Λ and μ are scalars, find the value of λ and the value of μ

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Discussion Overview

The discussion revolves around a geometry problem involving a rectangle and the determination of scalar values λ and μ based on the intersection of lines in a coordinate system. Participants explore various mathematical relationships and calculations related to the geometry of the rectangle, including ratios and coordinates.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • Participants suggest using given ratios to find CF and AD in terms of a and c, respectively.
  • One participant proposes setting up a Cartesian coordinate system to define points O, A, C, B, and F, and derives equations for lines OD and AF.
  • A participant calculates the coordinates of point D based on the ratio AD:DB = 2:3, leading to the conclusion that D is at (a, (2/5)c).
  • Another participant provides the equations for lines OD and AF and suggests finding their intersection to determine λ and μ.
  • Calculations for distances OX, OD, AX, and AF are presented, but no consensus on the values of λ and μ is reached.

Areas of Agreement / Disagreement

Participants generally agree on the setup of the problem and the use of coordinate geometry, but there is no consensus on the specific values of λ and μ, as calculations are ongoing and not finalized.

Contextual Notes

Some calculations involve assumptions about the coordinates and the relationships between the segments, which may depend on the accuracy of earlier steps. The discussion does not resolve the mathematical steps necessary to find λ and μ definitively.

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Figure shows a rectangle OABC in which OA = a and OC = c. F is the midpoint of CB and D is the point on AB such that AD : DB = 2:3
(a) Find
_ _ _ _ _ __ (i) CF in terms of a
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ (ii) AD in terms of c

The lines OD and AF intersect at the point X Given that
OX = λOD and AX = μAF, where λ and μ are scalars,

(b) find the value of λ and the value of μ

Given that OX : XD = n:1

(c) find the value of n

Given also that I a l = 12 cm and l c I = 12.5 cm,

(d) find the area, in cm2, of quadrilateral XDBF .
 
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skeeter said:
part (a) is straightforward ... you should be able to use the given ratios to determine CF and AD in terms of a and c respectively.

part (b) ... let O be the origin, (0,0), of a Cartesian grid, A = (a,0), F = (a/2,c), etc.

OD has slope $\dfrac{|AD|}{a} \implies \text{ segment OD has equation } y = \dfrac{|AD|}{a} \cdot x$, where |AD| is in terms of c found in part (a)

AF has slope $-\dfrac{2c}{a} \implies \text{ segment AF has equation } y = -\dfrac{2c}{a} (x - a)$

using the two linear equations, determine the intersection, which will allow you to determine the values of $\lambda$ & $\mu$

part (c) should be easy to determine from the intersection information in working part (b)

part (d) ... area of XDBF = area of ABCO - CFAO - ADX
Don't leave it half-way
Plz complete it.
 
I would set up a coordinate system so that O is at the origin, A is at (a, 0), and C is at (0, c). Then B is at (a, c).
F is at (a/2, c) so the line from A to F is y= -(2c/a)(x- a)= -(2c/a)x+ 2c. D is the point on AB such that AD : DB = 2:3 so if we take D to be (a, x), AD= c- x and DB= x. That is, AD/DB=x/(c- x)= 2/3. 3x=2(c- x)= 2c- 2x. 5x= 2c so x= (2/5)c. D is at (a, (2/5)c). The line from O to D is y= ((2/5)c/a)x. X is the intersection of the two lines so y= -(2c/a)x+ 2c= ((2/5)c/a)x. 2c= ((2/5+ 2)(c/a)x= (12/5)(c/a)x. x= (5/12)(a/c)(2c)= (5/6)a.
y= (2/5)(c/a)(5/6)a= c/3.

X is the point ((5/6)a, c/3).https://mathhelpboards.com/account/

OX= sqrt((25/36)a^2+ c^2/9)= sqrt{25a^2/36+ 4c^2/36}= sqrt(25a^2+ 4c^2)/6.
OD= sqrt(a^2+ 4c^2/25)= sqrt(25a^2+4c^2)/5.

AX= sqrt(a^2/36+ c^2/9)= sqrt((a^2/36+ 4c^2/36)= sqrt(a^2+ 4c^2)/6.
AF= sqrt(a^2/4+ c^2)= sqrt(a^2+ 4a)/2.
 
Country Boy said:
I would set up a coordinate system so that O is at the origin, A is at (a, 0), and C is at (0, c). Then B is at (a, c).
F is at (a/2, c) so the line from A to F is y= -(2c/a)(x- a)= -(2c/a)x+ 2c. D is the point on AB such that AD : DB = 2:3 so if we take D to be (a, x), AD= c- x and DB= x. That is, AD/DB=x/(c- x)= 2/3. 3x=2(c- x)= 2c- 2x. 5x= 2c so x= (2/5)c. D is at (a, (2/5)c). The line from O to D is y= ((2/5)c/a)x. X is the intersection of the two lines so y= -(2c/a)x+ 2c= ((2/5)c/a)x. 2c= ((2/5+ 2)(c/a)x= (12/5)(c/a)x. x= (5/12)(a/c)(2c)= (5/6)a.
y= (2/5)(c/a)(5/6)a= c/3.

X is the point ((5/6)a, c/3).https://mathhelpboards.com/account/

OX= sqrt((25/36)a^2+ c^2/9)= sqrt{25a^2/36+ 4c^2/36}= sqrt(25a^2+ 4c^2)/6.
OD= sqrt(a^2+ 4c^2/25)= sqrt(25a^2+4c^2)/5.

AX= sqrt(a^2/36+ c^2/9)= sqrt((a^2/36+ 4c^2/36)= sqrt(a^2+ 4c^2)/6.
AF= sqrt(a^2/4+ c^2)= sqrt(a^2+ 4a)/2.
Tnx
 
Tnx
Solved
 

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