Algebra Questions -- Please answer quickly! --

IDK how to solve this so yeah, have fun.

1)   First we need to find the slope of  l₁ . Let's get its equation into slope-intercept form.

5x + 8y  =  -9          Subtract  5x  from both sides of the equation.

8y  =  -5x - 9           Divide through by  8 .

y  =  -\(\frac58\)x - \(\frac98\)

Now we can see that the slope of  l₁  =  -\(\frac58\)  .

Line  l₂  is perpendicular to  l₁  , so the slope of  l₂  =  +\(\frac85\)  .

Line  l₂  has a slope of  \(\frac85\)  and passes through the point  (10, 10) .

So.... in point - slope form, the equation of  l₂  is

y - 10  =  \(\frac85\)(x - 10)             We want this in the form  y = mx + b  . Distribute the  \(\frac85\) .

y - 10  =  \(\frac85\)x - \(\frac85\)(10)

y - 10  =  \(\frac85\)x - 16               Add  10  to both sides.

y  =  \(\frac85\)x - 6

Now the equation for  l₂  is in the form  y = mx + b  , where  m = \(\frac85\)  and  b = -6 .

We can look at a graph to verify that  l₁  and  l₂  are perpendicular, and  l₂  passes through (10, 10) .

m + b   =  \(\frac85\) + -6   =   -4.4

2)

The midpoint of  (1, 2)  and  (7, 4)  \(=\,(\frac{1+7}{2},\frac{2+4}{2}) \,=\,(\frac82,\frac62)\)  =  (4, 3)

The slope between  (1, 2)  and  (7, 4)  \(=\,\frac{4-2}{7-1}\,=\,\frac26\,=\,\frac13\)

So the slope of the perpendicular bisector   =   -\(\frac31\)   =   -3

We want an equation of a line that passes through  (4, 3)  with a slope of  -3 .

In point-slope form, that is

y - 3  =  -3(x - 4)         This is the equation of the line. We need it in  y = mx + b  form.

                                   Distribute the  -3 .

y - 3  =  -3x + 12

                                   Add  3  to both sides.

y  =  -3x + 15