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#1 2008-10-28 16:10:35

WiZaRd
Real Member
Registered: 2006-09-22
Posts: 570

Help Me!!!!

Please solve this question!!

Find the Co-Ordinates of all points P on the X-axis so that  the line through A(1, 2 and P is perpendicular to the line through B(8, 3) and P.

Plaese Solve this and make it quick!! I am facing a little problem in it!down


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#2 2008-10-29 00:05:16

LuisRodg
Real Member
Registered: 2007-10-23
Posts: 322

Re: Help Me!!!!

So you have the points A and B given and arbitrary point P.

The line A to P can be represented as a vector from the point A to P.
v1 = P - A = <x-1, y-2>

The line B to P can be represented as a vector from the point B to P.
v2 = P - B = <x-8, y-3>

So if the two lines (vectors) are perpendicular, it means that their dot product is equal to 0. So we have:

v1 * v2 = 0
(x-1)(x-8) + (y-2)(y-3) = 0
x^2 - 8x - x + 8 + y^2 - 3y - 2y + 6 = 0
x^2 - 9x + y^2 - 5y = -14

Completing the square we get:

x^2 - 9x + (81/4) + y^2 - 5y + (25/4) = -14 + (81/4) + (25/4)
x^2 - 9x + (81/4) + y^2 - 5y + (25/4) = 25/2
(x - 9/2)^2 + (y - 5/2)^2 = 25/2

So all the points P that make the two lines perpendicular lie on the circle with center (9/2 , 5/2) and radius 5/sqrt(2).

Edit: I just noticed that you said that P needs to be in the x-axis. If P is in the x axis, then P will always have the form P(x,0) since y=0 in all the x-axis.

(x - 9/2)^2 + (y - 5/2)^2 = 25/2
(x - 9/2)^2 + (0 - 5/2)^2 = 25/2
(x - 9/2)^2 + 25/4 = 25/2
(x - 9/2)^2 = 25/4
x - 9/2 = +- 5/2
x = +- 5/2 + 9/2

x1 = 7
x2 = 2

So all the points P that make the two lines perpendicular and lie on the x-axis are the points P1(7,0) and P2(2,0)

Last edited by LuisRodg (2008-10-29 00:09:18)

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#3 2008-10-29 16:13:45

WiZaRd
Real Member
Registered: 2006-09-22
Posts: 570

Re: Help Me!!!!

Thank you Very Much.....!!
but can you solve it by using the line formula, not the vectors !!!


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#4 2008-10-31 19:46:48

WiZaRd
Real Member
Registered: 2006-09-22
Posts: 570

Re: Help Me!!!!

Please solve it by using the Co-Ordinet system which should include distance, slope formulas.!!
NOt vectors, NOt vectors, NOt vectors, please do it its really urgent!!!

thank you very much!


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#5 2008-11-01 03:49:33

LuisRodg
Real Member
Registered: 2007-10-23
Posts: 322

Re: Help Me!!!!

Why cant u use vectors? Anyways here is the solution using just basic algebra:

So we have:
A(1,2)
B(8,3)
P(x,0)   --> (x,0) because since P lies on the x-axis, its y-coord will always be 0.

So we have the line1 from A -> P and the line2 from B -> P. When you have two points, you can always find the slope of the line through this two points by doing rise/run. Or:

(y2-y1)/(x2-x1)

So we find the slope of line1:

m1 = (2-0)/(1-x) = 2/(1-x)

And we find the slope of line2:

m2 = (3-0)/(8-x) = 3/(8-x)

So we have the slopes of the two lines in unknown x. When two lines are perpendicular, m1 = -1/m2.

So we have: m1 = -1/m2

2/(1-x) = -(8-x)/3 = (x-8)/3

We cross multiply and we get:(1-x)(x-8) = 6
x-8-x^2+8x = 6
x^2 - 9x = -14

Completing the square we get:

x^2 - 9x + 81/4 = -14 + 81/4
(x - 9/2)^2 = 25/4
x - 9/2 = +- 5/2
x = +- 5/2 + 9/2

x1 = 7
x2 = 2

So since P(x,0) and we have x1 and x2 then it follows that the points P are (7,0) and (2,0).

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#6 2008-11-02 16:07:27

WiZaRd
Real Member
Registered: 2006-09-22
Posts: 570

Re: Help Me!!!!

Thank you Very Much Rodg!!!


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