Hi i was wondering if i could please get help with these two questions. I have attempted to do q8 but im not sure whether i am right.
In regards to q12 i was particularly confused on how to do parts a and b
Hey!
For Question 8, I think parts of your working are correct. Let me give you a hand interpreting!
Part A: The two lines are each parallel to \(x+2y=0\). Now if you rearrange this line, you get \(y=\frac{-x}{2}\), you did this yourself. So this line has a gradient of \(m=\frac{-1}{2}\). The lines L1 and L2 just need to be any lines with a gradient of this, aka, any line of the form \(y=\frac{-x}{2}+c\). Just pick two random values for \(c\)
Part B: Here, it is now easier to consider two general equations in gradient intercept form. Before we do though, the fact that the lines are perpendicular means that we can set the two gradients to be \(m\), and \(\frac{-1}{m}\). This is because, when two lines are perpendicular, \(m_1=\frac{-1}{m_2}\). So we can write the lines like this:
L1: \(y=mx+b_1\)
L2: \(y=\frac{-x}{m}+b_2\)
Note that the two intercepts are different, at least for now. In fact, both lines meet at the point (0, 4), which sits on the y-axis. So in actuality, both lines have a y-intercept of 4, so the equations are now:
L1: \(y=mx+4\)
L2: \(y=\frac{-x}{m}+4\)
Both of our conditions are now satisfied, so you can pick any value of \(m\) you like and substitute it here to get your answer
Part C: Just do a cheeky answer;
any two parallel lines will do. Do y=0 and y=1, those will definitely not intersect!
For Question 12 your equations look fine for Part B, that answer is correct! For Part A, your equation should be \(y=80x+100\), the $80 is the half-hourly rate and the $100 is the constant call out fee.
Be careful, you have defined \(x\) as half an hour here but one hour in Part B. This is okay as long as you are aware of it!
Try subbing with this new equation, you should get $740.00 as your answer for Part A