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3 years ago

Write the equation of a line with that passes through the points (-3, -2) and (6,1)?

Mathematics

1 answer:

True [87]3 years ago

5 0

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: Y = 1/3x - 1

M = y2 - y1/ x2-x1

M = 6 - (-3) / 1 - (-2)

Y = 1/3x + b

Then substitute one of the points into the equation

-2= 1/3(-3) + b

-2 + 2 + b = 1/3(-3) +b - b + 2

-b = 1/3(-3) + 2

-b = 1

b = -1

Y = 1/3x - 1

Then i plugged it into

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Verify that the points are the vertices of a parallelogram and find its area. (2,-1,1), (5, 1,4), (0,1,1), (3,3,4)

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Answer:

Area = 13.15 square units

Step-by-step explanation:

Let the given vertices be represented as follows:

A(2, -1, 1) = 2i - j + k

B(5, 1, 4) = 5i + j + 4k

C(0, 1, 1) = 0i + j + k

D(3, 3, 4) = 3i + 3j + 4k

(i) Let's calculate the vectors of all the sides:

$\\$ AB = B - A = (5i + j + 4k) - (2i - j + k)

AB = 5i + j + 4k - 2i + j - k [Collect like terms]

AB = 3i + 2j + 3k

BC = C - B = (0i + j + k) - (5i + j + 4k)

BC = 0i + j + k - 5i - j - 4k [Collect like terms]

BC = -5i + 0j - 3k

CD = D - C = (3i + 3j + 4k) - (0i + j + k)

CD = 3i + 3j + 4k - 0i - j - k [Collect like terms]

CD = 3i + 2j + 3k

DA = A - D = (2i - j + k) - (3i + 3j + 4k)

DA = 2i - j + k - 3i - 3j - 4k [Collect like terms]

DA = -i - 4j - 3k

AC = C - A = (0i + j + k) - (2i - j + k)

AC = 0i + j + k - 2i + j - k [Collect like terms]

AC = -2i + 2j

BD = D - B = (3i + 3j + 4k) - (5i + j + 4k)

BD = 3i + 3j + 4k - 5i - j - 4k [Collect like terms]

BD = -2i + 2j

(ii) From the results in (i) above, we can deduce that;

AB = CD This implies that AB || CD [AB is parallel to CD]

AC = BD This implies that AC || BD [AC is parallel to BD]

(iii) Therefore, ABDC is a parallelogram since opposite sides (AB and CD) are parallel. Hence, the points are vertices of a parallelogram

Now let's calculate the area

To find the area of the parallelogram, we find the magnitude of the cross product of any two adjacent sides.

In this case, we'll choose AB and AC

Area = |AB X AC|

Where;

$AB X AC = \left[\begin{array}{ccc}i&j&k\\3&2&3\\-2&2&0\end{array}\right]$

AB X AC = i(0 - 6) - j(0 + 6) + k(6 + 4)

AB X AC = - 6i - 6j + 10k

|AB X AC| = $\sqrt{(-6)^2 + (-6)^2 + (10)^2}$

|AB X AC| = $\sqrt{172}$

|AB X AC| = 13.15

Area = 13.15 square units.

PS: ACBD is also a parallelogram. The diagram has also been attached to this response.

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