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  1. Line FG contains points F (3, 7) and G (−4, −5). Line HI contains points H (−1, 0) and I (4, 6). Lines FG and HI are

parallel


perpendicular


neither

  1. Line AB contains points A (8, −4) and B (1, −5). The slope of line AB is

−7


negative 1 over 7


1 over 7


7

Verified Answer :

  1. >> neither
    2. >> m= 
    \(- \frac{1}{7}\) >> negative 1 over 7

Answer:

1. Neither

2. 1 over 7

Step-by-step explanation:

We know that the formula for slope joining the points \(( x_{1} ,y_{1} )\) and \(( x_{2} ,y_{2} )\) is given by \(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\).

1. We have the line FG having end points ( 3,7 ) and ( -4,-5 ). The slope of this line is given by,

\(m_{FG}=\frac{-5-7}{-4-3}\)

i.e. \(m_{FG}=\frac{-12}{-7}\)

i.e. \(m_{FG}=1.714\)

Also, the line HI is given with end points ( -1,0 ) and ( 4,6 ). Its slope is given by,

\(m_{HI}=\frac{6-0}{4+1}\)

i.e. \(m_{HI}=\frac{6}{5}\)

i.e. \(m_{HI}=1.2\)

Since, neither the slope of FG and HI are equal nor their product is -1.

Hence, FG and HI are neither parallel nor perpendicular respectively.

2. We have the line AB with end points ( 8,-4 ) and ( 1,-5 ). So, the slope of AB is,

\(m_{AB}=\frac{-5+4}{1-8}\)

i.e. \(m_{AB}=\frac{-1}{-7}\)

i.e. \(m_{AB}=\frac{1}{7}\)

Hence, slope of AB is 1 over 7.