Find the equation perpendicular to 10x – 9y + 2 = 0 and passing through point (3,3) is given by...
This is the answer given:
Eqn. of reqd line is of form 9x+10y+k = 0; k is to be determined.
(3,3)belongs to line=> 27+30+k = 0 or k = -57
9x+10y-57 = 0
Ans: C
I set up the point-slope :
y=10/9x + 2/9
m for the perp line is -9/10, but then how do I solve for the intercept?
y = -9/10x +c
but then I set
10y=-9x+10c
10y+9x=10c
c= y + 9/10x - that isn't right!
That makes no sense. Normally I don't have to set k=nc.
For example in this problem:
Write an equation of the line
perpendicular to 4y - x = 20 and
containing the point (2, -3).
y = 1/4x +20
slope perpendicular line = -4
y = -4x +b
Plug in x and y
-3 = -4(2) + b
-3 = -8 +b
5=b
Perp line is therefore
y = -4x +5
Or
4x + y - 5 = 0
there's no setting an arbitrary k = nc
Forget your second example. It was just a red herring.
Let me put it this way. The equation of the line can be written as either
or
They are the same. Yes? If you write it the first way, the constant term appears as If the second way, it appears as Does this now make sense?
Yes, but how do I get there from the slope point form?
Forget we have the answer.
Find the equation perpendicular to 10x – 9y + 2 = 0 and passing through point (3,3) is given by...
-9y =-10x -2
y = 10/9x +2/9
Now we have the slope of the perp line = -9/10
Perp Line y = -9/10x + b
So we can find the y int given the point (3,3)
3 = -9/30 +b
What, without having the solution, would tell us to set k = -10 b? (I don't even know where I got c )
We'd just solve:
b = 3 + 9/30
So that's what I'm asking, without that solution, how do we know to set some arbitrary k?
y - y1 = m (x - x1)m for the perp line is -9/10, but then how do I solve for the intercept?
y - 3 = -9/10 (x - 3)
y - 3 = -9x/10 + 27/10
y = -9x/10 + 57/10
When you know a point on a line and the slope of the line, just grind them into point-slope form for lines and get your line. No reason to get all wordy.