JEE Main 16 March 2021 Shift 2 Solved Paper

Given, points A(−1,1),B(3,4),C(2,0)
Equation of AC=‌

y−1

x+1

=‌

0−1

2+1

=‌

−1

3

⇒‌‌3y−3=−x−1
⇒‌‌x+3y=2...(i)
On solving Eq. (i) and y=mx, we get P(‌

2

3m+1

,‌

2m

3m+1

)
Equation of BC=‌

y−0

x−2

=‌

4−0

3−2

⇒‌‌y=4x−8 ....(ii)
Similarly, on solving Eq. (ii) and y=mx, we get Q(‌

8

4−m

,‌

8m

4−m

)Area of △ABC=3 Area of △PQC (given)
⇒‌‌13=‌

3

4+11m−3m2

×(52m2)
⇒15m2−11m−4=0
⇒‌‌m=1,‌

−4

15

 [but m>0 ]
⇒‌‌m=1