Find Co-ordinates of Point C in Problem Involving Straight Line Equations

[chwala]

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!

 

[andrewkirk]

Simpler approach:
To get from A to B we add 4 to x and 2 to y.
The segment from B to C is perpendicular to AB and the same length, so it must have absolute values of changes in x and y reversed, ie 2 and 4, and those changes must have opposite signs.
Hence, adding those changes to the coordinates of B = (1,4), we see it will be
(1 + 2, 4 - 4) = (3, 0)
OR
(1 - 2, 4 + 4) = (-1, 8)

 

[chwala]

Simpler approach:
To get from A to B we add 4 to x and 2 to y.
The segment from B to C is perpendicular to AB and the same length, so it must have absolute values of changes in x and y reversed, ie 2 and 4, and those changes must have opposite signs.
Hence, adding those changes to the coordinates of B = (1,4), we see it will be
(1 + 2, 4 - 4) = (3, 0)
OR
(1 - 2, 4 + 4) = (-1, 8)

Thanks, I had initially thought of this approach and found it ambiguous ...correct though...