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 | |
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 | Quote:
Step 1 : get some graph paper... Last edited by skipjack; January 1st, 2019 at 08:00 AM. | |
January 1st, 2019, 05:59 AM | #3 | |
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 | Quote:
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 |
Senior Member Joined: Oct 2009 Posts: 863 Thanks: 328 | 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 |
Global Moderator Joined: Dec 2006 Posts: 20,966 Thanks: 2215 |
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. Did you mean $\displaystyle y+4x-15=0$? |