# Help with linear equations homework (1 Viewer)

#### AmalJ

##### New Member
So I'm doing linear equations at school and we have this question
"Find the equation of the line that is parallel to y=5-x/2 and passes through (-1,6)"

I found my answer using a different method and I'm not sure that if I were to use it in an exam, I would get full marks

So the method I used was as follows:

y=5-x/2
y=-1/2x+5 P(-1,6)

6=-1/2(-1)+b
6=1/2+b
11/2 = b
so Y=-1/2x+11/2

The one we're taught is this one:

y=5-x/2
y=-1/2x+5 P(-1,6)

y-y=m(x-x)
y-6=-1/2(x-1)
y-6=-1/2x - 1/2
y-11/2 = -1/2x
y= -1/2x+11/2

I found method 2 to be more complicated and I spent some time figuring out how it even works so I'd rather not use it in an exam. Do I have to use it? and even if I don't, should I still practice doing it? (I checked the textbook and it just says that the answer is y=-1/2x+11/2)

• cossine

#### cossine

##### Active Member
The method used is interesting. You should get full marks.

the gradient formula is given by

(y1 - y2 ) / (x1 - x2) = m

=> m(x1 - x2) = y1 - y2 #Multiply by (x1-x2)

#Just small note there are some minor issues but steps 3-5 are correct
y - y = m(x - x)
y - 6 = -1/2(x - -1)
y - 6 = (-1/2)x - 1/2
y - 11/2 = (-1/2)x
y = (-1/2)x + 11/2

Good job.

• AmalJ

#### Eagle Mum

##### Well-Known Member
You’re essentially doing the steps in a different order.
The approach you’ve been taught is to i) set out the general equation of the line ii) substitute the negative reciprocal of the slope coefficient of the original equation & iii) substitute the point coordinates, whereas you have performed the steps in the order ii), iii), then i). I am not involved in HSC maths at all, but IMHO you ought not to lose any marks because your solution is valid. However, the way it’s been taught is the formally ‘correct’ way as it sets out the general equation first so that in the solution, it’s evident what processes (substitutions) are occurring.

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• AmalJ

#### quickoats

##### Well-Known Member
Just note that if you ever go onto further maths, the 2nd way is much neater when you extend to 3d etc.

• AmalJ

#### CM_Tutor

##### Moderator
Moderator
So I'm doing linear equations at school and we have this question
"Find the equation of the line that is parallel to y=5-x/2 and passes through (-1,6)"

I found my answer using a different method and I'm not sure that if I were to use it in an exam, I would get full marks

So the method I used was as follows:

y=5-x/2
y=-1/2x+5 P(-1,6)

6=-1/2(-1)+b
6=1/2+b
11/2 = b
so Y=-1/2x+11/2
Looking at the method you used, the part I don't like is that the pronumeral just suddenly appears. However, the reasoning is sound. I would tweak your approach as follows:

Please note, however, that the general method that you have been taught will occur regularly in HSC work in situations where you have the gradient of a desired line and a point on it. Your approach is adaptable to such situations but it may not always be quicker or easier, so it is worth learning and understanding the standard approach.

• AmalJ

#### AmalJ

##### New Member
Thank you for helping me! )

#### AmalJ

##### New Member
Oh I should've put it in but it just comes from y=mx+b

also wow that method looks so much better I'll make sure to learn it thank you!!
And you're right, I'll probably learn the other method as well since in the future it's probably going to be the easier option once the equations start getting more complicated.

#### CM_Tutor

##### Moderator
Moderator
@AmalJ, I realised that the comes from the gradient-intercept form.

The ability to see alternative approaches is something that stronger students often have. The best can also find wats to express their working in a way that is clear and persuasive. When I first read your answer, I had doubts because of the setting out until I recognised the reasoning you were using. I encourage you to work on presentation of your reasoning, it will benefit you in the longer term - and by that, I mean using both algebra and words.

• AmalJ