What is the answer to solvefor x,y-y^0=m(x-x^0) ?

The answer to solvefor x,y-y^0=m(x-x^0) is x=(y-1)/m+1;\quad m\ne 0

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solvefor x,y-y^0=m(x-x^0)

solve for

x, y - y^{0} = m (x - x^{0})

x = \frac{y - 1}{m} + 1; m \neq = 0

Solution steps

y - y^{0} = m (x - x^{0})

Switch sides

m (x - x^{0}) = y - y^{0}

Apply rule

a^{0} = 1, a \neq = 0

y^{0} = 1

m (x - x^{0}) = y - 1

Divide both sides by

m; m \neq = 0

x - 1 = \frac{y - 1}{m}; m \neq = 0

Move

1

to the right side

x = \frac{y - 1}{m} + 1; m \neq = 0

x, 2 x + 2 y = 9

3 x - 1 = 2 x + 6

x, 8 x - 2 y = 4

x^{2} - 1 = (x - 1) (x + 1)

x, 3 x - 4 y - 18 = 0

What is the answer to solvefor x,y-y^0=m(x-x^0) ?
The answer to solvefor x,y-y^0=m(x-x^0) is x=(y-1)/m+1;\quad m\ne 0