How do you write in standard form an equation of the line passing through the given point (-3,3) with the given slope 1?

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Answer 1 · 2015-04-01T19:24:18

Answer: $x - y + 6 = 0$ $x - y + 6 = 0$

Why?

Use the general standard form:
$a x + b y + c = 0$ $ax + by + c = 0$

Step1:
in gradient and one point form (also known as point-slope form) you have,
$y - 3 = 1 \cdot (x + 3)$ $y - 3 = 1*(x + 3)$

Step2:
arranging this you get,
$x - y + 6 = 0$ $x - y + 6 = 0$

Answer 2 · 2015-04-01T19:32:43

$- x + y = 0$ $-x+y=0$

Since $y = m x + n$ $y=mx+n$ can represent any line, we can use it.

$m$ $m$ is the slope of our line. The given slope is $1$ $1$ so $m = 1$ $m=1$ .

So our formula is formed as $y = 1 \cdot x + n$ $y=1*x+n$

We can now plug the point to this formula to find the value of $n$ $n$

$3 = 1 \cdot (- 3) + n$ $3=1*(-3)+n$
$n = 6$ $n=6$

So $y = x + 6$ $y=x+6$ represents the line. But we should reform this equality in order to be in the standart form:

$- x + y = 6$ $-x+y=6$ is the standart form ( $a x + b y = c$ $ax+by=c$ )