geometry help?

Question

1) find 2 points that has a distance of 5 and a slope of 3/4 to the point (2,3)

2) find the interior angles and area of the triangle with points (4,2), (0,1), and (6,-1)

I already tried to answer this problem but my angles has negative, so it wont be equal to 180 degrees

Answer 1

The above answer is perfect. Here is another way how solve the second problem:

2)

Let
A = (4, 2)
B = (0, 1)
C = (6, -1)
L = (0, -1)
M = (0, 2)
N = (6, 2)

First let us calculate the angles.

∡NAC = arctan(3/2) = 56.31°
∡MAB = arctan(1/4) = 14.04°
∡LCB = arctan(2/6) = arctan(1/3) = 18.43°

∡BAC = 180° - ∡NAC - ∡MAB = 109.65°
∡ABC = ∡MAB + ∡LCB = 32.47°
∡ACB = 180° - ∡BAC - ∡ABC = 37.87°

Now let us calculate the area of the triangle.

ABC = LCNM – ABM – ANC – BLC = 3*6 – 1*4/2 – 2*3/2 - 2*6/2 = 7

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Answer 2

1) a distance of 5 and a slope of 3/4 to the point (2,3)
Draw the point (2,3). Start from this point with the slope 3/4.
Rise is 3, run is 4. Move 4 units to the right and 3 units up. The new point is (6,6) This point and the point (2,3) form a 3-4-5 triangle. The distance is 5.
Do the opposite directions from (2,3) to find the other point.
Move 4 units to left and 3 down. The new point is (-2,0).

2) A(4,2),B (0,1), and C(6,-1)
Use the distance formula to find AB,BC,AC
then use Law of cosine to find all angles.