Drawing a sketch is always worth the time and effort involved Higher Unit 1.1 Straight Line Drawing a sketch is always worth the time and effort involved

Distance Formula Length of a straight line x y O Pythagoras’ Theorem B(x2, y2) AB 2 = AC 2 + CB 2 y2 – y1 A(x1,y1) x2 – x1 C AB 2 = (x2 – x1) 2 + (y2 – y1) 2 AB = √(x2 – x1) 2 + (y2 – y1) 2

Finding Mid-Point of a line x y O B(x2, y2) The mid-point of AB ● M A(x1,y1)

What You Should Know About GRADIENTS

All graphs are read from left to right m is +ve, when line slopes up m is –ve , when line slopes down m < 0

All graphs are read from left to right m is 0, when line is horizontal m = 0 Equation of line is y = a constant m is undefined or infinite , when line is vertical Equation of line is x = a constant m = ∞

Parallel lines have equal gradients More Gradient facts Parallel lines have equal gradients m1 = m2 The gradient = the tangent of the angle the line makes with the positive x - axis θ m = tan θ

Exact Values 1 1 2 tan xº cos xº sin xº 90º 60º 45º 30º 0º x 45º 30º 3 1 45º 2 Exact Values tan xº cos xº sin xº 90º 60º 45º 30º 0º x 1 1

All Sinners Take Care CAST 90° 90° to 180° sin positive 0° to 90° all positive S A 180 - 180° 0°,360° 180 + C 360 - T 180° to 270° tan positive 270° to 360° cos positive 270°

Angle must be between 90 and 180 Find the size of the angle between the positive x-axis and the lines with gradient: 60o x 150o x m = tan θ 90o  A S 0o 180o 360o  T C 270o Angle must be between 90 and 180

Points are said to be collinear if they lie on the same straight line Collinearity Points are said to be collinear if they lie on the same straight line To prove: A ● B C x y O show mAB = mBC Then state A, B and C are collinear since AB and BC are parallel through the common point B

Show mAB = mBC Collinearity Example Prove that the points A(– 2 , – 3), B(– 1 , 0) and C(1 , 6) Show mAB = mBC mAB = mBC  AB is parallel to BC So A, B and C are collinear since AB and BC are parallel through the common point B

Gradient of perpendicular lines Draw a line from O to (3 , 2) Now rotate the line 90o anticlockwise x y Work out the gradient of each line What do you notice about the gradients? m1 × m2 = – 1

Gradient of perpendicular lines If two lines with gradients m1 and m2 are perpendicular then m1 × m2 = –1 x y O Conversely: If m1 × m2 = –1 then the two lines with gradients m1 and m2 are perpendicular m1 m2 The converse is used to prove lines perpendicular 14

Gradient of perpendicular lines If two lines with gradients m1 and m2 are perpendicular then m1 × m2 = -1 You need to be able to work out NEGATIVE INVERSES m1 m2 m1 m2 15

Perpendicular and Parallel Lines Exercise 3 Page 7 Numbers 1, 2, 3, 6, 7, 9, 10 and 11 16

Straight line Facts y = mx + c y- intercept

ax + by + c = 0 ax + by + c = 0 is an alternative version of the straight line equation. When given the equation in this form it is often useful to rearrange it into the form y = mx + c, especially if you need to know the gradient of the line. Step 1: isolate y term on left hand side. by = – ax – c Step 2: divide by y term coefficient

Points on a line A point will only lie on a line if its coordinates satisfy the equation of the line. In other words if substituting the x and y coordinates makes the equation work out. Given the line with equation 3x – 2y = 4 (2 , 1) lies on the line since 3×2 – 2×1 = 4 As does (0 , – 2) since 3×0 – 2×(– 2) = 4 But (3 , 2) does NOT since 3×3 – 2×2 = 5

(2k , k) lies on the line  2×2k – k = 9 Points on a line Given the line with equation 2x – y = 9, find the value of k, so that (2k , k) lies on the line. (2k , k) lies on the line  2×2k – k = 9  3k = 9  k = 3

Positive direction x-axis Negative direction x-axis Angle between lines Find the angle between the lines with equations 6x – 2y + 6 = 0 and 2x + 5y – 10 = 0 Remember the connection between angles and lines is m = tan θ, where θ is the angle between the line and the positive x-axis However, when m is negative θ becomes the angle between the line and the negative x-axis. Positive direction x-axis Negative direction x-axis

Negative direction x-axis Positive direction x-axis Angle between lines Find the angle between the lines with equations 6x – 2y + 6 = 0 and 2x + 5y – 10 = 0 m = 0∙75 m = – 0∙75  tan θ = 0∙75  tan θ = – 0∙75 tan-1 0∙75 = 36∙9o tan-1 0∙75 = 36∙9o Negative direction x-axis Positive direction x-axis 36∙9o 36∙9o

