Key Concepts and Formulas

• Intersection of Lines: The coordinates of the intersection point of two lines $L_1: a_1x + b_1y + c_1 = 0$ and $L_2: a_2x + b_2y + c_2 = 0$ are found by solving their equations simultaneously.
• Section Formula (Internal Division): If a point $P(x, y)$ divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m:n$, then its coordinates are given by: $x = \frac{mx_2 + nx_1}{m+n} \quad \text{and} \quad y = \frac{my_2 + ny_1}{m+n}$
• Equation of a Line (Two-Point Form): The equation of a straight line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

Step-by-Step Solution

Let the vertices of the triangle be $A$, $B$, and $C$. The equations of sides AB and AC are given as:

• Side AB: $4x + y = 14$ (Equation 1)
• Side AC: $3x - 2y = 5$ (Equation 2)

Let the coordinates of vertex B be $(x_B, y_B)$ and vertex C be $(x_C, y_C)$. We are given that the point $P\left(2, -\frac{4}{3}\right)$ divides the side BC internally in the ratio $2:1$. This means that if we consider the segment from B to C, the point P is closer to C.

Step 1: Apply the Section Formula to establish relationships between the coordinates of B and C.

Since point $P\left(2, -\frac{4}{3}\right)$ divides BC internally in the ratio $2:1$, we use the section formula with $m=2$ and $n=1$, where $(x_B, y_B)$ and $(x_C, y_C)$ are the coordinates of B and C respectively.

For the x-coordinate of P: $2 = \frac{2 \cdot x_C + 1 \cdot x_B}{2+1}$ $2 = \frac{2x_C + x_B}{3}$ Multiplying by 3, we get a linear relationship between $x_B$ and $x_C$: $6 = 2x_C + x_B \quad \text{(Equation 3)}$ From this, we can express $x_B$ in terms of $x_C$: $x_B = 6 - 2x_C$

For the y-coordinate of P: $-\frac{4}{3} = \frac{2 \cdot y_C + 1 \cdot y_B}{2+1}$ $-\frac{4}{3} = \frac{2y_C + y_B}{3}$ Multiplying by 3, we get a linear relationship between $y_B$ and $y_C$: $-4 = 2y_C + y_B \quad \text{(Equation 4)}$ From this, we can express $y_B$ in terms of $y_C$: $y_B = -4 - 2y_C$

Step 2: Utilize the fact that vertices B and C lie on their respective sides.

• Vertex B$(x_B, y_B)$ lies on side AB. Therefore, its coordinates must satisfy the equation of line AB: $4x_B + y_B = 14 \quad \text{(Equation 5)}$
• Vertex C$(x_C, y_C)$ lies on side AC. Therefore, its coordinates must satisfy the equation of line AC: $3x_C - 2y_C = 5 \quad \text{(Equation 6)}$

Step 3: Solve the system of equations to find the coordinates of B and C.

We now have a system of four equations relating $x_B, y_B, x_C, y_C$. We can simplify this by substituting the expressions for $x_B$ and $y_B$ (from Step 1) into Equation 5: Substitute $x_B = 6 - 2x_C$ and $y_B = -4 - 2y_C$ into Equation 5: $4(6 - 2x_C) + (-4 - 2y_C) = 14$ $24 - 8x_C - 4 - 2y_C = 14$ $20 - 8x_C - 2y_C = 14$ Rearrange to isolate the variables: $-8x_C - 2y_C = 14 - 20$ $-8x_C - 2y_C = -6$ Dividing the entire equation by -2 to simplify: $4x_C + y_C = 3 \quad \text{(Equation 7)}$

Now we have a system of two linear equations solely in terms of $x_C$ and $y_C$:

1. $4x_C + y_C = 3$ (Equation 7)
2. $3x_C - 2y_C = 5$ (Equation 6)

We can solve this system using substitution or elimination. From Equation 7, express $y_C$ in terms of $x_C$: $y_C = 3 - 4x_C$ Substitute this expression for $y_C$ into Equation 6: $3x_C - 2(3 - 4x_C) = 5$ $3x_C - 6 + 8x_C = 5$ Combine like terms: $11x_C - 6 = 5$ $11x_C = 11$ $x_C = 1$

Now substitute the value of $x_C$ back into $y_C = 3 - 4x_C$ to find $y_C$: $y_C = 3 - 4(1)$ $y_C = 3 - 4$ $y_C = -1$ So, the coordinates of vertex C are $(1, -1)$.

