# Thread: A linear graph problem

1. ## A linear graph problem

Ok, so, there is a question I need to have answered before Tuesday afternoon.
I am at 5th grade of high school now.
As in the title I will post the problem and its solution.
I am seeking a second and shorter way to solve this.
There is! another way, cause our teacher told us.
I have been trying myself to solve this for many hours now, so my failure to find it out led me here.
NOTE: I am from greece, thus I am not experienced in writing Math in English, but I will do my best. (Please correct me or ask for clarification any time.)
Here it is:
Find all the points of the straight line e1: x-y+2=0, which have a distance of 1 from the straight line e2: 12x-5y+60=0

1st solution: Assuming M(Xo,Yo) as a point of e1, such as to be d(M,e2)=1

• Because M belongs to e1 (I tried to find this ε like the euro sign that means belongs to but I couldn't) it is Xo-Yo+2=0 (1)

• It is d(M,e2)=1 <=> |12Xo-5Yo+60|/(whole denominator on sqrtroot)12^2+(-5)^2=1 <=>

<=>|12Xo-5Yo+60|/13=1 <=> |12Xo-5Yo+60|=13 <=> 12Xo-5Yo+60=13 or 12Xo-5Yo+60=-13 <=> 12Xo-5Yo+47=0 or 12Xo-5Yo+73=0
As of this we have:
{ Xo-Yo=-2 | x (-12)
{
12Xo-5Yo=-47 | x 1

{-12Xo+12Yo=24
{
12Xo-5Yo=-47

{Xo-Yo=-2 (Returning to the first equation of the system)
{
7Yo=-23

{Xo=-37/7 (Replacing)
{
Yo=-23/7

So M(-37/7,-27/7)

Second Possibility of M:

{ Xo-Yo=-2 | x (-12)
{
12Xo-5Yo=-73 | x 1

{-12Xo+12Yo=24
{
12Xo-5Yo=-73

{ Xo-Yo=-2
{ 7Yo=-49

{ Xo=-9
{ Yo=-7

So M(-9,-7)

End of the 1st Solution.
2nd Solution: ?

Thank you in advance for helping.

2. ## Straight line graphs using similar triangles

Hello -

Find the equations of the lines parallel to 5y = 12x + 60, 1 unit away, using the intercept on the y-axis (y = 12) and similar triangles.

In the attached diagram, triangles ABC, DCA and FEA are all similar 5, 12, 13 triangles. AC = AE = 1 unit = 5/5 units. So AD = AF = 13/5.

A is at y = 12. So D and F are at y = 12 - 13/5 = 9.4 and 12 + 13/5 = 14.6.

So the equations of CD and EF are:

y = 12x/5 + 9.4 and y = 12x/5 + 14.6

These lines intersect y = x + 2 at the two points required.

i.e. where x + 2 = 12x/5 + 9.4 and x + 2 = 12x/5 + 14.6

i.e. where x = -37/7 and x = -9

So the points are (-37/7, -23/7) and (-9, -7).

That's a bit easier!

3. Originally Posted by Grandad
Hello -

Find the equations of the lines parallel to 5y = 12x + 60, 1 unit away, using the intercept on the y-axis (y = 12) and similar triangles.

In the attached diagram, triangles ABC, DCA and FEA are all similar 5, 12, 13 triangles. AC = AE = 1 unit = 5/5 units. So AD = AF = 13/5.

A is at y = 12. So D and F are at y = 12 - 13/5 = 9.4 and 12 + 13/5 = 14.6.

So the equations of CD and EF are:

y = 12x/5 + 9.4 and y = 12x/5 + 14.6

These lines intersect y = x + 2 at the two points required.

i.e. where x + 2 = 12x/5 + 9.4 and x + 2 = 12x/5 + 14.6

i.e. where x = -37/7 and x = -9

So the points are (-37/7, -23/7) and (-9, -7).

That's a bit easier!

Well, I don't understand very well this method, and this might be because 1) We haven't been taught similar triangles so thoroughly at school or 2) I wasn't paying enough attention when taught . Thank you very much for your post and secondly, if you could explain me a bit more or maybe show me this solution typed in mathematics because I suppose I will be able to understand it more then. I mean, my questions are how did u find the triangle values at the beginnings and what are the criterias that u used to prove that they are similar triangles.

4. ## Similar Triangles

Hello Aionios -

Originally Posted by Aionios
Well, I don't understand very well this method, and this might be because 1) We haven't been taught similar triangles so thoroughly at school or 2) I wasn't paying enough attention when taught . Thank you very much for your post and secondly, if you could explain me a bit more or maybe show me this solution typed in mathematics because I suppose I will be able to understand it more then. I mean, my questions are how did u find the triangle values at the beginnings and what are the criterias that u used to prove that they are similar triangles.
Two mathematical figures are said to be similar if they have exactly the same shape. (They're not necessarily the same size; if they are the same size as well, they are said to be identical, or congruent.)

