Points A(-2, 4), B(1, 3), C(4, -1) and D form a parallelogram. What are the coordinates of D?

2 answers:

8 0

Let point A(-2,4) = A (X1,Y1)
point B( 1,3 )= B (X2,Y2)
point C(4,-1) = C (X3,Y3)
and point D(?, ?) = D (X4,Y4) We have to find this point

To find X4 we have to use the formula:
X2-X1=X3-X4

Now just plug in the numbers that correspond to the letters provided:
(1)-(-2)=(4)-(X4) ----> we don't know what X4 is yet, so we have to solve for it!
1+2=4-X4
3=4-X4
3-4=-X4
-1=-X4 divide both sides by -1
X4=1

Now we have to find Y4 using this formula:
Y2-Y1=Y3-Y4

Therefore,
(3)-(4)=(-1)-(Y4)
-1=-1-Y4
-1+1=-Y4
0=-Y4
So,
Y4=0

Now we have found the coordinates of the point D, which is (1,0)

Hope this helped!

Basile [38]3 years ago

3 0

Answer: d- (1,0)

Step-by-step explanation:

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In ΔABC, the lengths of a, b, and c are 22.5 centimeters, 18 centimeters, and 13.6 centimeters, respectively.

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Given the values of the three sides of the triangle, we can apply the Cosine Law to find the angles of the triangle. Recall that for we can express the value of c through the equation below.

$c^{2} = a^{2} + b^{2} - 2abcosC$

Rearranging this equation, we can find the value ∠C as shown below.

$\cos C = \frac{a^{2}+b^{2}-c^{2}}{2ab}$
$C = cos^{-1} (\frac{a^{2}+b^{2}-c^{2}}{2ab})$

We can apply the same reasoning for finding the value of ∠B as shown.

$B = cos^{-1} (\frac{a^{2}+c^{2}-b^{2}}{2ac})$

Plugging in the values of the sides (see image attached) from the given. It will now be straightforward to compute for ∠B and ∠C.

$C = cos^{-1} (\frac{22.5^{2}+18^{2}-13.6^{2}}{2(22.5)(18)})$
$C \approx 37.19$

$B = cos^{-1} (\frac{22.5^{2}+13.6^{2}-18^{2}}{2(22.5)(13.6)})$
$B \approx 53.13$

Answer: ∠C = 37.19° and ∠B = 53.13°