Equations of straight lines

1 Equations of straightlinesmc-TY-strtlines-2009-1In this unit we find the equation of a straight line, when we aregiven some information aboutthe line. The information could be the value of its gradient,together with the co-ordinates of apoint on the line. Alternatively, the information might be the co-ordinates of two different pointson the line. There are several different ways of expressing the final equation, and some are moregeneral than order to master the techniques explained here it is vital that you undertake plenty of practiceexercises so that they become second reading this text, and/or viewing the video tutorial on this topic, you should be able to: find the equation of a straight line, given its gradient and its intercept on they-axis; find the equation of a straight line, given its gradient and one point lying on it; find the equation of a straight line given two points lying on it; give the equation of a straight line in either of the formsy=mx+corax+by+c= equation of a line through the origin with a given of a equation of a straight line with a given gradient, passingthrough a given equation of a straight line through two given most general equation of a straight mathcentre 20091.

2 IntroductionThis unit is about the Equations of straight lines . These Equations can take various formsdepending on the facts we know about the lines . So to start, suppose we have a straight linecontaining the points in the following are many more points on the line, but we have enough now to see a pattern. If we takeanyxvalue and add 2, we get the correspondingyvalue:0 + 2 = 2,1 + 2 = 3,2 + 2 = 4, andso on. There is a fixed relationship between thexandyco-ordinates of any point on the line,and the equationy=x+ 2is always true for points on the line. We can label the line using The equation of a line through the origin with a givengradientSuppose we have a line with equationy=x. Then for every point on the line, theyco-ordinatemust be equal to thexco-ordinate. So the line will contain points in the following :xy00112233xyy = xWe can find the gradient of the line using the formula for gradients,m=y2 y1x2 x1, mathcentre 2009and substituting in the first two sets of values from the table.

3 We getm=1 01 0= 1so that the gradient of this line is about the equationy= 2x? This also represents a straight line, and for all the pointsonthe line eachyvalue is twice the correspondingxvalue. So the line will contain points in thefollowing 2x:xy001224xyy = 2xy = xIf we calculate the gradient of the liney= 2xusing the first two sets of values in the table, weobtainm=2 01 0= 2so that the gradient of this line is take the equationy= 3x. This also represents a straight line, and for all the pointson theline eachyvalue is three times the correspondingxvalue. So the line will contain points in thefollowing 3x:xy001326xyy = 3xy = 2xy = xIf we calculate the gradient of the liney= 3xusing the first two sets of values in the table, weobtainm=3 01 0= 3so that the gradient of this line is mathcentre 2009We can start to see a pattern here. All these lines have Equations whereyequals some numbertimesx.

4 And in each case the line passes through the origin, and the gradient of the line isgiven by the number multiplyingx. So if we had a line with equationy= 13xthen we wouldexpect the gradient of the line to be 13. Similarly, if we had aline with equationy= 2xthenthe gradient would be 2. In general, therefore, the equationy=mxrepresents a straight linepassing through the origin with PointThe equation of a straight line with gradientmpassing through the origin is given byy=mx .3. They-intercept of a lineConsider the straight line with equationy= 2x+ 1. This equation is in a slightly different formfrom those we have seen earlier. To draw a sketch of the line, we must calculate some 2x+ 1:xy011325xyy = 2x + 1 Notice that whenx= 0the value ofyis 1. So this line cuts they-axis aty= about the liney= 2x+ 4? Again we can calculate some mathcentre 2009y= 2x+ 4:xy-120416xyy = 2x + 44 This line cuts they-axis aty= about the liney= 2x 1?

5 Again we can calculate some 2x 1:xy-1-30-111xyy = 2x - 1-1 This line cuts they-axis aty= general equation of a straight line isy=mx+c, wheremis the gradient, andy=cis thevalue where the line cuts they-axis. This numbercis called theintercepton PointThe equation of a straight line with gradientmand interceptcon they-axis isy=mx+c . mathcentre 2009We are sometimes given the equation of a straight line in a different form. Suppose we have theequation3y 2x= 6. How can we show that this represents a straight line, and findits gradientand its intercept value on they-axis?We can use algebraic rearrangement to obtain an equation in the formy=mx+c:3y 2x= 6,3y= 2x+ 6,y=23x+ now the equation is in its standard form, and we can see thatthe gradient is23and theintercept value on they-axis is can also work backwards. Suppose we know that a line has a gradient of15and has a verticalintercept aty= 1.

