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$ (III)y=-7x −1(IV)y=4−3x ⇔y=-71 x−1⇔y=-3x+4 $ It is now easier to identify the slopes.

Line | Equation | Slope |
---|---|---|

(I) | $y=-3x$ | $-3$ |

(II) | $y=7x+41 $ | $7$ |

(III) | $y=-71 x−1$ | $-71 $ |

(IV) | $y=-3x+4$ | $-3$ |

(V) | $y=7x+0.25$ | $7$ |

We can see that two slopes appear more than once. $ I and IV have slope-3II and V have slope7 $ Now, to be sure that they are parallel, they must not have the same $y$-intercept.

Lines | $y$-intercepts | Parallel? |
---|---|---|

I and IV | $0 =4$ | Yes |

II and V | $41 =0.25$ | No |

Since II and IV have the same slope, they are the same line. Thus, it's only I and IV that are parallel.

b

$m_{1}⋅m_{2}=-1$ In part A, we found the slopes of each line.

Line | Equation | Slope |
---|---|---|

(I) | $y=-3x$ | $-3$ |

(II) | $y=7x+41 $ | $7$ |

(III) | $y=-71 x−1$ | $-71 $ |

(IV) | $y=-3x+4$ | $-3$ |

(V) | $y=7x+0.25$ | $7$ |

To determine whether or not they are perpendicular, we calculate the product of their slopes. Any two slopes whose product equals $-1$ are opposite reciprocals. Thus, let's rewrite the slopes that are whole numbers as reciprocals to find which ones that are opposite.

Line | Slope |
---|---|

(I) | $-13 $ |

(II) | $17 $ |

(III) | $-71 $ |

(IV) | $-13 $ |

(V) | $17 $ |

Since the product equals $-1,$ line III is perpendicular to both line II and V.