_{2}= m

_{2}X + b

_{2}Perpendicular to L

_{1 : 3X + 4y = 13 }passing through ( 2, 7)

_{2 = }4/3

_{2: }Y

_{ = }4/3 X

_{ + }b

_{2}

_{2}

_{2 = 7 - 8/3 = 13/3}

_{2}: Y = 4/3 X + 13/3

Summer Math packet question. write the equation in slope intercept and standard form

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Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

Equation of Line L_{2}= m_{2}X + b_{2} Perpendicular to L_{1 : 3X + 4y = 13
}passing through ( 2, 7)

L1 :

y = -3/4 X + 13/4

then, m_{2 = }4/3

L_{2: }Y_{ = }4/3 X_{ + }b_{2}

7 = 4/3(2) + b_{2}

b_{2 = 7 - 8/3 = 13/3}

L_{2} : Y = 4/3 X + 13/3

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.

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.

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

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.

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