Write an equation (a) in slope-intercept form and (b) in standard form for the line passing through (1,7) and perpendicular to 3
x+7y=1
1 answer:
Slope intercept form & in standard form , passes through (1, 7) , perpendicular to 3x + 7y = 1
Turn 3x + 7y = 1 into slope-intercept form.
Remember, slope-intercept form is : y=mx+b where m=slope & b=y-intercept.
To turn out given equation into slope-intercept form, we must get x & y onto different sides.
So, subtract 3x from both sides.
7y = -3x + 1
Then, divide both sides by 7.
y = -3/7x + 1/7
Remember, when an equation is perpendicular to another equation, both equations have negative reciprocals.
m₁=-3/7 & m₂=7/3
Our new equation has these things in it :
A slope of 7/3 & it passes through (1, 7)
So, simply plug these into the slope-intercept equation.
y=mx+b
y = 7/3x + 7 → (a)Slope-intercept form
Now we must put this into (b)standard form.
Standard form is : Ax+Bx=C
So, we just use our slope-intercept form, but rearrange it :)
y = 7/3x + 7
In standard form, x & y are on the same side, so we simply subtract 7/3x from both sides.
-7/3x + y = 7 → (b)Standard Form
~Hope I helped!~
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