How to find the slope-intercept equation of a line perpendicular to x-7y=5 and containing the point (3,3) ?  

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Note that two lines are perpendicular if the slope of one of the line is negative reciprocal of the other line's slope. The formula that express the relationship of the slopes of perpendicular lines is:

  `m_2= -1/ m_1`

So to solve , determine the slope of the given line. To do so, express the equation in slope-intercept form y=mx + b.

`x-7y=5`

`-7y=-x + 5`

`y = (- x + 5) /(-7)` 

`y= 1/7x - 5/7`

Base on this equation, the slope of the given line is `m_1=1/7` .

Next, determine the equation of the second line. Since the second line is perpendicular to the given line, use the formula indicated above.

`m_2 = - 1/m_1 = -1/(1/7) = - 7`

So the slope of the line perpendicular to x-y=5 is -7.

Since the second line has a slope of -7 and contains the point (3,3), the next step is to use the slope-intercept formula of a line which is:

`y- y_1 = m(x-x_1)`

where m is the slope and (x1 , y1) is the given point. So,

`y-3=-7(x-3)`

`y-3=-7x + 21`

Then, isolate y to express the equation of the second line in slope-intercept form y=mx+b.

`y=-7x+25`

Hence, the equation of the line perpendicular to `x-7y=5` and containing the point (3,3) is `y=-7x+25` .

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