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First of all, is it Geometric?

Image of the drafted:

enter image description here

I need help solving this question, and I am completely lost on how can I solve this. Could anyone explain the way of solving this geometric question?

Here is a drafting of the two lines I and II. With 3 formulas (a), (b), (c).

(a) $y = 2x + 8$,

(b) $y = - 2x + 8$

(c) $y = x + 2$.

A. Every of the straights I and II, please find the correct formula out of the formulas (a) , (b) , (c), and explain your answer.

b. Find the points of intersection of the straight rates.

c. Find the straight formula that's going through the way of the point (5, 2) and parallels to the straight II.

How do I do this?

This is a translated version.

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  • $\begingroup$ Exactly what part of this are you having trouble with? Can you outline your attempt? $\endgroup$ – Ataraxia May 20 '13 at 12:49
  • $\begingroup$ @ZettaSuro: I think it is better if you explain to the OP why you think the "thank you so much" should be removed and ask them to remove it themselves, rather than removing it yourself. $\endgroup$ – user1729 May 20 '13 at 12:50
  • $\begingroup$ 73230, do you know anything about analytic geometry? about representing equations with graphs? about the equation of a straight line? If not, you need more help than we can give here; you need to find a text that covers these topics, and study it. But surely you would not have been given this as homework if these topics had not been covered somewhere? Maybe a talk with the teacher would be the best idea. $\endgroup$ – Gerry Myerson May 20 '13 at 12:55
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Hints: for the straight line $\,y=ax+b\,$:

$\bullet\;\;$ The line above is ascending iff $\,a>0\,$ , and it is constant iff $\,a=0\,$ ;

$\bullet\;\;$ the $\,y$-intersection of the above line is the point $\,(0,b)\,$ ;

$\bullet\;\;$ The line above intercepts the line $\,y=mx +n\,$ at the point $\,(x_0,y_0)\,$ which is the solution of the linear system

$$\begin{cases}y=ax+b\\{}\\y=mx+n\end{cases}$$

How to solve: compare $\,y$-s:

$$ax+b=mx+n\implies(a-m)x=n-b\implies \,\text{if}\;a\neq b\;,\;x=\frac{n-b}{a-m}$$

Iff $\,a=b\,$ the lines are parallel and thus they are the very same line or else they have no common point.

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