help would be great

\(\text{Both points are on the line so plug them in}\\ a(-5+4i) + b(-5-4i)=38\\ a(7+2i)+b(7-2i) = 38\\ (-5a-5b)+i(4a-4b)=38\\ (7a+7b)+i(2a-2b)=38\)

\(-5(a+b)-14(a+b)=-38\\ -19(a+b)=-38\\ a+b=2\)

\(-5(2) + i(4a-4(2-a))=38\\ i(8a-8)=48\\ i(a-1)=6\\ a-1 = -6i\\ a=1-6i\\ b=2-(1-6i) = 1+6i\)

\((a,b) = (1-6i,1+6i)\)

Hi Rom,

Thanks for your answer  

but...

I do not understand that initial plug in.  Why have you used the conjugates??

Could you explain a little more please. 

we're told that both points lie on the line

\(a z + b \bar{z} = 38\)

\(\text{So just plug both of these points into the formula there and we solve for }a \text{ and }b\\ \text{I use the complex conjugate because that's how they've specified the line}\)