Answer:
28°
Step-by-step explanation:
You're given that line DE and line FG are parallel and KL and FG are perpendicular. Then you can find out angle ∠BAC by using the vertical angles property: ∠BAC=62°. Then since KL and FG are perpendicular ∠ABC = 90°. So you find the angle ∠BCA by finding the sum of interior angles: 62+90+∠BCA=180, therefore ∠BCA is 28°. Finally, ∠x or ∠JCG = 28 because ∠JCG and ∠BCA are vertical angles and congruent.
The fourth vertex would be at point (-9,1).
The slope from (5,-4) to (3,4) is -8/2 or -4.
So I did that with (-7,-7) to the fourth vertex to get the answer.
Step-by-step explanation:
sol;
x+1=y...(1)
3y-7=2x....(2)
or, 3(x+1)-72x [from (1)]
or, 3x+3-7=2x
or, 3x-2x=7-3
x=4
now,
putting the value of x in (1)
y=x+1
=4+1
=5
PR and SQ are the diagonal of PQRS.
