What is the equation of the line (in point-slope form) that is parallel to y+2=−7(x+3) and passes through the point (1,4)?

y+1=−7(x−4)
y+4=7(x+1)
y−1=−7(x+4)
y−4=−7(x−1)
y−4=7(x−1)

  y−4=−7(x−1)

Step-by-step explanation:

The point-slope form of the equation for a line with slope m through point (h, k) is ...

  y -k = m(x -h)

Your given line has a slope of -7. The parallel line will have the same slope.

So, you want to fill in the above form with m=-7, (h, k) = (1, 4):

  y -4 = -7(x -1)

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

x> 6

step-by-step explanation:

8x+8> 7x+14

8x-7x> 14-8

x> 6