr-17

14. Typical Exam Questions
Find the equation of the line which passes through the point (-1, 3) and is perpendicular to the line with equation
Find slope of given line:
Find slope of perpendicular:
Find equation:

15. Typical Exam Questions
Find the equation of the straight line which is parallel to the line with equation
and which passes through the point (2, –1).
Find slope of given line:
Knowledge: Slopes of parallel lines are the same.
Therefore for line we want to find has slope
Find equation: Using y – b =m(x - a)

16. Exam Type Questions A and B are the points (–3, –1) and (5, 5).
Find the equation of
a) the line AB.
b) the perpendicular bisector of AB
Find slope of the AB:
Find equation of AB
Find mid-point of AB
Slope of AB (perp):
Use y – b = m(x – a) and point ( 1, 2) to obtain line of perpendicular bisector of AB we get

17. Typical Exam Questions
A triangle ABC has vertices A(4, 3), B(6, 1)
and C(–2, –3) as shown in the diagram.
Find the equation of AM, the median from B to C
Find mid-point of BC:
Find slope of median AM
Find equation of median AM

18. Typical Exam Questions
P(–4, 5), Q(–2, –2) and R(4, 1) are the vertices
of triangle PQR as shown in the diagram.
Find the equation of PS, the altitude from P.
Find gradient of QR:
Find slope of PS (perpendicular to QR)
Find equation of altitude PS

19. Solve p and q simultaneously for intersection
Exam Type Questions p q
Triangle ABC has vertices
A(–1, 6), B(–3, –2) and C(5, 2)
Find:
a) the equation of the line p, the median from C of triangle ABC.
b) the equation of the line q, the perpendicular bisector of BC.
c) the co-ordinates of the point of intersection of the lines p and q.
Find mid-point of AB
(-2, 2)
Find slope of p
Find equation of p
Find mid-point of BC
(1, 0)
Find slope of BC
Find slope of q
Find equation of q
Solve p and q simultaneously for intersection
(0, 2)

20. Exam Type Questions l2 l1 (5, 4) (7, 1)
Triangle ABC has vertices A(2, 2), B(12, 2) and C(8, 6).
a) Write down the equation of l1, the perpendicular bisector of AB
b) Find the equation of l2, the perpendicular bisector of AC.
c) Find the point of intersection of lines l1 and l2.
Mid-point AB
Perpendicular bisector AB
Find mid-point AC
(5, 4)
Find slope of AC
Slope AC perp.
Equation of perp. bisector AC
Point of intersection
(7, 1)

21. Slope of perpendicular AD
Exam Type Questions
A triangle ABC has vertices A(–4, 1), B(12,3) and C(7, –7).
a) Find the equation of the median CM.
b) Find the equation of the altitude AD.
c) Find the co-ordinates of the point of intersection of CM and AD
Mid-point AB
Slope CM (median)
Equation of median CM using y – b = m(x – a)
Slope BC
Slope of perpendicular AD
Equation of AD using y – b = m(x – a)
Solve simultaneously for point of intersection
(6, -4)

22. Hence AB is perpendicular to BC, so B = 90°
Exam Type Questions M
A triangle ABC has vertices A(–3, –3), B(–1, 1) and C(7,–3).
a) Show that the triangle ABC is right angled at B.
b) The medians AD and BE intersect at M.
i) Find the equations of AD and BE. ii) Find find the co-ordinates of M.
Slope AB
Slope BC
Hence AB is perpendicular to BC, so B = 90°
Mid-point BC
Slope of median AD
Equation AD
Mid-point AC
Slope of median BE
Equation BE
Solve simultaneously for M, point of intersection