Angle between lines = 180o – (21∙8o + 71∙6o) Find the angle between the lines with equations 6x – 2y + 6 = 0 and 2x + 5y – 10 = 0 5y = – 2x + 10 – 2y = – 6x – 6 y = – 2/5x + 2 y = 3x + 2  tan θ = – 2/5  tan θ = 3 tan-1 0∙4 = 21∙8o x y tan-1 3 = 71∙6o Angle between lines = 180o – (21∙8o + 71∙6o) = 86∙6o 71∙6o 21∙8o

Another Equation of the Straight Line The equation of the line through the point (a , b) with gradient m is x y O (a , b) ● m y – b = m(x – a)

Using y – b = m(x – a) All questions require you to find the gradient, m and the coordinates (a , b) The gradient may be given or found from two points on the line found from angle line makes with x-axis found by changing given line equation into y = mx + c then parallel  m1 = m2 perpendicular  m1 × m2= – 1

Use one of the given points as (a , b) Using y – b = m(x – a) All questions require you to find the gradient, m and the coordinates (a , b) Find the equation of the line through the points (– 5 , 3) and ( 7 , 7) Use one of the given points as (a , b) × 3

× 3 Parallel m1 = m2 Using y – b = m(x – a) All questions require you to find the gradient, m and the coordinates (a , b) Find the equation of the line parallel to 2x + 3y = 12 passing through the point ( 1 , 5) Rearrange 2x + 3y = 12 to find m. × 3 Parallel m1 = m2

Using y – b = m(x – a) All questions require you to find the gradient, m and the coordinates (a , b) Find the equation of the line perpendicular to 2x + 3y = 12 passing through the point ( 1 , 5) Rearrange 2x + 3y = 12 to find m. × 2 Perpendicular m1 × m2 = – 1

Past Paper Multiple Choice Questions

The line with equation y = ax + 4 is perpendicular to the line with equation 3x + y + 1 = 0. What is the value of a? A. – 3 B. – 1/3 C. 1/3 D. 3 Perpendicular m1 × m2 = – 1

The line 2y = 3x + 6 meets the y-axis at C The line 2y = 3x + 6 meets the y-axis at C. The gradient of the line joining C to A (4,-3) is: A. 9/4 B. 2/3 C. -2/3 D. -3/2 E. -9/4 2y = 3x + 6 cuts y-axis at x = 0  2y = 6  y = 3 C(0 , 3)

A.(3/2,1) B.(3/2,-1) C.(1,3/2) D.(1,-3/2) P(2 , 0) R(0 , – 3) The line 3x - 2y – 6 = 0 cuts the x-axis at P and the y-axis at R. The mid-point of PR is: A.(3/2,1) B.(3/2,-1) C.(1,3/2) D.(1,-3/2) 3x - 2y – 6 = 0 cuts x-axis at y = 0  3x – 6 = 0  x = 2 P(2 , 0) 3x - 2y – 6 = 0 cuts y-axis at x = 0  – 2y – 6 = 0  y = – 3 R(0 , – 3)

Given that the line joining the points (2 , 3) and (8 , k) is perpendicular to the line 2y – 3x + 5 = 0, then the value of k is: A. –1 B. –2 C. –7 D. 1 2y – 3x + 5 = 0,  2y = 3x – 5  y = 3/2 x – 5/2  m = 3/2 Perpendicular m1 × m2 = – 1  m = – 2/3

For varying values of p the equation y – 1 = p(x – 1) is the equation of a straight line. All such lines A. have the same gradient B. cut the x axis at the same point. C. pass through a fixed point not on the axis. D. cut the y axis at the same point.  y – 1 = p(x – 1) From y – b = m(x – a) Varies with p m = p Line passes through (1 , 1) Cuts x-axis at y = 0 Cuts y-axis at x = 0 Varies with p Varies with p

A straight line passes through the points P(−5, −2) and Q(−2, −1) A straight line passes through the points P(−5, −2) and Q(−2, −1). What is the equation of the straight line which passes through P and is perpendicular to PQ ? Perpendicular m1 × m2 = – 1

The medians are CONCURRENT, they all pass through the same point A median is a line joining a vertex to the midpoint of the opposite side ● ● ● The medians are CONCURRENT, they all pass through the same point

The altitudes are CONCURRENT, they all pass through the same point An altitude is a line from a vertex to meet the opposite side at right angles The altitudes are CONCURRENT, they all pass through the same point

Perpendicular Bisector A perpendicular bisector is a line which cuts the given line in half at right angles ●

Intersecting Lines The point of intersection of two lines is found by solving simultaneous equations Three or more lines are said to be concurrent if they all pass through a common point. To prove concurrency solve one pair of simultaneous equations and show that the point of intersection lies on the other line.

