Question

Help? big question...

It's kind of a big question but i would love help if possible.
Thank-you so much if you do

Answer 1

F(3, 1), G(5, 2), H(2, 4), J(1, 6).

The formula of a slope:

`m=(y_2-y_1)/(x_2-x_1)`

Part A)

FG:

`slope\ m_(FG)=(2-1)/(5-3)=(1)/(2)`

Part B)

HJ:

`slope\ m_(HJ)=(6-4)/(1-2)=(2)/(-1)=-2`

Part C)

`\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2,\ \text{then}\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-(1)/(m_1)\\\\l\ \parallel\ k\iff m_1=m_2`

We have

`m_1=(1)/(2),\ m_2=-2\\\\m_1\\eq m_2\\\\m_1m_2=\left((1)/(2)\right)\left(-2\right)=-1`

Therefore the lines are perpendicular.

Answer 2

Part A and Part B

Check out the attached image to see how I got each slope. I used the slope formula.

The answer to part A is 1/2. The answer to part B is -2.

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Part C

Using the results from parts A and B, we can see that the slopes aren't the same so they cannot be parallel lines. Parallel lines always have equal slopes. Note how we can multiply the two slopes to get (1/2)*(-2) = -1 which proves that the two lines are perpendicular to one another.

note: One slope is the negative reciprocal of the other. You flip the fraction and the sign (eg: 1/2 ---> -2/1) to go from one slope to the other.