HELP ASAP WILL GIVE BRAINLIEST!! Write the equation in point-slope form for the line that contains the points
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Answer:
The answer is below
Step-by-step explanation:
In triangle DEF, ∠D = 30°, ∠F = 60° and EF = 6. Find DE + DF.
Solution:
A triangle is a polygon with three sides. There are different types of triangles such as isosceles triangle, scalene triangle equilateral triangle, right angled triangle.
In triangle DEF:
∠D + ∠E + ∠F = 180° (sum of angles in a triangle)
30 + ∠E + 60 = 180
∠E + 90 = 180
∠E = 180 - 90
∠E = 90°
This is a right angled triangle since one angle = 90° (∠E).
Using sine rule:
Hence DE + DF = 12 + 10.39 = 22.39
The standard form for a parabola is (x - h)2 = 4p (y - k), where the focus is (h, k + p) and the directrix is y = k - p. If the parabola is rotated so that its vertex is (h,k) and its axis of symmetry is parallel to the x-axis, it has an equation of (y - k)2 = 4p (x - h), where the focus is (h + p, k) and the directrix (d)
is x = h - p.
So directrix is: y = k - p and the focus is at:
(h, k+p)
Since our focus is: (1, 3) and directrix is: y = 1,
thus h = 1, k+p = 3, and k-p = 1
Therefore k = 3-p, 3-p-p = 1, k = 3-p = 3-1 = 2
3-2p = 1, -2p = -3+1, -2p = -2, p = 1
Now we plug p, k, & h into standard form:
(x - h)2 = 4p (y - k)
y = 1/4 (x-1)^2 + 2
is x = h - p.
So directrix is: y = k - p and the focus is at:
(h, k+p)
Since our focus is: (1, 3) and directrix is: y = 1,
thus h = 1, k+p = 3, and k-p = 1
Therefore k = 3-p, 3-p-p = 1, k = 3-p = 3-1 = 2
3-2p = 1, -2p = -3+1, -2p = -2, p = 1
Now we plug p, k, & h into standard form:
(x - h)2 = 4p (y - k)
y = 1/4 (x-1)^2 + 2