Answer:
4x-y+1=0
Step-by-step explanation:
here,given equation of a line id
4x-y-2=0.. eqn(i)
equation of any line parallel to line (i) is
4x-y+k=0...eqn(ii)
since, the line(ii) passes through (1,5)[replacing x=1 and y=5 in eqn(ii), we get]
4*1-5+k=0
or, 4-5+k=0
or,-1+k=0
•°•k=1
substituting the value of k=1 in eqn(ii),
4x-y+1=0 is the required equation of the line.
Answer: 28/r+4
Step-by-step explanation: Divide each term by r + 4 and simplify
(x, y) --> (x + 5, y - 1)
A(3, -1) --> A'(3 + 5, -1 - 1) --> A'(8, -2)
B(6, 1) --> B'(6 + 5, 1 -1) --> B'(11, 0)
C(2, 4) --> C'(2 + 5, 4 - 1) --> C'(7, 3)
D(-1, 3) --> D'(-1 + 5, 3 - 1) --> D'(4, 2)
Community property of addition because -6 + 6 would equal zero
