CAIE IGCSE Additional Math 7 Straight Line Graphs Topic Practice

$m_{A B}=\frac{-9-3}{6--4}$ oe isw or $-\frac{6}{5}$$\left[m_{\perp}=\right] \frac{5}{6}$ or $\frac{-1}{\text { their } m_{A B}}$
FT their $m_{A B}$ providing that
$\frac{\text { difference in } y}{\text { difference in } x}$ attempted
[Midpoint] $\left(\frac{-4+6}{2}, \frac{3-9}{2}\right)$ oe isw or (1,-3)
Finds equation of perpendicular bisector:
$$y \text {-their }(-3)=-\frac{1}{\text { their }\left(-\frac{6}{5}\right)}(x-\text { their } 1) \mathrm{oe}$$
FT their perpendicular gradient providing gradient is $\frac{-1}{\text { their } m_{A B}}$ and using their midpoint

Their midpoint must not be any of A, B, C or D$11 x-75=\frac{5}{6} x-\frac{23}{6}$
Eliminates one variable
dep on previous $\mathbf{M}$ mark
11 x-75= their perp bisector equation
E(7,2)
[Area C D E= ] 40
Dependent on all previous marks

$\ln y=mx+c$ soi
Correct method to find m, e.g. $m=\frac{\ln 75-\ln 15}{2-1}$ oe, or $-m=\ln 15-\ln 75$ oe, or $m=\ln 5$ oe.
M1 FT their c if necessary.
Correct method to find c, e.g. $\ln 15=(\text{their }\ln 5)(1)+c$ oe, $\ln 75=(\text{their }\ln 5)(2)+c$ oe, or $c=\ln 3$ oe, soi.
M1 FT their m if necessary.
$y=3\times 5^x$ nfww, mark final answer

$\sqrt{y}=m x^{3}+c$ soi
$m=\frac{21-5}{10-2}$ oe or 2
c=5-2(2) oe or 1
FT their m

$y=\left(2 x^{3}+1\right)^{2}$ oe, isw
Alternative method

$\sqrt{y}=m x^{3}+c$ soi
21=10 m+c and 5=2 m+c and solves to find
m=2 or c=1
Finds the other unknown
FT their m or c

$y=\left(2 x^{3}+1\right)^{2}$ oe, isw