Find the equation of the line that is parallel to 4x - 8y equals 9 and passes through (-3, 6)

Mathematics

1 answer:

vazorg [7]3 years ago

6 0

For this case we have the following equation:

$4x-8y = 9$

We manipulate algebraically to bring the equation to the slope-intersection form y = mx + b

Where:

m: It's the slope

b: It is the cut-off point with the y axis

$8y = 4x-9\\y =\frac {4} {8} x- \frac {9} {8}\\y =\frac {1} {2} x- \frac {9} {8}$

By definition, if two lines are parallel then their slopes are equal.

Thus, the line is of the form:

$y = \frac {1} {2} x + b$

We find the cut-off point by replacing the given point:

$6 = \frac {1} {2} (- 3) + b\\6 = - \frac {3} {2} + b\\b = 6 + \frac {3} {2}\\b = \frac {12 + 3} {2}\\b = \frac {15} {2}$

Thus, the line is of the form:

$y = \frac {1} {2} x + \frac {15} {2}$

Answer:

$y = \frac {1} {2} x + \frac {15} {2}$

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A rhombus ABCD has AB = 10 and m∠A = 60°. Find the lengths of the diagonals of ABCD.

melisa1 [442]

Three important properties of the diagonals of a rhombus that we need for this problem are:
1. the diagonals of a rhombus bisect each other
2. the diagonals form two perpendicular lines
3. the diagonals bisect the angles of the rhombus

First, we can let O be the point where the two diagonals intersect (as shown in the attached image). Using the properties listed above, we can conclude that ∠AOB is equal to 90° and ∠BAO = 60/2 = 30°.

Since a triangle's interior angles have a sum of 180°, then we have ∠ABO = 180 - 90 - 30 = 60°. This shows that the ΔAOB is a 30-60-90 triangle.

For a 30-60-90 triangle, the ratio of the sides facing the corresponding anges is 1:√3:2. So, since we know that AB = 10, we can compute for the rest of the sides.

$\overline{OB}:\overline{AB} = 1:2$
$\overline {OB}:10 = 1:2$
$\overline{OB} = \frac{1}{2}(10) = 5$

Similarly, we have

$\overline{AO}:\overline{AB} = \sqrt{3}:2$
$\overline {AO}:10 = \sqrt{3}:2$
$\overline{AO} = \frac{\sqrt{3}}{2}(10) = 5\sqrt{3}$

Now, to find the lengths of the diagonals,

$\overline{AD} = 2(\overline{AO}) = 10\sqrt{3}$
$\overline{BC} = 2(\overline{OB}) = 10$

So, the lengths of the diagonals are 10 and 10√3.

Answer: 10 and 10√3 units