So all squares will be similar, but not all rectangles. That's because all four sides of a square are the same length, but a rectangle can be long and thin, or short and fat.

You can always tell when you have two similar triangles: if their angles are the same as each other. For instance, if in triangle ABC, A = 90 deg, B = 60 deg and C = 30 deg, while in triangle PQR, Q = 90 deg, P = 60 deg and R = 30 deg, then the triangles are similar.

If you look at the diagram I attached with my first posting, you'll see some similar triangles. Triangle ABC is right-angled at A, and triangle ACD is right-angled at C. But that alone doesn't make them similar. So what about the other angles? Well, can you see why angle CAB = angle ADC? It's because when you add each one to angle DAC you get 90 degrees. Now if in two triangles, we have two pairs of equal angles, then the third pair must be equal, because all three angles in a triangle add up to 180 deg.

So, triangles ABC and DCA are similar. So what? Well, this means that their sides are in the same ratio. What do I mean by this? I'll come to that in a minute.

First, look at the original line, y = 12x/5 + 12. This has a gradient of 12/5, which means that as we move along the line, for every 5 units we move in the x-direction, we shall move 12 units in the y-direction. Now the line EAC is at right-angles to this line, so as we move along EAC, for every 12 units we move in the x-direction we shall move -5 units in the y-direction. So the ratio AB:BC = 12:5. By Pythagoras Theorem, this means that triangle ABC is a 5, 12, 13 triangle.

What this also means is that any other triangle that is similar to triangle ABC is also a 5, 12, 13 triangle. Aha! that means that triangle ACD is also a 5, 12, 13 triangle. So AD:AC = 13:5. But AC is 1 unit long, because we drew the lines DC and EF 1 unit away from y = 12x/5 +12.

So AD = 13/5 = 2.6 units. And so is AF, by the same reasoning.

Now use y = mx + c to find the equations of the lines DC and EF, and you're home and dry!

Hope that helps.

5. Originally Posted by Grandad
Hello Aionios -

Two mathematical figures are said to be similar if they have exactly the same shape. (They're not necessarily the same size; if they are the same size as well, they are said to be identical, or congruent.)

So all squares will be similar, but not all rectangles. That's because all four sides of a square are the same length, but a rectangle can be long and thin, or short and fat.

You can always tell when you have two similar triangles: if their angles are the same as each other. For instance, if in triangle ABC, A = 90 deg, B = 60 deg and C = 30 deg, while in triangle PQR, Q = 90 deg, P = 60 deg and R = 30 deg, then the triangles are similar.

If you look at the diagram I attached with my first posting, you'll see some similar triangles. Triangle ABC is right-angled at A, and triangle ACD is right-angled at C. But that alone doesn't make them similar. So what about the other angles? Well, can you see why angle CAB = angle ADC? It's because when you add each one to angle DAC you get 90 degrees. Now if in two triangles, we have two pairs of equal angles, then the third pair must be equal, because all three angles in a triangle add up to 180 deg.

So, triangles ABC and DCA are similar. So what? Well, this means that their sides are in the same ratio. What do I mean by this? I'll come to that in a minute.

First, look at the original line, y = 12x/5 + 12. This has a gradient of 12/5, which means that as we move along the line, for every 5 units we move in the x-direction, we shall move 12 units in the y-direction. Now the line EAC is at right-angles to this line, so as we move along EAC, for every 12 units we move in the x-direction we shall move -5 units in the y-direction. So the ratio AB:BC = 12:5. By Pythagoras Theorem, this means that triangle ABC is a 5, 12, 13 triangle.

What this also means is that any other triangle that is similar to triangle ABC is also a 5, 12, 13 triangle. Aha! that means that triangle ACD is also a 5, 12, 13 triangle. So AD:AC = 13:5. But AC is 1 unit long, because we drew the lines DC and EF 1 unit away from y = 12x/5 +12.

So AD = 13/5 = 2.6 units. And so is AF, by the same reasoning.

Now use y = mx + c to find the equations of the lines DC and EF, and you're home and dry!

Hope that helps.

havent read it all already but I think I found something that looks like a typo? "This has a gradient of 12/5, which means that as we move along the line, for every 5 units we move in the x-direction, we shall move 12 units in the y-direction" x-direction and y-direction reversed in the sentence? And second, "Triangle ABC is right-angled at A" If i am not wrong right-angled at an angle means that angle is 90 degrees. Well, the 90 angle in that triangle is B. These found yet. Thanks in advance for clarifying.

6. ## Typos

Hi -

Originally Posted by Aionios
havent read it all already but I think I found something that looks like a typo? "This has a gradient of 12/5, which means that as we move along the line, for every 5 units we move in the x-direction, we shall move 12 units in the y-direction" x-direction and y-direction reversed in the sentence? And second, "Triangle ABC is right-angled at A" If i am not wrong right-angled at an angle means that angle is 90 degrees. Well, the 90 angle in that triangle is B. These found yet. Thanks in advance for clarifying.
Yes and no!

