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 January 1st, 2019, 05:05 AM #1 Member   Joined: Aug 2018 From: Nigeria Posts: 73 Thanks: 2 What do I do here???? Find the minimum distance of the line HM to origin O, given that H is on the point (2,7) and M is on the point (4,-1). Last edited by skipjack; January 1st, 2019 at 07:58 AM. January 1st, 2019, 05:34 AM   #2
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 Originally Posted by Harmeed Find the minimum distance of the line HM to origin O, given that H is on the point (2,7) and M is on the point (4,-1).

Step 1 : get some graph paper...

Last edited by skipjack; January 1st, 2019 at 08:00 AM. January 1st, 2019, 05:59 AM   #3
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Quote:
 Originally Posted by Harmeed Find the minimum distance of the line HM to origin O, given that H is on the point (2,7) and M is on the point (4,-1).
I can think of two ways to solve this, one primarily reliant on algebra and one on calculus. Which are you studying?

In any case, it will be helpful to have an equation representing the line segment HM. Can you find that?

Last edited by skipjack; January 1st, 2019 at 08:00 AM. January 1st, 2019, 06:53 AM #4 Member   Joined: Aug 2018 From: Nigeria Posts: 73 Thanks: 2 Someone said I should use my knowledge on vector, sir. Last edited by skipjack; January 1st, 2019 at 08:00 AM. January 1st, 2019, 07:02 AM   #5
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 Originally Posted by Harmeed Someone said I should use my knowledge on vector, sir.
Yes, you probably should.

Last edited by skipjack; January 1st, 2019 at 08:01 AM. January 1st, 2019, 08:05 AM #6 Senior Member   Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 Equation of the line is $\displaystyle y+4x+15=0$ The point is $\displaystyle O(0,0)$ Search in google for “distance of line to point” , find the formula and apply it January 1st, 2019, 08:16 AM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra The direction of the line is the direction of $\vec{HM}$. The vector representation of any point on the line $P(t)$ is $\vec{OQ} + t\vec{HM}$ for any value of $t$ where $Q$ is any given point on the line (you have been given two). The shortest distance between a point (the origin) and a line is given by the perpendicular vector from that point to the line. That is, you want to find $\vec{OP(t)}$ such that $\vec{OP(t)}$ is perpendicular to the direction of the line, $\vec{HM}$. $\vec{OP(t)}$ and $\vec{HM}$ are perpendicular if $\vec{OP(t)} \cdot \vec{HM} = 0$ (the scalar product of the two vectors is equal to zero). This fact will give you an equation that can be solved for the particular value of $t$ that you need. You then determine the length of the vector $\vec{OP(t)}$. January 1st, 2019, 08:52 AM   #8
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Triangle HMO has area 15 (by Pick's theorem or by a vector method or some other means).

HM has length 2√17 (by Pythagoras).

Hence a perpendicular from O to the line through H and M has length 15/√17.

It remains to be shown that this is the required minimum distance.

Quote:
 Originally Posted by idontknow Equation of the line is $\displaystyle y+4x+15=0$
Did you mean $\displaystyle y+4x-15=0$? ### linear algebra forum

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