Maths in Focus - Margaret Grove - ch7

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Linear Functions TERMINOLOGY Collinear points: Two or more points that lie on the same straight line

Interval: A section of a straight line including the end points

Concurrent lines: Two or more lines that intersect at a single point

Midpoint: A point lying exactly halfway between two points

Gradient: The slope of a line measured by comparing the vertical rise over the horizontal run. The symbol for gradient is m

Perpendicular distance: The shortest distance between a point and a line. The distance will be at right angles to the line

Chapter 7 Linear Functions

INTRODUCTION IN CHAPTER 5, YOU STUDIED functions and their graphs. This chapter looks at the linear function, or straight-line graph, in more detail. Here you will study the gradient and equation of a straight line, the intersection of two or more lines, parallel and perpendicular lines, the midpoint, distance and the perpendicular distance from a point to a line.

DID YOU KNOW? Pierre de Fermat (1601–65) was a lawyer who dabbled in mathematics. He was a contemporary of Descartes, and showed the relationship between an equation in the form Dx = By, where D and B are constants, and a straight-line graph. Both de Fermat and Descartes only used positive values of x, but de Fermat used the x-axis and y-axis as perpendicular lines as we do today. De Fermat’s notes Introduction to Loci, Method of Finding Maxima and Minima and Varia opera mathematica were only published after his death. This means that in his lifetime de Fermat was not considered a great mathematician. However, now he is said to have contributed as much as Descartes towards the discovery of coordinate geometry. De Fermat also made a great contribution in his discovery of differential calculus.

Class Assignment Find as many examples as you can of straight-line graphs in newspapers and magazines.

Distance The distance between two points (or the length of the interval between two points) is easy to find when the points form a vertical or horizontal line.

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Maths In Focus Mathematics Preliminary Course

EXAMPLES Find the distance between 1. ^ -1, 4 h and ^ -1, -2 h

Solution

Counting along the y-axis, the distance is 6 units. 2. ^ 3, 2 h and ^ -4, 2 h

Solution

Counting along the x-axis, the distance is 7 units.

When the two points are not lined up horizontally or vertically, we use Pythagoras’ theorem to find the distance.

Chapter 7 Linear Functions

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EXAMPLE Find the distance between points ^ 3, -1 h and ^ -2, 5 h.

Solution

BC = 5 and AC = 6 By Pythagoras’ theorem,

You studied Pythagoras’ theorem in Chapter 4.

c =a +b AB 2 = 5 2 + 6 2 = 25 + 36 = 61 2

2

2

` AB = 61 Z 7.81

DID YOU KNOW? Pythagoras made many discoveries about music as well as about mathematics. He found that changing the length of a vibrating string causes the tone of the music to change. For example, when a string is halved, the tone is one octave higher.

The distance between two points _ x 1, y 1 i and _ x 2, y 2 i is given by d=

2 2 _ x2 - x1 i + _ y2 - y1 i

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Maths In Focus Mathematics Preliminary Course

Proof

If points A and B were changed around, the formula would be d =

(x 1 - x 2 ) + (y 1 - y 2 ) , 2

2

which would give the same answer.

Let A = _ x 1, y 1 i and B = _ x 2, y 2 i Length AC = x 2 - x 1 and length BC = y 2 - y 1 By Pythagoras’ theorem AB 2 = AC 2 + BC 2 d 2 = _ x 2 - x 1 i2 + _ y 2 - y 1 i2 `d=

2 2 _ x2 - x1 i + _ y2 - y1 i

EXAMPLES 1. Find the distance between the points ^ 1, 3 h and ^ -3, 0 h.

Solution Let ^ 1, 3 h be _ x 1, y 1 i and ^ -3, 0 h be _ x 2, y 2 i d=

2 2 _ x2 - x1 i + _ y2 - y1 i

= ] -3 - 1 g2 + ] 0 - 3 g2 = ] -4 g2 + ] -3 g2 = 16 + 9 = 25 =5 So the distance is 5 units. 2. Find the exact length of AB given that A = ^ -2, -4 h and B = ^ -1, 5 h .

