ed84c Posted November 7, 2007 Share Posted November 7, 2007 Ok so I have 2 points A = 2 + 3j C = 4 + 7j on our argand diagram they make an equailateral triangle with a point (or more precisely either of 2 points) which I'll call z Now I know; [math] ABC = arg ( \lambda ) = arg (\frac{a - b}{c-b}) [/math] so [math] arg(\lambda) = \frac{\pi}{3} [/math] and [math] \lambda = \frac{a-b}{c-b} [/math] [math] \lambda = \frac{(z-(4+7j))-((4+7j)-(2+3j))}{(z-(2+3j))-((4+7j)-(2+3j))} [/math] [math] \lambda = \frac{z-6-11j}{z-4-j} [/math] No where do I go from here; solve simultaneously, or am I using the completely wrong method? I've attatched a sketch of the argand diagram. Link to comment Share on other sites More sharing options...
Country Boy Posted November 10, 2007 Share Posted November 10, 2007 Ok so I have 2 points A = 2 + 3j C = 4 + 7j on our argand diagram they make an equailateral triangle with a point (or more precisely either of 2 points) which I'll call z Now I know; [math] ABC = arg ( \lambda ) = arg (\frac{a - b}{c-b})[/math] I'm not familiar with this. "ABC" is a real number. What number? The cosine of the angle between sides? I would be more inclined to use modulus than arg: |A- C|= |-2- 4j|= [math]\sqrt{4+ 16}= \sqrt{20}[/math]. Since this is an equilateral triangle, we must have also |z- A|= |z- C|=[math]\sqrt{20}[/math] also. Letting z= x+ yj, [math]\sqrt{(x-2)^2+ (y-3)^2}= \sqrt{20}[/math] and [math]\sqrt{(x-4)^2+ (y-7)^2}= \sqrt{20}[/math]. Of course, those immediately give [math](x- 2)^2+ (y-3)^2= 20[/math] and [math](x- 4)^2+ (y- 7)^2= 20[/math]. You should be able to see that, after you multiply those out, the [math]x^2[/math] and [math]y^2[/math] terms cancel and you have two linear equations to solve for x and y. so [math] arg(\lambda) = \frac{\pi}{3} [/math] and [math] \lambda = \frac{a-b}{c-b} [/math] [math] \lambda = \frac{(z-(4+7j))-((4+7j)-(2+3j))}{(z-(2+3j))-((4+7j)-(2+3j))} [/math] [math] \lambda = \frac{z-6-11j}{z-4-j} [/math] No where do I go from here; solve simultaneously, or am I using the completely wrong method? I've attatched a sketch of the argand diagram. Link to comment Share on other sites More sharing options...
Freirec Posted November 12, 2007 Share Posted November 12, 2007 if 2a+7x(95z^2)=67 how do u find a? Link to comment Share on other sites More sharing options...
timo Posted November 12, 2007 Share Posted November 12, 2007 By asking for z and x. Link to comment Share on other sites More sharing options...
Freirec Posted November 13, 2007 Share Posted November 13, 2007 no, like in the form a=?z +/- ?x Link to comment Share on other sites More sharing options...
timo Posted November 13, 2007 Share Posted November 13, 2007 1) substract the 2nd addend on the left-hand side from both sides 2) divide both sides by 2 Link to comment Share on other sites More sharing options...
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