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perpendicular to 3x+4y=13 and through (2,7)

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2 Answers

Equation of Line L2= m2X + b2 Perpendicular to L1 : 3X + 4y = 13 passing through ( 2, 7)
 
 
    L1 :
             y = -3/4 X + 13/4
 
 
         then, m2 = 4/3
 
 
       L2:  Y = 4/3 X + b2
              7 = 4/3(2) + b2
 
              b2 = 7 - 8/3 = 13/3
 
        L2 : Y = 4/3 X + 13/3    
 

Comments

This is easy way of getting the answer, without going through problems of unnecessary calculation and making a mistake. By using all feature of Linear equation at hand.
3x+4y=13
 
first we want to write this in standard form which is: y=mx+b       (where the slope is m and y-intercept is b)
 
so 4y=13-3x  and so y = (13-3x)/4 = 13/4 - 3/4 x  or  Y =(-3/4) X + (13/4)
 
The slope of this line is -3/4
 
For a line to be perpendicular to that, obviously it has to have the opposite sign. It turns out that a the slope of a perpendicular line is negative reciprocal  (inverse) of the slope of the original line. So we can say:
 
Y' = (4/3) X +b'          wher b' is the y intercept of the perpendicular line.
 
Since the line passes through (2,7), we have:
 
7 = (4/3) X + 2  
 
So  (4/3)X=5     and X = 5*(4/3) =20/3
 
So Y' = (4/3) X + (20/3)
 
Note: I have denoted the perpendicular line as Y' since I have already defined the original line as Y.

Comments

I may be incorrect, but I get a different answer, and I've checked it numerous times, even using an online calculator.

The perpendicular line equation is y=(4/3)x+b, as Seyed states. Here's where my calculations differ:

If I solve using the point-slope formula (y-y1)=m(x-x1), I get y-7=(4/3)(x-2) => y-7=(4/3)x-8/3 => y=(4/3)x-8/3+7 => y=(4/3)x+13/3.

If I solve instead to find b, I get 7=(4/3)2+b => 7-8/3=b => b=13/3; thus, y=(4/3)x+13/3.

Forgive me if I've stepped on your toes, Seyed.
Hi -
 
I'll second the comment from Shannon.
 
I believe the first answer had it correct up to:
 
    For a line to be perpendicular to that, ... . So we can say:

    Y' = (4/3) X +b' wher b' is the y intercept of the perpendicular line.

But it goes wrong at:
 
    Since the line passes through (2,7), we have:

    7 = (4/3) X + 2
 
We are given (x=2, y=7); therefore we should be solving for b'. Thus, it should be:
 
    7 = (4/3) * 2 + b'
 
So the method of the first method is fundamentally fine, it just got off track a bit.
 
Best wishes.
yes! My bad. I should have put the two for the x and not for b. Messed up on that!
yes! My bad. I should have put the two for the x and not for b. Messed up on that!