Answer to Question #100638 in Analytic Geometry for Sandra

Question

the equations [1,2] + t[3,2] and [7,6] + t[-6,-4] give the same line, as(7,6) is a point on the first line,and the direction vectors are scalar multiples of each other. give a third equation that generates the same line.write another two different ewuations that give the same line as each other. include a langraph of your lines

Answer 1

Finding a third equation that generates the same line

From the given the given data it is said that the two vectors are scalar multiples of each other.two vectors are parallel when direction vectors scalar multiple of each other

when t=2 the first equation

$[1,2] + t[3,2] = [7,6]$

when t=0 the second equation

$[7,6] + t[-6,-4] =[7,6]$

therefore the two lines are coincident lines.

lets consider first equation

substituting t=t'+2

$[1,2]+(t'+2)[3,2] \newline =[7,6]+t'[3,2] \to(1)$

solving this for t=0 gives the point $(7,6)$

Writing two other equations that gives the same line

lets consider first equation

substituting t=2t'+2 gives

$[1,2]+(2t'+2)[3,2] \newline =[7,6]+t'[6,4] \to(2)$

substituting t=3t'+2 to the first equation gives

$[1,2]+(3t'+2)[3,2] \newline =[7,6]+t'[9,6] \to(3)$

and for both of them, the direction vector is a scalar multiple of each other

Since the lines are coincident and parallel the lines are as shown in the following figure ,where for various t values the lines follow the vector shown in the figure.