- #1

chwala

Gold Member

- 2,727

- 382

- 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! heeey!

Last edited: