Scholar
Higher Mathematics
Homework Session
Thursday 29th January 7:30pm
You will need a pencil, paper and a calculator for some of
the activities
Margaret Ferguson
SCHOLAR online tutor for Maths
and
Author of the new SCHOLAR
National 5 & Higher Maths courses
Tonight’s Revision Session
will cover
The Straight Line
and
The Circle
The Straight Line
things you show know
(x2 - x1 )2 - (y2 - y1 )2
The distance formula is D =
The gradient formula is m = y - y
x -x
y
A horizontal line has gradient 0
The equation of a horizontal straight line which b
passes through the point (a,b) is y = b
A vertical straight line has an undefined gradient
The equation of a vertical straight line which passes
through the point (a,b) is x = a
The equation of a straight line with gradient m and
y-intercept (0,c) is y = mx + c
The equation of a straight line with gradient m and
passes through the point (a,b) is y - b = m(x - a)
2
1
2
1
a
x
The Straight Line
more things you must know
The general form of the equation is Ax + By + C = 0
To be able to identify the gradient or y-intercept:
1.
2.
the equation of a straight line must be in the form y = mx + c
if the equation is not in that form it must be re-arranged
Parallel lines have equal gradients and distinct y-intercepts
A set points are Collinear if they all lie on a single straight line:
1.
2.
3.
find the gradients of the lines joining pairs of points
equal gradients indicate parallel lines
a common point determines collinearity
m = tan θ where θ is the angle between a line and the positive
direction of the x-axis
Lines with gradients m1 & m2 are perpendicular if m1 × m2 = -1
æ x2 + x1 y2 + y1 ö
The midpoint formula is çè 2 , 2 ÷ø
y
What is the equation of the line l2
which passes through the point (8,-2)
and is perpendicular to the line l1.
p
4
??
x
l1
1. Find the gradient of the line l1?
What is the given angle in degrees? 45°
What is the angle between l1 and the positive direction of the x-axis?
What is the gradient of the line l1?
ml1 = tan135° = -1
135°
2. Find the equation of the line l2?
What is the gradient of l2?
Why?
What is the equation of the line l2?
ml2 = 1
for perpendicular lines ml1 x ml2 = -1
y + 2 = 1(x – 8)
y = x - 10
B
Triangles: A Median
A
C
A median of a triangle is a line from a vertex to the midpoint on
the opposite side
The medians of a triangle are concurrent and the point of
intersection is called the centroid.
If the points M(a, b), N(c, d) and P(e, f) are the vertices of a
triangle then the coordinates of the centroid are æçè a + c + e , b + d + f ö÷ø
3
3
The centroid lies two thirds of the distance along each median
measured from its vertex.
Altitudes and Perpendicular Bisectors
B
An altitude of a triangle is a line from a vertex, which
A
is perpendicular to the opposite side.
C
The altitudes of a triangle are concurrent and
the point of intersection is called the
orthocentre
C
The perpendicular bisector of a side in a triangle is
the line which is perpendicular through the midpoint
of the side.
A
B
The perpendicular bisectors of the sides of a
triangle are concurrent and the point of intersection
is called the circumcentre.
You must know which is which : median, altitude and perpendicular bisector.
The Triangle ABC has vertices A(-1,6), B(-3,-2) and C(5,2).
(a) What is the equation of the line p, the median from C?
What is a Median?
What are the coordinates of R, the midpoint of AB? R(-2,2)
What is the gradient of CR? mCR = 0, CR is a horizontal line
Look at the coordinates of C and R to identify the equation of p (the median from C).
y=2
(b) What is the equation of the line q, the altitude from A?
What is an Altitude?
What is the gradient of BC? mBC = ½
What is the gradient of the line perpendicular to BC? mperp = -2
Now find the equation of the line q (the altitude from A).
y - 6 = -2(x + 1)
y = -2x + 4
(c) What are the coordinates of the point of intersection of the lines p and q?
Use simultaneous equations with your answers to the previous parts of this question.
2 = -2x + 4
-2 = -2x
x=1
The point of intersection of the lines p and q is (1,2)
A line l has equation 3y + 2x = 6.
