# Write the equation of the line perpendicular tothe line 2x + 5y = 12 and passing through the point (-10, 3) Put the equation in standard form.

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Given the line 2x+5y=12, we are asked to write the equation of the line perpendicular to the given line passing through the point (-10,3):

The slope of a line given in standard form Ax+By=C is `m=-A/B` , so the slope of the given line is `m=-2/5` .

** Alternatively you can rewrite in slope-intercept form and read off the slope: 2x+5y=12 ==> 5y=-2x+12 so `y=-2/5x+12/5` **

Given the slope of a line, the slope of a line perpendicular to the given line is the opposite reciprocal of the given slope. (Or `m_1*m_2=-1` )

The slope of the line we are seeking is `5/2` .

We now have the slope, `m=5/2` , and a point (-10,3) so we can use the point-slope form to find an equation for the line.

`y-3=5/2(x-(-10))`

`y-3=5/2x+25`

`y=5/2x+28` Multiply both sides by 2:

`2y=5x+56`

-5x+2y=56

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The equation of the line we seek is -5x+2y=56 or 5x-2y=-56

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