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What is the equation in standard form of
the line that passes through the points
(3, 5) and (-7, 2)?

Sagot :

amysun2025

Answer:

3x-10y=-41

Step-by-step explanation:

"standard form of the line" is ax+by=c, where a, b, and c are free coefficients

first, we need to find the slope (m) of the line

that is calculated with the formula (y2-y1)/(x2-x1)

we have the points (3,5) and (-7,2)

label the points:

x1=3

y1=5

x2=-7

y2=2

substitute into the equation

m=(2-5)/(-7-3)

m=-3/-10

m=3/10

the slope is 3/10

before we put a line into standard form, we need to put it into another form first-- like slope-intercept form (y=mx+b, where m is the slope and b is the y intercept)

we already know the slope

here's our line so far:

y=3/10x+b

we need to find b; since the line will pass through the points (3,5) and (-7,2) we can use either one of them to find b

let's use (3,5) as an example. Substitute into the equation

5=3/10(3)+b

5=9/10+b

41/10=b

b is 41/10

this is the equation:

y=3/10x+41/10

now we can find the equation in standard form. Subtract 3/10x from both sides

-3/10x+y=41/10

a (the number in front of x cannot be negative OR less than one. First, let's multiply both sides by -1)

3/10x-y=-41/10

multiply both sides by 10 to clear the fraction

3x-10y=-41

^^ is the equation

hope this helps!

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