What is an equation of the line that passes through the point (1,-7) and is parallel to the line 3x +y = 3? pleasee help 15 points?​

The slope of the line, 3x-2y=8 is m=3/2.

So the parallel line will be y=(3/2)x+b.

Since the parallel line is passing (-1, -7), to determine b, put the number x=-1 and y=-7.

-7 = (3/2)(-1) +b

b=-11/2

Therefore, the parallel line to 3x-2y=8 and passing the point (-1, -7) is y=(3/2)x-(11/2)

Step-by-step explanation:

x=4

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)