AP EAPCET 04-Jul-2022 Shift 2 Paper

Let a,b and c are the position vectors of the vertices A,B and C, respectively.
Also d,e and f be the position vectors of points D,E and F, respectively
d=‌

2c+3b

5

,e=‌

2c+a

3

,f=‌

3b+a

4

Equation of line BE is r=b+(c−b), .
‌‌r‌=b+k(‌2c+a∕3−b)
⇒‌‌r=b+k[‌2c+a−3b∕3]
⇒‌‌r‌=‌k∕3a+b[1−k]+‌2k∕3c

Equation of line CF is
r‌=c+λ(f−c)
‌=c+λ(‌

3b+a

4

−c)=c+λ[‌

3b+a−4c

4

]
‌=‌

λa

4

+‌

3λb

4

+c(1−λ)
CF and BE intersect at point P
‌

ka

3

+b(1−k)+‌

2k

3

c=‌

λa

4

+‌

3λb

4

+c(1−λ)
On comparing,
‌

k

3

‌=‌

λ

4

⇒‌‌k‌=‌

3λ

4

Also,
‌1−k=‌

3λ

4

⇒1−k=k
‌⇒‌‌2k=1⇒k=‌

1

2

‌‌ and ‌
‌‌

1

2

=‌

3λ

4

‌λ=‌

2

3

Position vector of P=‌

a

6

+‌

b

2

+‌

c

3

=‌

a+3b+2c

6

AP=‌x1AB+y1AC
‌(‌

a+3b+2c

6

−a)=x1(b−a)+y1(c−a)
⇒‌‌‌

−5a+3b+2c

6

=a(−x1−y1)+x1b+y1c
∴‌‌−x1−y1=‌

−5

6

‌x1+y1=‌

5

6