 # Scholar Higher Mathematics Homework Session

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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 The distance formula is D = The gradient formula is m = A horizontal line has gradient 0 The equation of a horizontal straight line which 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) y b x a

more things you must know
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: 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: 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 The midpoint formula is

1. Find the gradient of the line l1?
y What is the equation of the line l2 which passes through the point (8,-2) and is perpendicular to the line l1. ?? x 1. Find the gradient of the line l1? l1 What is the given angle in degrees? What is the angle between l1 and the positive direction of the x-axis? What is the gradient of the line l1? 45° 135° ml1 = tan135° = -1 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 The centroid lies two thirds of the distance along each median measured from its vertex.

Altitudes and Perpendicular Bisectors
An altitude of a triangle is a line from a vertex, which is perpendicular to the opposite side. A C The altitudes of a triangle are concurrent and the point of intersection is called the orthocentre A B C The perpendicular bisector of a side in a triangle is the line which is perpendicular through the midpoint of the side. 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? What is the gradient of CR? Look at the coordinates of C and R to identify the equation of p (the median from C). (b) What is the equation of the line q, the altitude from A? What is an Altitude? What is the gradient of BC? What is the gradient of the line perpendicular to BC? Now find the equation of the line q (the altitude from A). (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. R(-2,2) mCR = 0, CR is a horizontal line y = 2 mBC = ½ mperp = -2 y - 6 = -2(x + 1) y = -2x + 4 2 = -2x + 4 -2 = -2x x = 1 The point of intersection of the lines p and q is (1,2)

Remember gradients of parallel lines are equal!!
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 -⅔ C D 2 Solution: 3y + 2x = 6 3y = -2x + 6 Remember gradients of parallel lines are equal!!

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 =

Lines & Circles The relationship between a line and a circle can be found by substituting the equation of the line in the equation of the circle collecting like terms to obtain a quadratic equation 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 A B C It is useful to remember that the angle in a semi-circle is a right angle.

Intersecting circles Circles which do not touch can be:
one inside the other or completely apart Circles can touch: internally 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 D 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 = so r = = √100 = 10

Substitute the equation of the line into x2 + y2 + 14x + 4y – 19 = 0
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).

The equation of the larger circle is x2 + y2 + 14x + 4y – 19 = 0
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) (-1,4) 2 2 6 (-7,-2) 6 and the radius of the larger circle is The radius of the smaller circle is ⅓ of the larger circle Use the diagram to find the coordinates of the centre C. (1,6) Double check the distance from P to C. Hence the equation of the smaller circle is (x - 1)2 + (y – 6)2 = 8

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