To write the equations of the lines x+y+z−1=0 and 4x+y−2z+2=0 in symmetrical form, we need to rearrange the equations to isolate the variables on one side.

For the first equation, x+y+z−1=0, we can rearrange it as:
x+y+z=1

For the second equation, 4x+y−2z+2=0, we can rearrange it as:
4x+y−2z=-2

Now, let's express the equations in terms of the symmetrical form, which is of the form (x-x1)/a = (y-y1)/b = (z-z1)/c.

For the first equation, x+y+z=1, we can choose any point on the line, let's say (1,0,0). We can substitute this point into the equation to find the values of a, b, and c:
(1-1)/a = (0-0)/b = (0-0)/c
0/a = 0/b = 0/c

This means that a, b, and c can take any non-zero value. So, the symmetrical form of the first equation is:
(x-1)/a = (y-0)/b = (z-0)/c

For the second equation, 4x+y−2z=-2, we can choose any point on the line, let's say (0,-2,0). We can substitute this point into the equation to find the values of a, b, and c:
(0-0)/a = (-2-0)/b = (0-0)/c
0/a = -2/b = 0/c

This means that a and c can take any non-zero value, but b must be equal to -2. So, the symmetrical form of the second equation is:
(x-0)/a = (y+2)/(-2) = (z-0)/c

Therefore, the symmetrical form of the equations x+y+z−1=0 and 4x+y−2z+2=0 is:
(x-1)/a = (y-0)/b = (z-0)/c
(x-0)/a = (y+2)/(-2) = (z-0)/c

Question

To write the equations of the lines x+y+z−1=0 and 4x+y−2z+2=0 in symmetrical form, we need to rearrange the equations to isolate the variables on one side.

For the first equation, x+y+z−1=0, we can rearrange it as:
x+y+z=1

For the second equation, 4x+y−2z+2=0, we can rearrange it as:
4x+y−2z=-2

Now, let's express the equations in terms of the symmetrical form, which is of the form (x-x1)/a = (y-y1)/b = (z-z1)/c.

For the first equation, x+y+z=1, we can choose any point on the line, let's say (1,0,0). We can substitute this point into the equation to find the values of a, b, and c:
(1-1)/a = (0-0)/b = (0-0)/c
0/a = 0/b = 0/c

This means that a, b, and c can take any non-zero value. So, the symmetrical form of the first equation is:
(x-1)/a = (y-0)/b = (z-0)/c

For the second equation, 4x+y−2z=-2, we can choose any point on the line, let's say (0,-2,0). We can substitute this point into the equation to find the values of a, b, and c:
(0-0)/a = (-2-0)/b = (0-0)/c
0/a = -2/b = 0/c

This means that a and c can take any non-zero value, but b must be equal to -2. So, the symmetrical form of the second equation is:
(x-0)/a = (y+2)/(-2) = (z-0)/c

Therefore, the symmetrical form of the equations x+y+z−1=0 and 4x+y−2z+2=0 is:
(x-1)/a = (y-0)/b = (z-0)/c
(x-0)/a = (y+2)/(-2) = (z-0)/c

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