2005 paper 2 solution
2005 paper 2 solution
2005 paper 2 solution
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
3. (a) m AB =<br />
( y<br />
( x<br />
2<br />
2<br />
− y1)<br />
− x )<br />
1<br />
midpoint = (3,2)<br />
(4 − 0)<br />
=<br />
(5 −1)<br />
4<br />
= 4<br />
y – b = m(x – a)<br />
y – 2 = –1(x – 3)<br />
= 1 y – 2 = –x + 3<br />
m chord = –1 as m 1 .m 2 = –1 for lines x + y = 5<br />
(b) m T =<br />
1<br />
− (c) Perpendicular bisector of AB (answer (a))<br />
3<br />
m r = 3 and radius CA will intersect at centre, C.<br />
y – b = m(x – a) x + (3x – 3) = 5<br />
y – 0 = 3(x – 1) 4x = 8<br />
y = 3x – 3 x = 2<br />
y = 3<br />
C(2,3)<br />
r =<br />
r =<br />
2<br />
( xC<br />
− xA<br />
) + ( yC<br />
− y A<br />
2<br />
( 2 − 1) + (3 − 0)<br />
2<br />
)<br />
2<br />
r =<br />
2<br />
( 1) + (3)<br />
2<br />
r = 10<br />
Equation of circle: (x – 2) 2 + (y – 3) 2 = 10<br />
Solutions by MAitchison www.mathsroom.co.uk
Change language