How do you find a standard form equation for the line with #A (-1,4)#; Slope: #2/5#?

Answer 1

#y-2/5x=4frac(2)(5)# which leads to

#2x-5y =-22#

The standard form of a line is expressed in the following form #Ax+By=C#, where #A,B and C # are integers
To find this form use the formula #(y-y_1)=m(x-x_1)#

substitute in the point and slope (m)

#(y-4)=2/5(x-(-1))" "# or #" "(y-4)=2/5(x+1)#
#y-4=2/5x+2/5#
#y=2/5x+2/5+4#
#y=2/5x+4frac(2)(5)" "# now put into standard form
Multiply by #5# to clear the denominators
#5y-(cancel5xx2)/cancel5x=cancel5xx22/cancel5#
#5y-2x=22#
# rArr 2x-5y = -22#
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Answer 2

To find the standard form equation of a line with slope 2/5 passing through point (-1, 4), use the point-slope form equation y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the given point. Then, rearrange the equation to standard form (Ax + By = C).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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