Calculate the distance from the point (3,5) to the line y = x + 4. HELP PLEASE

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Calculate the distance from the point (3,5) to the line y = x + 4. HELP PLEASE

Mathematics
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sqrt(2)
HUH?
\[D=\sqrt{(x-3)^2+(y-5)^2}=\sqrt{(x-3)^2+((x+4)-5)^2}\]

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\[D=\sqrt{(x-3)^2+(x-1)^2}\]
y= x+4 To get a distance, you need to find the perpendicular line from the point to the line to make a perpendicular line to y=x+4, one needs a slope of 1/-m, or -1. Therefore, the line has an equation of y = -x +b, and it goes through (3,5). Therefore, b = 8 to find the point where the two lines intersect, make y = y, or x+4 = -x +8 Therefore, x = 2 and y = 6 the distance from (3,5) to (2,6) is sqrt(2)
oh square root. k
if you draw a line from the point to the line, it will be perpendicular. so that means that the slope will be the negative inverse of the slope of the line. the slope of the line is 1, so the slope of the line from the point to the original line will be -1. Find the equation of this new line. Use the point slope formula. It will be y-5 = -1(x-3), which simplifies to y = -x+8. set this equation equal to your first equation and solve for x. you should get 2. plug this back into either of your equations. you should get 6. so the point at which these two lines intersect is (2,6). now you just need to find the distance between (3,5) and (2,6)...use the distance formula ... sqrt( (5-6)^2 + (3-2)^2 ) = sqrt(2)
(3,5) has a different distance from all points on the line y=x+4
you're finding the shortest distance.
are we finding the smallest distance?
most likely
if we are looking for the shortest distance we could do this by taking the derivative of D^2 let's call D^2=d(x) d(x)=(x-3)^2+(x-1)^2=x^2-6x+9+x^2-2x+1=2x^2-8x+10 so d'(x)=4x-8 critial numbers can be found by setting d'(x)=0 so we have 4x-8=0 so x=2 so we have a minimum distance for x=2 since it is decreasing before x=2 and increasing after x=2 so the D(2)=sqrt[(2-3)^2+(2-1)^2]=sqrt(1+1)=sqrt(2) is the smallest distance

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