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Determining the equation of a line Linear equation Exercises involving straight lines Circles

PreCalculus for Engineers

Dr David J.J. Devlin

Lecture 3
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Points, distances and midpoints

The length of this segment of straight line (which is the distance betweenthe two points) is

d=p

(x2 x1)2 + (y2 y1)2,

by direct application of Pythagoras theorem.

The midpoint M(AB) of the segment AB is the centre of the segment. Itscoordinates are

M(AB) =x1+x2

2 ,

y1+y22

Example: What is the distance d between the points A = (4,2)andB= (6, 7), and what is the midpoint M(AB)?

d=p

(4 6)2 + (2 7)2 = 4 + 81 = 85,and

M(AB) =

4 + 6

2 ,2 + 7

2

=

5,

5

2
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Equation of a line

A curve in the (x, y)plane is specified by some equation obeyed by thepoints (x, y) lying on the curve.

An important specific curve is the straight line. A standard form of the equation of a straight line is

y=mx+b,

O

y=b

(x,y)

y

x

slope m = (yb)/x

distance = yb

x

y=mx+b

distance = x

Figure: Illustration of a straight line.
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Equation of a line

Such a line (illustrated in figure 2passes through the point (0, b), and anyother point lying on the line satisfies

y bx

=m,

where m is a constant called the slope or gradient of the line. Note: This

also corresponds to the tangent of the angle indicated in figure2. Special cases:

A horizontal line y=bhas a slope m= 0.

A vertical line does not have a slope (it can be seen as being infinite). Theequation for such lines are

x=c

where c is a constant.

If the slope m>0, then the value for y increases as we increase x.Conversely, ifm

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Equation of a line

Illustrations are provided in figure3.

y

(a) (b)

(d)(c)

y=b

x=c

m>o

m

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Determining the equation of a line

A line can be specified by its slope m and one point (a, b) it passesthrough, see figure4. The equation of the line is then

y b=m(x a),or, equivalently

y=mx+ (b ma),see figure4.

xa

O

(x,y)

y

x

slope m = (yb)/(xa)

(a,b)

yb

Figure: Illustration of a straight line with slope m, passingthroughthepoint(a,b).

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Example:

Find the equation of the straight line of gradient 3, passing through thepoint (1, 2).

Answer:We have

y 2 = 3 (x 1),y= 3x 1.

Note: A wise precaution is to check that the point (1,2) lies indeed on theline; here 2-2=3(1-1).

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Another way to specify a line is to give two points (x1, y1) and (x2, y2) it

passes through (in Euclidean geometry).

slope m = (y2y1)/(x2x1)

O

(x2,y2)

y

x

y2y1

x2x1

(x1,y1)

x1

y1

y2

x2

Figure: Line specified by two points it passes through.

The geometry is illustrated in figure 5.

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The slope m of the line is given by the ratio of the increase in y for agiven increase in x. So here, knowing that when x goes from x1 to x2, ygoes from y1 to y2,

m= y2 y1x2 x1 .

Then a point (x, y) along the line satisfies

(y y1) = y2 y1x2 x1 (x x1),

or equivalently

(y

y2) =

y2 y1

x2 x1(x

x2),

whether you choose the point (x1, y1) or (x2, y2) as reference point the linepasses through. Obviously either choice is right, and you will end up withthe same equation in the end!

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Proof: The first equation reads

y= y2 y1x2 x1 x

y2 y1x2 x1 x2+y1,

y= y2 y1x2 x1 x

(y2 y1)x2+ (x2 x1)y1x2 x1 ,

y= y2

y1

x2 x1 xy2x2

y1x2+x2y1

x1y1

x2 x1 .while the second reads

y= y2 y1x2

x1x y2 y1

x2

x1x1+y2,

y= y2 y1x2 x1 x

(y2 y1)x1+ (x2 x1)y2x2 x1 ,

y= y2 y1x2 x1 x

y2x1 y1x1+x2y2 x1y2x2 x1 .

which is the same.

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For example:

Consider the line passing through the two points (1, 3) and (2, 4).The gradient of the line is

m= 3 41 2 , equivalently

4 32 (1) =

1

3,

Now using the fact that the line passes through (

1, 3) we have

y 3 = 13

(x (1)) = 13

(x+ 1).

y= x

3+

10

3 .

We can check that the line passes through the two points

3 =1

3 +

10

3 ,

4 = 2

3+

10

3 .

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We could have used the fact that the line passes through the point (2 , 4)instead to determine the equation of the line

(y 4) = 13

(x 2),

y = x

3 2

3+ 4 =

x

3+

10

3 .

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Linear equation

A linear equation in x and y is a equation of the form

ax+by+c= 0.

Its solution is (graphically) a straight line. Ifb= 0 but a = 0, the solutionis the vertical linex= c

a.

Ifb= 0, the solution is the standard straight line

y =

a

bx

c

b.

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Parallel lines

Parallel lines

Two lines are parallel if they share the same slope.

Example: Find the equation for the line L1 passing through the point(5, 1), parallel to the line L

2joining the points (1, 1) and (5, 3)?

