Question

How do you write y+1=4/5(x-3) in standard form?

Answer 1

Answer:

The standard form of the given expression is $- (\frac{4}{5})x + y + (\frac{17}{5}) = 0$

Step-by-step explanation:

Here, the given expression is  $y + 1 = \frac{4}{5} (x-3)$

Now, the Standard Form of the equation is ax + by + c = 0

Solving this, we get:

$y + 1 = \frac{4}{5} (x-3) \implies y + 1 = (\frac{4}{5})x - \frac{12}{5}$

⇒$y + 1 - (\frac{4}{5})x + (\frac{12}{5}) = 0$

or, $y - (\frac{4}{5})x + (\frac{12}{5} + 1 )= 0 \implies y - (\frac{4}{5})x + (\frac{12+ 5}{5}) = 0$

⇒$y - (\frac{4}{5})x + (\frac{17}{5}) = 0$

Hence, the standard form of the given expression is $- (\frac{4}{5})x + y + (\frac{17}{5}) = 0$