Question 447396
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I need help writing the slope-intercept equation for the line that passes through (-3, -15) 
and is perpendicular to -6x + 8y = 3. Thank you.
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        The  " solution "  in the post by @mananth is  INCORRECT.

        Speaking in soft manner,  it is absurdist,  since he identifies the values  -4/3  and  -1.33,

        although they are absolutely different,  as any  5th grade student knows.


        Therefore,  I will re-write his solution to present it in correct form.



<pre>
Your starting equation is

    -6x + 8y = 3.


Find the slope of this line

    m1 = {{{6/8}}} = {{{3/4}}}.


The slope of a line perpendicular to the above line will be the negative reciprocal m2 = {{{-4/3}}}.


Now we want to write an equation of a straight line with the slope m2 = {{{-4/3}}} through point (-3,-15).


Write it in the form

    y = {{{(-4/3)x}}} + b.


We need to find a constant 'b'.

Find 'b' by plugging the values of x = -3,  y = -15 

    -15 = {{(-4/3)*(-3)}}} + b

    -15 = 4 + b

    b = -19


The required equation is y = {{{(-4/3)x}}} - 19.    <U>ANSWER</U>
</pre>

Solved correctly.


In this kind of problems, @mananth runs his computer code, &nbsp;which generates 
incorrect solutions every time, &nbsp;from problem to problem.


I just tired to disprove them.