Answer: The slope m of the line L2 is m= (3 1)/(5 1) = 1/2. So theequation of the line L1 is

y 1 = 12

(x 5),

y = x2 3

2.

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Perpendicular lines

Two lines are perpendicular (or orthogonal) if the product of their slopeequals1. In other words, consider the lines L1 of slope m1, and L2 ofslope m2, then the lines are perpendicular if

m1m2 = 1. Note: Special case. Recall that a horizontal line and a vertical line are

perpendicular. A vertical line has an equation of the form x=cwhile ahorizontal line has an equation of the form y=b, with slope 0. You cansee a vertical line as a line with a slope that tends to

.

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Perpendicular lines

Illustration

Dy

x

y

(yb)=m(xa)

(a,b)

(x,y)

Dx = (xa)

Dy = (yb)

Dx

m=Dy/Dx

(yb)=(xa)/m

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Exercises involving straight lines

Example 1Find where the line passing through the points (1, 1) and (5,4)intersects the xaxis.AnswerThe gradient of the line is

m=4 1

5 1 = 5

4.

So the equation of the line (considering it passes through the point (1, 1))is

(y 1) = 54

(x 1),

y =

5

4

x+9

4

.

Along thexaxisy= 0, so setting y= 0 in the equation of the line above,5

4x=

9

4,

x= 9

5.

The point is then (x, y) = (9/5, 0).

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Example 2:Find the line passing through (1, 2) and perpendicular to the line passingthrough (3, 4) and (5,2). Where do these lines intersect?Answer

The gradient of the line passing through (3, 4) and (5,2) is

m1 = 4 (2)

3 5 = 3

So the equation of the line is

(y 4) = 3(x 3),y = 3x+ 13.

The line perpendicular to this line has a slope m2 = 1/3, as m1m2 = 1.Since it passes through the point (1, 2), its equation is

(y 2) = 13

(x 1),

y= 1

3x+

5

3,

or

3y=x+ 5.

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At the intersection, wee need to satisfy both equations,

3x+y= 13 (1),

x 3y = 5 (2).To solve that, we can combine for example 3*(1) + (2) (to eliminate y)

9x+ 3y+x 3y = 39 5,10x= 34, 5x= 17, x= 17/5

To find y, we can use (20)

y=

3x+ 13 =

3

17/5 + 13 = 14/5

so the intersection is (17/5, 14/5).

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Circles

EquationIf we consider the circle of radius rcentred at the point (a, b), the

condition for the point (x, y) to lie on the circle is that the distancebetween (a, b) and (x, y) equals r, which is conveniently expressed by

(x a)2 + (y b)2 =r2.

y

a

b

r

x

Figure: Illustration of a circle of radius rcentred at (a,b).

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Example 1

Find the equation of the circle of radius 2, centred at (1, 2).

Answer:(x 1)2 + (y 2)2 = 22 = 4.

Example 2

Find the equation of the circle of radius

3, centred at (,

2)

Answer:(x )2 + (y

2)2 = 3.

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Example 3

Find the equation of the circle for which the segment on the straight linejoining the points (x1, y1) = (1, 3) and the point (x2, y2) = (5,1) is adiameter.

Answer:

The centre (a, b) of the circle is the midpoint of the diameter, so

(a, b) =

1 + 5

2 ,

3 12

= (3, 1)

We know now the centre of the circle. The only other bit of information

we need to find the full equation of the circle is its radius r. There are 3different, yet equivalent ways, to proceed.

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Recall that (a) the radius is the distance between the centre and a point lying on the

circle. (b) a diameter dof the circle is twice a radius r.

Using statement (a), and the information given, we may say that the radiusis the distance between points (a, b) the centreto the point (x1, y1) lyingon the circle - option (a.1); or equivalently the distance between points

(a,b) the centreto the point (x2, y2) lying on the circle - option (a.2) Option (a.1) gives (for the square of the radius)

r2 = (a x1)2 + (b y1)2 = (3 1)2 + (1 3)2 = 22 + 22 = 8.Hence r=

8.

Option (a.2) gives

r2 = (a x2)2 + (b y2)2 = (3 5)2 + (1 + 1)2 = 22 + 22 = 8.

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The last option is (b), in which we state, that

d=r/2,

squaring,d2 =r2/4.

d is the length of the diameter, so

d2 = (x2 x1)2 + (y2 y 1)2 = (5 1)2 + (3 + 1)2 = 42 + 42 = 32,so r2 =d2/4 = 8.

Obviously one need to use only one of the three alternatives above, asthey are mathematically equivalent.

Finally, we have for the circle

(x 3)2 + (y 1)2 = 8.

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Illustration of a circle from example 3.

(x,y)

(1,3)

(5,1)

(3,1)

r = 82

x

y

O

Figure: Illustration of a circle of radius inexample3.

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Example 4

Find the equation of the circle of centre (2,3), passing through the point(5,4).

Answer:

The square of radius of the circle is

r2 = (5 2)2 + (4 3)2 = 10.So the equation is

(x

2)2 + (y

3)2 = 10
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