Solution Let ^ -2, -4 h be _ x 1, y 1 i and ^ -1, 5 h be _ x 2, y 2 i d=

You would still get 82 if you used (- 2, - 4) as (x 2 , y 2 ) and (-1, 5) as (x 1 , y 1 ).

2 2 _ x2 - x1 i + _ y2 - y1 i

=

6 -1 - ^ -2 h @ 2 + 6 5 - ^ -4 h @ 2

= = =

12 + 92 1 + 81 82

Chapter 7 Linear Functions

7.1 Exercises 1.

Find the distance between points (a) ^ 0, 2 h and ^ 3, 6 h (b) ^ -2, 3 h and ^ 4, -5 h (c) ^ 2, -5 h and ^ -3, 7 h

2.

Find the exact length of the interval between points (a) ^ 2, 3 h and ^ -1, 1 h (b) ^ -5, 1 h and ^ 3, 0 h (c) ^ - 2, -3 h and ^ - 4, 6 h (d) ^ -1, 3 h and ^ -7, 7 h

3.

4.

Find the distance, correct to 2 decimal places, between points (a) ^ 1, -4 h and ^ 5, 5 h (b) ^ 0, 4 h and ^ 3, -2 h (c) ^ 8, -1 h and ^ -7, 6 h Find the perimeter of D ABC with vertices A ^ 3, 1 h, B ^ -1, 1 h and C ^ -1, -2 h .

5.

Prove that the triangle with vertices ^ 3, 4 h, ^ -2, 7 h and ^ 6, -1 h is isosceles.

6.

Show that AB = BC, where A = ^ -2, 5 h, B = ^ 4, -2 h and C = ^ -3, -8 h .

7.

Show that points ^ 3, -4 h and ^ 8,1 h are equidistant from point ^ 7, -3 h .

8.

A circle with centre at the origin O passes through the point _ 2 , 7 i . Find the radius of the circle, and hence its equation.

9.

Prove that the points X _ 2 , -3 i, Y _ -1, 10 i and Z _ - 6 , 5 i all lie on a circle with centre at the origin. Find its equation.

10. If the distance between ^ a, -1 h and ^ 3, 4 h is 5, find the value of a. 11. If the distance between ^ 3, -2 h and ^ 4, a h is 7 , find the exact value of a.

12. Prove that A ^ 1, 4 h, B ^ 1, 2 h and C _ 1 + 3 , 3 i are the vertices of an equilateral triangle. 13. If the distance between ^ a, 3 h and ^ 4, 2 h is 37 , find the values of a. 14. The points M ^ -1, -2 h, N (3, 0), P ^ 4, 6 h and Q ^ 0, 4 h form a quadrilateral. Prove that MQ = NP and QP = MN. What type of quadrilateral is MNPQ? 15. Show that the diagonals of a square with vertices A ^ -2, 4 h, B ^ 5, 4 h, C ^ 5, -3 h and D ^ -2, -3 h are equal. 16. (a) Show that the triangle with vertices A ^ 0, 6 h, B ^ 2, 0 h and C ^ -2, 0 h is isosceles. (b) Show that perpendicular OA, where O is the origin, bisects BC. 17. Find the exact length of the diameter of a circle with centre ^ -3, 4 h if the circle passes through the point ^ 7, 5 h . 18. Find the exact length of the radius of the circle with centre (1, 3) if the circle passes through the point ^ -5, -2 h . 19. Show that the triangle with vertices A ^ -2, 1 h, B ^ 3, 3 h and C ^ 7, -7 h is right angled. 20. Show that the points X ^ 3, -3 h, Y ^ 7, 4 h and Z ^ - 4, 1 h form the vertices of an isosceles right-angled triangle.

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Midpoint The midpoint is the point halfway between two other points.

The midpoint of two points _ x 1, y 1 i and _ x 2, y 2 i is given by M=e

x1 + x2 y1 + y2 o , 2 2

Proof

Can you see why these triangles are similar?

Find the midpoint of points A _ x 1, y 1 i and B _ x 2, y 2 i. Let M = ^ x, y h Then D APQ