Yes, you're right, triangle ABC is right-angled at B, not A, as I said. Sorry!

But no, a gradient of 12/5 means (increase in y)/(increase in x) = 12/5. So for every 5 units you move in the x-direction you do move 12 in the y-direction.

7. ## Last clarification thanks!

Sorry for not posting my questions earlier and I am wasting your time watching my posts again and again, but my connection sucked badly for some strange reason the past 12 hours. Another question too. What formula did u use to derive these?: " So D and F are at y = 12 - 13/5 = 9.4 and 12 + 13/5 = 14.6." And if you could, explain a bit further this last part: "These lines intersect y = x + 2 at the two points required.

i.e. where x + 2 = 12x/5 + 9.4 and x + 2 = 12x/5 + 14.6

i.e. where x = -37/7 and x = -9

So the points are (-37/7, -23/7) and (-9, -7)."
Also, tell me if I understood this well: "So the ratio AB:BC = 12:5. By Pythagoras Theorem, this means that triangle ABC is a 5, 12, 13 triangle." So you put the value 5 as the BC side and the value 12 at the AB side and by P.T. you derive it. But what about the next triangle that is similar but bigger than that? Why is it looking smaller? Because when saying its one side is 12/5 that looks smaller than the 12 side of the in fact smaller triangle ABC?

Thanks a lot for all you interest already in my thread!

8. ## Ratios and Similar Triangles

Hello Aionios -

Originally Posted by Aionios
What formula did u use to derive these?: " So D and F are at y = 12 - 13/5 = 9.4 and 12 + 13/5 = 14.6." And if you could, explain a bit further this last part: "These lines intersect y = x + 2 at the two points required.

i.e. where x + 2 = 12x/5 + 9.4 and x + 2 = 12x/5 + 14.6

i.e. where x = -37/7 and x = -9

So the points are (-37/7, -23/7) and (-9, -7)."
Also, tell me if I understood this well: "So the ratio AB:BC = 12:5. By Pythagoras Theorem, this means that triangle ABC is a 5, 12, 13 triangle." So you put the value 5 as the BC side and the value 12 at the AB side and by P.T. you derive it. But what about the next triangle that is similar but bigger than that? Why is it looking smaller? Because when saying its one side is 12/5 that looks smaller than the 12 side of the in fact smaller triangle ABC?

Thanks a lot for all you interest already in my thread!
You need to understand that when we talk about a 5, 12, 13 triangle, we are talking about the ratios of the sides, not their actual lengths. In other words, the lengths of the sides are in the ratio 5:12:13.

This means, for instance, that the shortest side is 5/12 of the middle side and 5/13 of the longest side. It doesn't mean that the shortest side is 5 units long, the next 12 units and the longest 13 units. So if, for example, the longest side is 26 cm, the shortest will be 5/12 x 26 = 10 cm long, and the middle sized side will be 12/13 x 26 = 24 cm long.

Turning these fractions the other way up tells you that the middle side is 12/5 times the length of the shortest side, and that the longest one is 13/5 times the shortest. So, for example, if the shortest side is 20 cm long, the middle one will be 12/5 x 20 = 48 cm, and the longest 13/5 x 20 = 52 cm.

Or, if the shortest side is 1 cm long, the longest one will be 13/5 x 1 = 2.6 cm.

Now in the diagram I drew, AC = 1 unit (because the lines were 1 unit apart), so AD = 13/5 x 1 = 2.6 units. And, of course, AF = 2.6 units, also, for the same reason.

Next, I used y = mx + c to get the equations of the two lines EF and DC. I hope you know that, in this equation, the value of m gives the gradient, and the value of c the intercept with the y-axis. (If you don't really know this, you need to start co-ordinate geometry from the beginning again!)

Now in the original line y = 12x/5 + 12, m = 12/5 and c = 12. So the value of m is also 12/5 for the lines EF and DC (because these lines are both parallel to y = 12x/5 + 12) and the value of c is 12 + 2.6 = 14.6 for EF and 12 - 2.6 = 9.4 for DC. So the lines are:

y = 12x/5 +14.6 and y = 12x/5 + 9.4.

Finally, you need to solve some simultaneous equations to find where these lines meet y = x + 2. For EF, we eliminate y to get:

x + 2 = 12x/5 + 14.6

Multiply both sides by 5: 5x + 10 = 12x + 47

Therefore: -37 = 7x

Therefore: x = -37/7.

Now y = x + 2, so when x = -37/7, y = -37/7 + 2 = - 23/7.

I'll leave it to you to solve

{ y = 12x/5 + 9.4
{ y = x + 2

in the same way.