Answer:
<u>Given rhombus ABCD with</u>
- m∠EAD = 67°, CE = 5, DE = 12
<u>Properties of a rhombus:</u>
- All sides are congruent
- Diagonals are perpendicular
- Diagonals are angle bisectors
- Diagonals bisect each other
<u>Solution, considering the above properties</u>
- 1. m∠AED = 90°, as angle between diagonals
- 2. m∠ADE = 90° - 67° = 23° as complementary of ∠EAD
- 3. m∠BAE = 67°, as ∠BAE ≅ ∠EAD
- 4. AE = CE = 5, as E is midpoint of AC
- 5. BE = DE = 12, as E is midpoint of BD
The equation of the straight line is y=mx+q where q is the number on the y-axis where the line passes, as you can see it is -3. It turns into:
y=mx+(-3) -> y=mx-3
Then consider a point on the line and take the coordinates, such as the point with coordinates (-2;-4), so now you know that:
x=-2 and y=-4
At this point you put these values into the equation:
y=mx-3
-4=m(-2)-3
then solve:
-4=-2m-3
-2m=+3-4
-2m=-1
m=+1/2
Put the value of m into the equation and you found it:
y=1/2x-3
Answer: use fotomath
Step-by-step explanation:
