Presentation on theme: "Scholar Higher Mathematics Homework Session"— Presentation transcript:
1Scholar Higher Mathematics Homework Session Thursday 29th January 7:30pmYou will need a pencil, paper and a calculator for some of the activities
2Margaret FergusonSCHOLAR online tutor for Maths and Author of the new SCHOLAR National 5 & Higher Maths courses
3Tonight’s Revision Session will cover The Straight Line and The Circle
4The Straight Linethings you show knowThe 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)ybxa
5more things you must know The Straight Linemore things you must knowThe general form of the equation is Ax + By + C = 0To be able to identify the gradient or y-intercept:the equation of a straight line must be in the form y = mx + cif the equation is not in that form it must be re-arrangedParallel lines have equal gradients and distinct y-interceptsA set points are Collinear if they all lie on a single straight line:find the gradients of the lines joining pairs of pointsequal gradients indicate parallel linesa common point determines collinearitym = tan θ where θ is the angle between a line and the positive direction of the x-axisLines with gradients m1 & m2 are perpendicular if m1 × m2 = -1The midpoint formula is
61. Find the gradient of the line l1? yWhat is the equation of the line l2 which passes through the point (8,-2) and is perpendicular to the line l1.??x1. Find the gradient of the line l1?l1What 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° = -12. Find the equation of the line l2?What is the gradient of l2?Why?What is the equation of the line l2?ml2 = 1for perpendicular lines ml1 x ml2 = -1y + 2 = 1(x – 8)y = x - 10
7BTriangles: A MedianACA 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 areThe centroid lies two thirds of the distance along each median measured from its vertex.
8Altitudes and Perpendicular Bisectors An altitude of a triangle is a line from a vertex, which is perpendicular to the opposite side.ACThe altitudes of a triangle are concurrent and the point of intersection is called the orthocentreABCThe 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.
9The 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 liney = 2mBC = ½mperp = -2y - 6 = -2(x + 1)y = -2x + 42 = -2x + 4-2 = -2xx = 1The point of intersection of the lines p and q is (1,2)
10Remember 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 nowAB -⅔CD 2Solution:3y + 2x = 63y = -2x + 6✓Remember gradients of parallel lines are equal!!
11The Equation of the Circle The equation of a circle with centre (0,0) and radius r isx2 + y2 = r2The equation of a circle with centre (a, b) and radius r is(x - a)2 + (y - b)2 = r2The general equation of the circle isx2 + y2 + 2gx + 2fy + c = 0with centre = (-g, -f)and radius =
12Lines & CirclesThe relationship between a line and a circle can be found bysubstituting the equation of the line in the equation of the circlecollecting like terms to obtain a quadratic equationevaluating and interpreting the discriminantif b2 – 4ac = 0 then the line is a tangent to the circleif b2 – 4ac > 0 then there are 2 points of intersectionif b2 – 4ac < 0 then the line does not meet the circle at allAny points of intersection can be found by solving the quadraticABCIt is useful to remember that the angle in a semi-circle is a right angle.
13Intersecting circles Circles which do not touch can be: one inside the otheror completely apartCircles can touch:internallyor externallyCircles may intersectThe distance between the centres of circles and the sum of their radii can help to determine the relationship between the circlesThe journey between the coordinates of centres or points of contact can help in problem solving situations
14A Circle has equation x2 + y2 + 8x + 6y – 75 = 0. What is the radius of the circle?Vote for the correct answer nowA 5B 10CDThe general equation of the circle takes the form x2 + y2 + 2gx + 2fy + c = 02g = 8 so g = 42f = 6 so f = 3radius =so r == √100 = 10✓
15Substitute 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 = 0x2 + (3 – x)2 + 14x + 4(3 – x) – 19 = 0x2 + (9 – 6x + x2) + 14x + (12 – 4x) – 19 = 0Collect like terms to obtain a quadratic equation2x2 + 4x + 2 = 0Evaluate and interpret the discriminantb2 – 4ac = 42 – 4(2)(2)= 16 – 16= 0Since b2 – 4ac = 0 the line y = 3 – x is a tangent to the circle.Find the coordinates of the point of contact2x2 + 4x + 2 = 02(x2 + 2x + 1) = 0(x +1)(x +1) = 0x = -1Let x = -1 in the equation of the tangenty = 3 – (-1) = 4Hence P has coordinates (-1,4).
16The 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 isx2 + y2 + 14x + 4y – 19 = 0so the centre is (-7,-2)(-1,4)226(-7,-2)6and the radius of the larger circle isThe radius of the smaller circle is ⅓ of the larger circleUse 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
17Question TimeIf you have any questions about tonight’s session please askI can only answer one question at a time so take turnsThe next session will be on 12th February at 7:30pmThe topics covered will be Integration & VectorsCarol will provide a link for you to give us feedback