+144 Suppose you graphed every single point of the form \((2t + 3, 3-3t)\). For example, when \(t = 2\), we have \(2t + 3 = 7\) and \(3-3t = -3\), so \((7,-3)\) is on the graph. Explain why the graph is a line, and find an equation whose graph is this line.
I'm incredibaly confused if someone could help that would be AMAZING!!
Thanks!
+118677 What you have here is a locus (set) of points such that
x=2t+3 and y=3-3t for t is any real value.
so what you need to do it make t the subject of both these equations
| x=2t+3 x-3=2t (x-3)/2 = t | y=3-3t y-3=-3t 3t=3-y t=(3-y)/3 |
So now you know that
\(t=\frac{x-3}{2}=\frac{3-y}{3}\\ \frac{x-3}{2}=\frac{3-y}{3}\\ 6 \cdot \frac{x-3}{2}=6 \cdot \frac{3-y}{3}\\ 3(x-3)=2(3-y)\\ 3x-9=6-2y\\ 3x+2y-15=0 \qquad \text{which is a line}\\ or\\ 2y=-3x+15\\ y=\frac{-3x}{2}+\frac{15}{2}\)
which is still the same line. gradient = -1.5 and y intercept = -7.5