What is the gradient of any line parallel to l?
Vote for the correct answer now
A
B
C
D
-2
-⅔
1.5
2
✓
Solution:
3y + 2x = 6
3y = -2x + 6
2
6
y=- x+
3
3
Remember gradients of parallel lines are
The Equation of the Circle
The equation of a circle with centre (0,0) and radius r is
x2 + y2 = r2
The equation of a circle with centre (a, b) and radius r is
(x - a)2 + (y - b)2 = r2
The general equation of the circle is
x2 + y2 + 2gx + 2fy + c = 0
with centre = (-g, -f)
and radius = g2 + f 2 - c
Lines & Circles
The relationship between a line and a circle can be found by
1 substituting the equation of the line in the equation of the circle
2 collecting like terms to obtain a quadratic equation
3 evaluating and interpreting the discriminant
• if b2 – 4ac = 0 then the line is a tangent to the circle
• if b2 – 4ac > 0 then there are 2 points of intersection
• if b2 – 4ac < 0 then the line does not meet the circle at all
Any points of intersection can be found by solving the quadratic
B
It is useful to remember that the angle in a
semi-circle is a right angle.
C
A
Intersecting circles
Circles which do not touch can be:
1. one inside the other
2. or completely apart
Circles can touch:
1. internally
2. or externally
Circles may intersect
The distance between the centres of circles and the sum of their radii
can help to determine the relationship between the circles
The journey between the coordinates of centres or points of contact
can help in problem solving situations
A Circle has equation x2 + y2 + 8x + 6y – 75 = 0.
What is the radius of the circle?
Vote for the correct answer now
A
5
B
10
C
75
D
175
The general equation of the circle takes the form
✓
x2 + y2 + 2gx + 2fy + c = 0
2g = 8 so g = 4
2f = 6 so f = 3
radius =
g2 + f 2 - c
so r = 4 + 3 - (-75)
= √100 = 10
2
2
Show that the line with equation y = 3 – x is a tangent to the circle
with equation x2 + y2 + 14x + 4y – 19 = 0 and find the coordiantes
of the point of contact P.
Substitute the equation of the line into x2 + y2 + 14x + 4y – 19 = 0
x2 + (3 – x)2 + 14x + 4(3 – x) – 19 = 0
x2 + (9 – 6x + x2) + 14x + (12 – 4x) – 19 = 0
Collect like terms to obtain a quadratic equation
2x2 + 4x + 2 = 0
Evaluate and interpret the discriminant
b2 – 4ac = 42 – 4(2)(2)
= 16 – 16
=0
Since b2 – 4ac = 0 the line y = 3 – x is a tangent to the circle.
Find the coordinates of the point of contact
2x2 + 4x + 2 = 0
2(x2 + 2x + 1) = 0
(x +1)(x +1) = 0
x = -1
Let x = -1 in the equation of the tangent
y = 3 – (-1) = 4
Hence P has coordinates (-1,4).
Relative to a suitable set of coordinate axes, the diagram below shows the
circle from the previous question an a smaller circle with centre C.
The line y = 3 – x is a common tangent at the point P.
The radius of the larger circle is 3 times that of the smaller circle.
Find the equation of the smaller circle.
The equation of the larger circle is
x2 + y2 + 14x + 4y – 19 = 0
so the centre is (-7,-2)
and the radius of the larger circle is
(-1,4)
(-7,-2)
2 2
6
6
72 + 22 - (-19) = 72 = 6 2
6 2
=2 2
3
Use the diagram to find the coordinates of the centre C. (1,6)
Double check the distance from P to C. 22 + 22 = 8 or 2 2
Hence the equation of the smaller circle is (x - 1)2 + (y – 6)2 = 8
The radius of the smaller circle is ⅓ of the larger circle
Question Time
• If you have any questions about tonight’s session please ask
• I can only answer one question at a time so take turns
• The next session will be on 12th February at 7:30pm
• The topics covered will be Integration & Vectors
• Carol will provide a link for you to give us feedback