Question 226490
You have to get the equation in the form {{{y = mx + b}}}
where {{{m}}} is the slope, so
{{{3x - 5y = 11}}}
Subtract {{{3x}}} from both sides
{{{-5y = -3x + 11}}}
Divide both sides by {{{-5}}}
{{{y = (3/5)*x - 11/5}}}
Now you can see that the slope is {{{m = 3/5}}}
ANY line which is perpendicular to this line
will have a slope which is {{{-(1/m)}}}, so
{{{-(1/m) = -(1/(3/5))}}}
{{{-(1/(3/5)) = -(5/3)}}}
So now I know that the line I want will look mlike
{{{y = -(5/3)x + b}}}
I'm told the perpendicular goes through (-1,7)
I can use the formula
{{{(y - 7)/(x - (-1)) = -5/3}}}
{{{(y - 7)/(x + 1) = -5/3}}}
Multiply both sides by {{{3*(x+1)}}}
{{{3*(y - 7) = -5*(x + 1)}}}
{{{3y - 21 = -5x - 5}}}
{{{3y = -5x + 16}}}
{{{y = -(5/3)x + 16/3}}} answer
I'll plot both equations
{{{ graph( 600, 600, -12, 12, -10, 10, (3/5)*x - 11/5, -(5/3)x + 16/3) }}}
This looks right, because if {{{y = 0}}} for both equations,
{{{x=3.67}}} for one and {{{x = 3.2}}} for the other