Lines are concurrent since they all pass through (-1 , -2) Show that the lines x + y + 3 = 0, 2x – y = 0 and 3x – 2y = 1 are concurrent. Add Put x = – 1 and y = – 2 in 3x – 2y Put x = – 1 in x + y = – 3 Lines are concurrent since they all pass through (-1 , -2)

Questions from SQA Past Papers

Use y = mx + c to find gradient of given line: Find the equation of the line which passes through the point (-1, 3) and is perpendicular to the line with equation Use y = mx + c to find gradient of given line: Use m1 × m2 = -1 to find gradient of perpendicular: Use y – b = m(x – a) to find equation:

and which passes through the point (2, –1). Find the equation of the straight line which is parallel to the line with equation and which passes through the point (2, –1). Use y = mx + c to find gradient of given line: Parallel lines have equal gradients

Find gradient of the line: Use table of exact values Find the size of the angle a° that the line joining the points A(0, -1) and B(33, 2) makes with the positive direction of the x-axis. y Find gradient of the line: B(3√3, 2) ● ao Use m = tan θ O x A(0, -1) Use table of exact values a = 30

A and B are the points (–3, –1) and (5, 5) 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 y To find equation of AB (5, 5) ● (a , b) (-3, -1) ● Using y – b = m(x – a) O

A and B are the points (–3, –1) and (5, 5) 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 y To find perpendicular bisector (5, 5) ● (a , b) (-3, -1) ● O Using y – b = m(x – a)

The line AB makes an angle of π/3 radians with the y-axis, as shown in the diagram. Find the exact value of the gradient of AB. The angle between AB and x-axis is equal to the angle between the line and a parallel line x y O B π/3 π/6 From m = tan θ A m = tan π/6 m = 1/√3

Find gradient of median AM 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 A. x y O Find mid-point of BC: B(6 , 1) A(4 , 3) C(-2 , -3) M(2 , -1) Find gradient of median AM ● M Equation of AM

Find gradient of PS (perpendicular to QR) 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. x y O P(-4 , 5) R(4 , 1) Q(-2 , -2) Find gradient of QR: Find gradient of PS (perpendicular to QR) S Equation of altitude PS

The lines y = 2x + 4 and x + y = 13 make angles of a and b with the positive direction of the x-axis, as shown in the diagram. a) Find the values of a and b, and the acute angle between the lines. y x O y = 2x+4 x + y = 13 bo ao Gradient of y = 2x + 4 is 2 tan ao = 2  a = tan-1 2 = 63o 135o Gradient of x + y = 13 is -1 tan bo = -1  a = tan-1 -1 = 135o 45o 63o Acute angle = 180 – (63 + 45) Acute angle = 72o

C and M have same y-coordinate Solving to find intersection, (0 , 2) 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. x y O C and M have same y-coordinate A(-1 , 6) B(-3 , -2) C(5 , 2) y = 2 M(-2 , 2) Mid point of BC is (1 , 0) ● y – 0 = – 2(x – 1) 2x + y = 2 Solving to find intersection, (0 , 2)

y ● x 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. M(7 , 2) A(2 , 2) B(12 , 2) C(8 , 6) y x l1 is vertical through M l1 is x = 7 N(5 , 4) ● N(5 , 4) mAC = 4/6 = 2/3 ● M ml2 = -3/2 m1 × m2 = – 1 l2 is y – 4 = -3/2( x – 5) when x = 7, y = 1 l1 and l2 intersect at (7 , 1) l2 is 3x + 2y = 23

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 1 of 3 M(½(12 – 4), ½(1 + 3)) M(4 , 2) Using C in y – b = m(x – a)

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 2 of 3 m1 × m2 = – 1

Point of intersection is (6 , – 4) 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 3 of 3 Sub in 2y + x = – 2 Point of intersection is (6 , – 4)

So lines are perpendicular and ABC is right angled 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 the co-ordinates of M. 1 of 4 mAB = 4/2 = 2 mBC = -4/8 = -1/2 mAB × mBC = –1 So lines are perpendicular and ABC is right angled

• A triangle ABC has vertices A(–3, –3), B(–1, 1) and C(7,–3). 2 of 4 b) The medians AD and BE intersect at M. i) Find the equations of AD and BE. ii) Find the co-ordinates of M. 2 of 4 D(3 , –1) mAD = 2/6 = 1/3 y + 3 = 1/3(x + 3) • D 3y + 9 = x + 3 3y – x = –6

• A triangle ABC has vertices A(–3, –3), B(–1, 1) and C(7,–3). 3 of 4 b) The medians AD and BE intersect at M. i) Find the equations of AD and BE. ii) Find the co-ordinates of M. 3 of 4 E(2 , –3) mBE = -4/3 y – 1 = -4/3(x + 1) 3y – 3 = –4x – 4 3y + 4x = –1 • E

A triangle ABC has vertices A(–3, –3), B(–1, 1) and C(7,–3). b) The medians AD and BE intersect at M. i) Find the equations of AD and BE. ii) Find the co-ordinates of M. 4 of 4 3y + 4x = –1 3y – x = –6 Subtract D 5 x = 5 M x = 1 (1 , -5/3) y = – 5/3 E