IB Maths AI HL Sl 2 1 Straight Lines Questions

2 A1

attempt to substitute their part (a) and point (3,-1) into the slope-intercept form or point-slope form of an equation
$-1=2 \times 3+c \quad$ OR $\quad y+1=2(x-3)$
y=2 x-7
Note: Equation must be in the form y=m x+c for A1 to be awarded.

METHOD 1
attempt to show that P does not lie on $L_{2}$
e.g. $-\frac{1}{2}(3)-\frac{5}{2} \quad \mathbf{OR}$ graph showing $L_{2}$ and P in approximate correct locations
$-1 \neq-\frac{1}{2}(3)-\frac{5}{2}(-1 \neq-4) \mathbf{OR}(3,-1)$ does not lie on the graph of $L_{2}$
hence $L_{2}$ is not the normal line to f(x) at point P

METHOD 2
attempt to find the equation of the normal line at (3,-1)
$\left(-1=-\frac{1}{2}(3)+c \quad\right.$ OR $\left.\quad y+1=-\frac{1}{2}(x-3)\right)$
the normal line is $y=-\frac{1}{2} x+\frac{1}{2}$
hence $L_{2}$ is not the normal line to f(x) at point P

METHOD 3
attempt to find the intersection of $L_{1}$ and $L_{2}$

Intersection of y=2 x-7 and $y=-\frac{1}{2} x-\frac{5}{2}$ is (1.8,-3.4)
$x=1.8 \neq 3 \quad$ OR $\quad y=-3.4 \neq-1$
hence $L_{2}$ is not the normal line to f(x) at point P

Note: Accept equivalent written arguments provided values are seen.
Methods 1 and 2 are independent of the answers in (a) and (b) but FT marks can be given for Method 3.