Let $A(-1,1), B(3,4)$ and $C(2,0)$ be given three points. A line $y=m x, m>0$, intersects lines $A C$ and BC at point P and Q respectively. Let $\mathrm{A}_{1}$ and $\mathrm{A}_{2}$ be the area of $\triangle \mathrm{ABC}$ and $\triangle PQC$ respectively, such that $A_{1}=3 A_{2}$, then the value of $m$ is equal to :

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Straight Lines

Let

A (- 1, 1), B (3, 4)

and

C (2, 0)

be given three points. A line

y = m x, m > 0

, intersects lines

A C

and BC at point P and Q respectively. Let

A_{1}

and

A_{2}

be the area of

△ ABC

and

△ P QC

respectively, such that

A_{1} = 3 A_{2}

, then the value of

m

is equal to :

Answer: (D)

Since

P

lies on

C A

and

Q

C B

, the triangles share the angle at

C

and

\frac{A _{2}}{A _{1}} = \frac{C P}{C A} \cdot \frac{C Q}{C B} = \frac{1}{3}

Line

A C

x + 3 y = 2

and line

B C

y = 4 x - 8

. Solving each with

y = m x

x_{P} = \frac{2}{1 + 3 m}, x_{Q} = \frac{8}{4 - m}

Comparing

x

-coordinates along each line from

C (2, 0)

\frac{C P}{C A} = \frac{x _{P} - 2}{- 1 - 2} = \frac{2 m}{1 + 3 m}, \frac{C Q}{C B} = \frac{x _{Q} - 2}{3 - 2} = \frac{2 m}{4 - m}

\frac{4 m ^{2}}{( 1 + 3 m ) ( 4 - m )} = \frac{1}{3} \Rightarrow 15 m^{2} - 11 m - 4 = 0 \Rightarrow (m - 1) (15 m + 4) = 0

The condition

m > 0

rules out

- \frac{4}{15}

, so

m = 1

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option