6 What would its equation be?To find the equation we just substitute the correct values into the general formulay=mx+ ,mis15andcis 1, so the equation isy=15x+ 1. If we want to remove the fraction, wecan also give the equation in the form5y=x+ 5, or5y x 5 = Determine the gradient andy-intercept for each of the straight lines in the table 3x+ 2y= 5x 2y= 2x+ 4y= 12xy=12x 232y 10x= 8x+y+ 1 = 02. Find the equation of the lines described below (give the equation in the formy=mx+c):(a) gradient 5,y-intercept 3;(b) gradient 2,y-intercept 1;(c) gradient 3, passing through the origin; (d) gradient13passing through(0,1);(e) gradient 34, mathcentre 20094. The equation of a straight line with given gradient,passing through a given pointExampleSuppose that we want to find the equation of a line which has a gradient of13and passes throughthe point(1,2). Here, whilst we know the gradient, we do not know the value start with the general equation of a straight liney=mx+ know the gradient is13and so we can substitute this value formstraightaway.

7 This givesy=13x+ now use the fact that the line passes through(1,2). This means that whenx= 1,ymustbe 2. Substituting these values we find2 =13(1) +cso thatc= 2 13=53So the equation of the line isy=13x+ can work out a general formula for problems of this type by using the same method. Weshall take a general line with gradientm, passing through the fixed pointA(x1, y1).We start with the general equation of a straight liney=mx+ now use the fact that the line passes throughA(x1, y1). This means that whenx=x1,ymust bey1. Substituting these values we findy1=mx1+cso thatc=y1 mx1So the equation of the line isy=mx+y1 can write this in the alternative formy y1=m(x x1)This then represents a straight line with gradientm, passing through the point(x1, y1). So thisgeneral form is useful if you know the gradient and one point on the PointThe equation of a straight line with gradientm, passing through the point(x1, y1), isy y1=m(x x1).

8 Mathcentre 2009 For example, suppose we know that a line has gradient 2and passes through the point( 3,2).We can use the formulay y1=m(x x1)and substitute in the values straight away:y 2 = 2(x ( 3))= 2(x+ 3)= 2x 6y= 2x Find the equation of the lines described below (give the equation in the formy=mx+c):(a) gradient 3, passing through(1,4);(b) gradient 2, passing through(2,0);(c) gradient25, passing through(5, 1); (d) gradient 0, passing( 1,2);(e) gradient 1, passing through(1, 1).5. The equation of a straight line through two given pointsWhat should we do if we want to find the equation of a straight line which passes through thetwo points( 1,2)and(2,4)?Here we don t know the gradient of the line, so it seems as though we cannot use any of theformul we have found so far. But we do know two points on the line, and so we can use themto work out the gradient. We just use the formulam= (y2 y1)/(x2 x1).

9 We getm=4 22 ( 1)= the gradient of the line is23. And we know two points on the line, so we can use one of themin the formulay y1=m(x x1). If we take the point(2,4)we gety 4 =23(x 2)3y 12 = 2x 43y= 2x+ 8y=23x+ before, it will be useful to find a general formula that can be used for examples of this suppose the general line passes through two pointsA(x1, y1)andB(x2, y2). We shall let ageneral point on the line beP(x, y).xyP(x, y)A(x1,y1)B(x2,y2) mathcentre 2009 Now we know that the gradient ofAPmust be the same as the gradient ofAB, as all threepoints are on the same line. But the gradient ofAPismAP=y y1x x1,whereas the gradient ofABismAB=y2 y1x2 , so we must havey y1x x1=y2 y1x2 this formula is fairly complicated, but it is easier to remember if all the terms involvingyare on one side, and all the terms involvingxare on the other. If we manipulate the formula, weget firsty y1= (x x1)y2 y1x2 x1and theny y1y2 y1=x x1x2 might help you to remember this formula if you notice that the pattern on the left-hand side,involvingy, is just the same as the pattern on the right-hand side, PointThe equation of a straight line passing through the two points(x1, y1)and(x2, y2)isy y1y2 y1=x x1x2 we can use this formula for an example.

10 Suppose that we want to find the equation of thestraight line which passes through the two points(1, 2)and( 3,0). We just substitute intothe formula, and rearrange. The various steps arey ( 2)0 ( 2)=x 1 3 1y+ 22=x 1 4y+ 2 =x 1 2= 12(x 1) 2y 4 =x 1 2y=x+ 3y= 12x mathcentre 2009So the line has gradient 12and its intercept on they-axis is 32. We can also rearrange theequation a little further to obtain2y= x 3, or2y+x+ 3 = Find the equation of the lines described below (give the equation in the formy=mx+c):(a) passing through(4,6)and(8,26), (b) passing through(1,1)and(4, 8),(c) passing through(3,4)and(5,4),(d) passing through(0,2)and(4,0),(e) passing through( 2,3)and(2, 5).6. The most general equation of a straight lineThere is one more form of the equation for a straight line thatis sometimes needed. This is theequationax+by+c= have written Equations in this form for some of our examples.