##### point slop from equation is 5x-8y-9=0 and it passes thru (-1,6)

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First, we solve the given equation for y in order to find its slope. So we would have the following:

5x - 8y - 9 = 0 (adding 8y from both sides) ------------> 5x - 8y + 8y - 9 = 0 + 8y ------------> 5x - 9 = 8y

5x - 9 = 8y (dividing by 8 from both sides) -----------> 5x/8 - 9/8 = 8y/8 ------------> 5/8x - 9/8 = y

So finally, we have: **y = 5/8x - 9/8. **And let's remember the slope-intercept form is: y = mx + b (where m is the slope and b is the y-intercept). Then we have:

If we compare y = mx + b with y = 5/8x + b we can see that the slope is **m = 5/8**. Ok, then if it says that the lines** **are parallel we use the same slope. By the way, let's remember that the slope-point form is:

y - y1 = m(x - x1) ; where: x1 and y1 are the coordinates of the given point.

Since the given point is: (-1 , 6) then x1 = -1 and y1 = 6 and we already know that the slope is m = 5/8. So we enter them into the formula and we have:

y - y1 = m(x - x1) ---------> y - 6 = 5/8(x -(-1) ) ------------> **y - 6 = 5/8(x + 1) This is the point-slope form.**

We could leave it to the slope-intercept form by solving for y like this:

y - 6 = 5/8(x + 1) --------> y - 6 = 5/8x + 5/8 ---------> y - 6 + 6 = 5/8x + 5/8 + 6 ---------> y = 5/8x + 5/8 + 6*8/8

y = 5/8x + 5/8 + 48/8 ---------> **y = 5/8x + 53/8 This is the intercept-slope form.**

If it says that the lines are perpendicular then you will need to find the perdicular slope respect to the slope 5/8. For that, we flip the fraction and we multiply by -1. So we have the following:

5/8 ---------> 8/5 ----------> (-1)(8/5) -------------> m = -8/5 (perpendicular slope).

Then we use the same formula (remember x1 = -1 and y1 = 6 from the given point):

y - y1 = m(x - x1) -----------> y - 6 = -8/5(x - (-1) ) -------> **y - 6 = - 8/5(x + 1) ****This is the point-slope form.**

y - 6 = -8/5x - 8/5 ----------> y - 6 + 6 = -8/5x - 8/5 + 6 ----------> **y = -8/5x + 22/5**

Please let me know if you have any doubt or question.

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