The equations of two sides AB and AC of a triangle ABC are 4 x+y=14 and 3 x-2 y=5 , respectively. The point (2,-(4)/(3)) divides the third side BC internally in the ratio 2: 1 . the equation of the side BC is

Correct answer: A. x+3 y+2=0 aligned & 2 x_2+x_1 3 =2, 2( 3 x_2-5 2 )+(14-4 x_1) 3 =(-4)/(3) & 2 x_2+x_1=6,3 x_2-4 x_1=-13 & x_2=1, x_1=4 aligned So, C(1,-1), B(4,-2) m=(-1)/(3) Equation of BC: y+1=(-1)/(3)(x-1) aligned & 3 y+3=-x+1 & x+3 y+2=0 aligned

JEE Main 20248 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Straight Lines

JEE Main 2024 · 8 April · Morning Shift · Mathematics · Straight Lines

The equations of two sides AB and AC of a triangle ABC are 4 x+y=14...

The equations of two sides

\mathrm{AB}

and

\mathrm{AC}

of a triangle

\mathrm{ABC}

are

4 x+y=14 and 3 x-2 y=5

, respectively. The point

\left(2,-\frac{4}{3}\right)

divides the third side

\mathrm{BC}

internally in the ratio

2: 1

. the equation of the side

\mathrm{BC}

The equations of two sides AB and AC of a triangle ABC are 4 x+y=14 and 3 x-2 y=5 , respectively. The point

Options

A $x+3 y+2=0$
B $x-6 y-10=0$
C $x-3 y-6=0$
D $x+6 y+6=0$

Correct answer

A. $x+3 y+2=0$

Step-by-step solution

\begin{aligned} & \frac{2 \mathrm{x}_2+\mathrm{x}_1}{3}=2, \frac{2\left(\frac{3 \mathrm{x}_2-5}{2}\right)+\left(14-4 \mathrm{x}_1\right)}{3}=\frac{-4}{3} \\ & 2 \mathrm{x}_2+\mathrm{x}_1=6,3 \mathrm{x}_2-4 \mathrm{x}_1=-13 \\ & \mathrm{x}_2=1, \mathrm{x}_1=4 \end{aligned}

So,

\mathrm{C}(1,-1), \mathrm{B}(4,-2) \mathrm{m}=\frac{-1}{3}

Equation of

\mathrm{BC}: \mathrm{y}+1=\frac{-1}{3}(\mathrm{x}-1)

\begin{aligned} & 3 y+3=-x+1 \\ & x+3 y+2=0\end{aligned}

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