- #1
chwala
Gold Member
- 2,697
- 369
- Homework Statement
- See attached
- Relevant Equations
- straight line equations
Find the question here; My interest is on question ##3(c)## only.
My approach, Let the co ordinates of ##C##= ##(x,y)## then considering points ##B## and ##C##. We shall have the gradient given by;
##\dfrac {y-4}{x-1}##=##-2##
also from straight line equation, considering points ##A## and ##C##, we shall have;
##(x+3)^2+(y-2)^2=40##
we know that, ##y=-2x+6## from the given equations above, then we shall have,
##(x+3)^2+(-2x+6-2)^2=40##
##(x+3)^2+(-2x+4)^2=40##
##5x^2-10x-15=0##
##x^2-2x-3=0##
therefore possible co ordinates of ##C## are ##(3,0)## and ##(-1,8)##
I am seeking a much simpler approach...of course i assume the reader is conversant with my approach...because of time i cannot show step by step...but shout out to me if an equation is not clear. Bingo!
heeey!
My approach, Let the co ordinates of ##C##= ##(x,y)## then considering points ##B## and ##C##. We shall have the gradient given by;
##\dfrac {y-4}{x-1}##=##-2##
also from straight line equation, considering points ##A## and ##C##, we shall have;
##(x+3)^2+(y-2)^2=40##
we know that, ##y=-2x+6## from the given equations above, then we shall have,
##(x+3)^2+(-2x+6-2)^2=40##
##(x+3)^2+(-2x+4)^2=40##
##5x^2-10x-15=0##
##x^2-2x-3=0##
therefore possible co ordinates of ##C## are ##(3,0)## and ##(-1,8)##
I am seeking a much simpler approach...of course i assume the reader is conversant with my approach...because of time i cannot show step by step...but shout out to me if an equation is not clear. Bingo!
Last edited: