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

[Help seeker]

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 .

 

[Help seeker]

D

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.

 

[HOI]

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.

 

[Help seeker]

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

 

[Help seeker]

Tnx
Solved