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

Question

Abigail Allen · Accepted Answer

#-x+y=0# Since #y=mx+n# can represent any line, we can use it. #m# is the slope of our line. The given slope is #1# so #m=1#. So our formula is formed as #y=1*x+n# We can now plug the point to this formula to find the value of #n# #3=1*(-3)+n# 
#n=6# So #y=x+6# represents the line. But we should reform this equality in order to be in the standart form: #-x+y=6# is the standart form (#ax+by=c#)

Eva Abramson · Answer

Answer: # x - y + 6 = 0# Why? Use the general standard form:
# ax + by + c = 0# Step1:
in gradient and one point form (also known as point-slope form) you have,
# y - 3 = 1*(x + 3)# Step2:
arranging this you get,
# x - y + 6 = 0#

Eva Adamson · Answer

undefined To write the equation of the line passing through the point (-3,3) with a slope of 1 in standard form, you can use the point-slope form of a linear equation, which is y - y1 = m(x - x1). Substituting the given point (-3,3) and slope 1 into the equation, you get y - 3 = 1(x - (-3)). Simplify this to y - 3 = x + 3. Then, rearrange the equation to get it in standard form, which is Ax + By = C, so the equation becomes x - y = -6.

Julia Adams · Answer

undefined The standard form of the equation of a line is $Ax + By = C$, where $A$, $B$, and $C$ are constants. To write the equation of the line passing through the point $(-3,3)$ with slope $1$ in standard form, you can follow these steps:

1. Use the point-slope form of a linear equation: $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the given point.
2. Substitute the given values into the equation: $y - 3 = 1(x - (-3))$.
3. Simplify the equation: $y - 3 = x + 3$.
4. Rearrange the equation into standard form by moving all terms to one side: $x - y = -6$.

So, the standard form of the equation of the line passing through the point $(-3,3)$ with slope $1$ is $x - y = -6$.