Find the distance from the point of intersection of the lines 2x+3y=10 and 3x-y=4 to the line 5x-6y=1

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Find the distance from the point of intersection of the lines 2x+3y=10 and 3x-y=4 to the line 5x-6y=1

Mathematics
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First find the point set both your equations equal and solve for (x,y) (you could probably even get it by guessing) Then use the distance formula from a point to a line
The point is 2,2 but I think I'm using the formula for the distance from a point to a line incorrectly... Could you show me how you'd do it?
Okay, the point is (2,2) and the line is 5x-6y=1 So the formula you can use is just \[\sqrt{ax+by+c} \div \sqrt{a ^{2}}+b ^{2}\] Where the point is (x,y) and the line is ax+by+c=0

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oops.. the ax by and c terms chould all be squared under the root
actually looking again.. they aren't squared. the first formula is right
(without the square root)
yeah That's the formula I have
so the equation is actually just |ax +by+c| / sqrt(a^2 + b^2 )
But I'm supposed to use the absolute value, right?
yeah
and that's what I think was making me do the question wrong
so it should be 3 right?
not -3
Yeah, but the answer is 0.38
So does the absolute value apply to the denominator too?
Well both terms are squared so it won't matter for the bottom
So... The bottom should be ... \[\sqrt{61}\] right?
yeah
Hallelujah the right answer! Finally :) Thanks for your help :)
awwyeah :)

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