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 August 22nd, 2018, 03:37 AM #1 Member   Joined: Aug 2018 From: Nigeria Posts: 73 Thanks: 2 Equation of a line Help!!! I need a mathematician to solve this... The line 5x+Ky-15=0 is perpendicular to the line 2x+5y-3=0,find the value of K. Last edited by Harmeed; August 22nd, 2018 at 03:40 AM. Reason: to erase
August 22nd, 2018, 04:35 AM   #2
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 Originally Posted by Harmeed Help!!! I need a mathematician to solve this... The line 5x+Ky-15=0 is perpendicular to the line 2x+5y-3=0,find the value of K.
A few questions.

The slope of line A has a slope of m. Line B is perpendicular to line A. What is the slope of line B? The answer to that question should be in your text or lecture notes, but you may not have seen its relevance.

Its relevance is this: if you determine the slope of 2x + 5y - 3 = 0, you can then find the slope of 5x + ky - 15 = 0. And then you can find k.

So what is the slope of 2x + 5y - 3 = 0?

 August 22nd, 2018, 09:02 AM #3 Member   Joined: Aug 2018 From: Nigeria Posts: 73 Thanks: 2 The slope of 5x+Ky-15=0 is -5/k and 2x+5y-3=0 is -2/5...Help me guy how are my going to find constant k Thanks from o2jam
 August 22nd, 2018, 09:12 AM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,312 Thanks: 1224 the product of the slopes of two perpendicular lines is -1
August 24th, 2018, 04:53 PM   #5
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 Originally Posted by Harmeed The slope of 5x+Ky-15=0 is -5/k and 2x+5y-3=0 is -2/5...Help me guy how are my going to find constant k
the opposite reciprocal of -2/5x is 5/2x, so the first equation's slope has to equal this value, and the only way to do that is by letting k be 2 and then dividing 5x by 2 to get 5/2x once you first rewrite the equations in slope-intercept form. Please correct me if i'm wrong I would like to know the correct solution.

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