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 www.mathsrevision.com

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) www.mathsrevision.com

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

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

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

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

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 www.mathsrevision.com 21-Apr-17

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 www.mathsrevision.com 21-Apr-17

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

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.

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 )

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. www.mathsrevision.com 21-Apr-17

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

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:

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)

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

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

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

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)

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)

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)

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