Need help understanding: The equation of a straight line....

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sujoy

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Q: The equation of a straight line through the point of intersection of two given lines:

. . .[a<sub>1</sub>x ] + [b<sub>1</sub>y] + c<sub>1</sub> = 0

. . .[a<sub>2</sub>x] + [b<sub>2</sub>y] + c<sub>2</sub> = 0

...is given by :

. . .((a<sub>1</sub>x) + (b<sub>1</sub>y) + c<sub>1</sub>) + k((a<sub>2</sub>x) + ( b<sub>2</sub>y) + c<sub>2</sub>) = 0

It's the bolded part above that I don't understand.

For this question, I do not have any difficulty in deriving the co-ordinates of the point of intersection, (x<sub>1</sub>, y<sub>1</sub>). They are:

. . .x<sub>1</sub> = ( b<sub>1</sub>c<sub>2</sub> - b<sub>2</sub>c<sub>1</sub>) / ( a<sub>1</sub>b<sub>2</sub> - a<sub>2</sub>b<sub>1</sub>)

. . .y<sub>1</sub> = (c<sub>1</sub>a<sub>2</sub> - c<sub>2</sub>a<sub>1</sub>) / (a<sub>1</sub>b<sub>2</sub> - a<sub>2</sub>b<sub>1</sub>)

But how does the bolded part work?

Regards,
Sujoy
 
The intersection of two lines is just a point. There are infintely-many other lines through that intersection point. What do you mean by "the" line through the intersection point?

Please reply with the full text of the original exercise, including definitions of a<sub>1</sub>, b<sub>1</sub>, etc, and how they relate to the lines and points in question.

Thank you.

Eliz.
 
Re: Need help understanding: The equation of a straight line

sujoy said:
Q: The equation of a straight line through the point of intersection of two given lines:

. . .[a<sub>1</sub>x ] + [b<sub>1</sub>y] + c<sub>1</sub> = 0

. . .[a<sub>2</sub>x] + [b<sub>2</sub>y] + c<sub>2</sub> = 0

...is given by :

. . .((a<sub>1</sub>x) + (b<sub>1</sub>y) + c<sub>1</sub>) + k((a<sub>2</sub>x) + ( b<sub>2</sub>y) + c<sub>2</sub>) = 0

It's the bolded part above that I don't understand.

[snipped]

But how does the bolded part work?

Regards,
Sujoy
Click to expand...
First, for any fixed k, the equation in bold is of a straight line since it has the form of ax + by + c = 0 after collecting coefficients.

Second, the line of the bold equation goes through the point of intersection of the first two lines because when (x,y) satisfies the first and second equations, it will satisfy the third also.
 
equation of a st. line is given by ax +by +c = 0 whert a,b,&c are constants. IT comes from the fact that that if we have 3 points whose co-ordinatews are known,then we may derive the equation to the the area of a triangle . Now this would obviously be 0 if the points lie on a st.line.A this point if we seperate one point(x ,y) ,take this common in the equation of the triangle so formed {which is 0}
what we get is of the form
constant1(a) + constant2(b) +constant3 = 0 , the general equation of a st.line.
Now from this we get that a_1,b_1 ....a_3,b_3,c_3 are all constants different from each other however vthe point in common is [x ,y]
what teacher JakeD wrote
First, for any fixed k, the equation in bold is of a straight line since it has the form of ax + by + c = 0 after collecting coefficients.

Second, the line of the bold equation goes through the point of intersection of the first two lines because when (x,y) satisfies the first and second equations, it will satisfy the third also.
Click to expand...
is exactly what is given,
what I fail to understand is the intutive thought of writing the equation in this way
it is given that the equation in question
. . .((a<sub>1</sub>x) + (b<sub>1</sub>y) + c<sub>1</sub>) + k((a<sub>2</sub>x) + ( b<sub>2</sub>y) + c<sub>2</sub>) = 0
is "ANY" line passing through the point
(x ,y)
this one is a difficulty of my own .
Regards
Sujoy
 
Sujoy, here is how I look at this. Maybe it will help your intuition.

Code: 
+                       *
  +                   *
    +               *  Line A point (x,y1)
      +           *
        +       *
          +   *
# # # # # # # # # # # # #  
          *   +        Line C point (x,(y1+ky2)/(1+k))
        *       +
      *           +
    *               +  Line B point (x,y2)
  *                   +
*                       +

If we take the equations of two lines in slope/intercept form (just for convenience)

\(\displaystyle \L \text{Line A: }y = m_1 x + d_1\)
\(\displaystyle \L \text{Line B: }y = m_2 x + d_2\)

and add equation A plus \(\displaystyle \L k\) times equation B we get

\(\displaystyle \L \text{Line C: }(1+k)y = (m_1 x + d_1 ) + k(m_2 x + d_2 ),\) or
\(\displaystyle \L \text{Line C: }y = \frac{(m_1 x + d_1 )}{(1+k)} +
\frac{k(m_2 x + d_2 )}{(1+k)}.\)

The points on Line C are combined just like the equations. That is, if \(\displaystyle \L (x,y_1 )\) is a point on Line A and \(\displaystyle \L (x,y_2 )\) is a point on Line B, then the corresponding point on Line C is \(\displaystyle \L (x,\frac{y_1}{1+k} + \frac{k y_2}{1+k} ) .\)

This is straightforward to show because we have

\(\displaystyle \L \text{Line A: }y_1 = m_1 x + d_1\)
\(\displaystyle \L \text{Line B: }y_2 = m_2 x + d_2\)

so that

\(\displaystyle \L \text{Line C: }y = \frac{(m_1 x + d_1 )}{(1+k)} + \frac{k(m_2 x + d_2 )}{(1+k)}
= \frac{y_1}{1+k} + \frac{k y_2}{1+k} .\)

By varying \(\displaystyle \L k\), we can vary the slope of Line C and so make it any line going through the intersection of Lines A and B.
 
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