In Exercises 1–8, parametric equations and a value for the parame...

Question

In Exercises 1–8, parametric equations and a value for the parameter t are given. Find the coordinates of the point on the plane curve described by the parametric equations corresponding to the given value of t.

x = 7 − 4t, y = 5 + 6t; t = 1

Accepted Answer

Welcome back everyone. In this problem, we want to determine the coordinates of the point on the plane curve represented by the parametric equations. X equals 11 minus six U and Y equals eight plus three U where U equals five for our answer choices A is the coordinate 41 23. B is negative 90 negative seven C is 41 negative seven and D is negative 1923. Now, what do we already know? Well, we know that these are parametric equations. So basically what it is telling us is that we have some, probably some fancy looking graph where every point on the graph, every card in it is related by that parameter you. So if we can substitute that parameter U into our equations, then we can find the X and Y values of the coordinate. So let's do that. Let me label the equation X equals 11 minus six, U as equation one and Y equals A plus three U as equation two. So solving for X in equation one, then I can write my equation X equals 11 minus six. U. Now remember U is equal to five. So that would be 11 minus six, multiplied by five, six, multiplied by five is 30. So that's 11 minus 30 and 11 minus 30 equals negative 19. So that is my value for X. Next in equation two, serving for A Y Y equals eight plus three, U so that's eight plus three, multiplied by five three, multiplied by five is 15 and eight plus equals 23. Therefore, X is equal to 19, Y equals 23 which means our answer is choice. D thanks a lot for watching guys. I hope this video helped.

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In Exercises 1–8, parametric equations and a value for the parameter t are given. Find the coordinates of the point on the plane curve described by the parametric equations corresponding to the given value of t. x = 7 − 4t, y = 5 + 6t; t = 1