 Now the job is to kind of use the reverse from what we were dealing with in the previous problems We were finding the missing coordinates now. We need to create the drawing right? We need to position and label the triangle on a coordinate plane So let's start by drawing the drawing the figure itself and then adding in the coordinates later So we have an isosceles triangle RST so first let's draw an isosceles triangle So here's an isosceles triangle. I see it's got a base of RS And so that means I could put R here. Let's say S and I know T would be the vertex So if RS is the base that means ST and RT must be the congruent sites Now let's follow Kind of follow these instructions use the origin as a vertex or centerline of the figure Well, isosceles triangles have centerlines right up down the middle So we can use that as one of the axes So if I've got my y-axis like this, I'd want to position it over the vertex of that isosceles triangle So there I have the y-axis Next it says place at least one side of the polygon on an axis While we've already placed the y-axis through the centerline of the triangle So number two that suggestion says keep that other axis Well, if we've got this horizontal axis, we should line that up on one of the sides And so we could set that on as the x-axis here. There's the x and y axes Now we need to fill in this information that the base is for a units So in other words this entire distance here is for a So since the isosceles triangle is split down that y-axis For a there's going to be two a units on the right side and two a units on the left side And so the coordinates of point s. I know it's x-coordinate would be to a and The x-coordinate of r would be negative to a Again since we're moving in the positive x direction and then the negative x direction Now since both points are and s lie on the axis Though the x-axis if their y-coordinates must be zero and then finally point t Well, I don't know exactly how tall this distance is But I know that it might not be related to the number a so I'll just pick something else Point t lies on the y-axis. So it's x-coordinate is zero and then oh, I don't know Be let's say so there we go Let's try another So here we are to position and label an isosceles right triangle. So an isosceles right triangle So an isosceles right triangle We want its Positioning to fit so that the axes sorry the sides are on axes. Let me just change the color of this thing a bit So we have our isosceles right triangle So these segments would be congruent and we have a right angle now in terms of x and y axes It should be pretty straightforward that the x and the y-axis would fit in the corner Or in the right angle corner of This triangle so there's our x-axis and the y-axis. We're gonna fit Right on that segment length So we have triangle DEF and we see its legs are both e units long so this length Must be e and this length Must also be e And so that gives us enough information to figure out the x and y coordinates of these points So I know that this point down here is going to be e units to the right and zero units up Similarly this point up top is going to be zero units to the left or right and e units up and then of course the origin That's a nice straightforward point. That's just 00 and then lastly we've got triangle DEF is the name so it doesn't really matter where we put DE and F So I'll just put them in those spots. So there we have the isosceles right triangle All right, let's do one more Oh good golly an equilateral triangle. It's called eq I We have a vertex and its sides are to be so first off draw yourself an equilateral triangle So we know it's vertex is called Q So Vertex we've got a lot of vertices, but let's call this up top here is Q I don't know eq I so maybe point e and I Now this is similar to that isosceles triangle that we were dealing with earlier We want the x or the y axis one to fit on a center line and one to fit on a side length So if we're dealing with the vertical axis, I would line that up right over that vertex and Then in terms of a x-axis, I think it makes sense to set that on one of the side lengths There we go So now we have our x and y coordinates or sorry our x and y axes We need to fill in what the coordinates of our points will be So there's enough information in the in the directions to help us out. We already know Q Q is at zero and then square root of three times B and The side lengths are to be units. So all of these are to be units long Since that last side length E I is straddling the y-axis I know if I've got a full length of two B units there's going to be B units on one side and B units on the other side of the y-axis and that tells me the Y coordinate or sorry the x-coordinate of point I is going to be B and Since it sits on the x-axis its y-coordinate is zero the x-coordinate of point E will be negative B since we're moving to the left of the origin negative B zero