m=2
XXXXXXXX in XXXXXXX is y=mx+b where b XX XXXXXXXXX
XXX XXXXXXXXXXX it XXX XX use XXXXX B(6,11)
So XX XXXX XXXX y=XX XXX x=X lets put in the XXXXXXXX
X=mx+b
11=X×6+b
b=-1
so equation is = X − 1
XXX doing so XXXXX XXXXXXXXX slope XX XXX perpendicular XXXXXXXX
XXXXXXXX of XX XXX (
,
)=(X,X)
XXXXX XX XXX XX XXX calculate from XXXXX points XX
= = 2
and XXXX XX the Perpendicular XXXXXXXX XX− XX XX so will XX -
XXX XX XXX XXXXXXXXX the XXXXXXXXX using (X, 9) and m XX -1/X
XX get
9=− ×X+b
and y =-12hence XXXXX XXX XXX XXXXXXXXXX XX point C
b=XX.X
now equation XX
X= − + 11.5 is
(3ks)
Let’s find the XXXXXXXX XX XXXX AC XXXXX XX perpendicular to line AB having equation XXXXX is y=2x-X perpendicular line XX one XXXX XXX a XXXXXXXX, XXXXXXXXXX XXXXX to XXXXXXX.
So XX XXXX have XXXXXXXX y=-1/2x-X
As y will XX XXXX for the point c XXXXXX we XXX equation of XX or XXXXXXXX Ac so we can XXX XXXX XXXXX
-X/XX-X=-XX+XX
2.5x=XX
X=92/5