Question

Write an equation (a) in slope-intercept form and (b) in standard form for the line passing through (1,7) and perpendicular to 3x+7y=1

Answer 1

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!~