Next, find the coordinates of vertex B using the expressions derived in Step 1: $x_B = 6 - 2x_C = 6 - 2(1) = 6 - 2 = 4$ $y_B = -4 - 2y_C = -4 - 2(-1) = -4 + 2 = -2$ So, the coordinates of vertex B are $(4, -2)$.

Step 4: Find the equation of side BC.

We have the coordinates of B$(4, -2)$ and C$(1, -1)$. We can use the two-point form of a straight line. First, calculate the slope $m$ of BC: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - (-2)}{1 - 4} = \frac{1}{-3} = -\frac{1}{3}$ Now, use the point-slope form $y - y_1 = m(x - x_1)$ with point B$(4, -2)$ and slope $m = -1/3$: $y - (-2) = -\frac{1}{3}(x - 4)$ $y + 2 = -\frac{1}{3}(x - 4)$ To eliminate the fraction, multiply both sides by 3: $3(y + 2) = -1(x - 4)$ $3y + 6 = -x + 4$ Rearrange the terms to get the standard form $Ax + By + C = 0$: $x + 3y + 6 - 4 = 0$ $x + 3y + 2 = 0$

However, the correct answer is given as (A) $x+6y+6=0$. Let's re-examine the section formula application. The point P(2,-4/3) divides BC in the ratio 2:1. This means BP:PC = 2:1.

Therefore, $2 = \frac{1*x_B + 2*x_C}{1+2}$ $6 = x_B + 2x_C$ $x_B = 6 - 2x_C$

$-4/3 = \frac{1*y_B + 2*y_C}{1+2}$ $-4 = y_B + 2y_C$ $y_B = -4 - 2y_C$

Substituting in $4x_B + y_B = 14$ $4(6-2x_C) + (-4-2y_C) = 14$ $24 - 8x_C - 4 - 2y_C = 14$ $20 - 8x_C - 2y_C = 14$ $6 = 8x_C + 2y_C$ $3 = 4x_C + y_C$

$3x_C - 2y_C = 5$ Multiply the first equation by 2: $6 = 8x_C + 2y_C$ Adding this to the second equation: $3x_C - 2y_C = 5$ We get: $11x_C = 11$ $x_C = 1$ Then, $3 = 4(1) + y_C$, so $y_C = -1$.

So $C = (1, -1)$. $x_B = 6 - 2(1) = 4$ $y_B = -4 - 2(-1) = -2$ So $B = (4, -2)$.

The slope of BC is $\frac{-1 - (-2)}{1 - 4} = \frac{1}{-3} = -\frac{1}{3}$. $y - (-2) = -\frac{1}{3}(x - 4)$ $y + 2 = -\frac{1}{3}x + \frac{4}{3}$ $3y + 6 = -x + 4$ $x + 3y + 2 = 0$.

Let us assume that the point (2,-4/3) divides BC externally in the ratio 2:1. Then BP:PC = 2:1, and P is outside the segment BC. Then, $2 = \frac{2x_C - x_B}{2-1}$, so $2 = 2x_C - x_B$ and $x_B = 2x_C - 2$ $-4/3 = \frac{2y_C - y_B}{2-1}$, so $-4/3 = 2y_C - y_B$ and $y_B = 2y_C + 4/3$ Then, $4(2x_C - 2) + 2y_C + 4/3 = 14$ $8x_C - 8 + 2y_C + 4/3 = 14$ $8x_C + 2y_C = 22 - 4/3 = 62/3$ $24x_C + 6y_C = 62$ (1) $3x_C - 2y_C = 5$, so $9x_C - 6y_C = 15$ (2) Adding (1) and (2) gives $33x_C = 77$, so $x_C = 7/3$. $3(7/3) - 2y_C = 5$, so $7 - 2y_C = 5$ and $2y_C = 2$ so $y_C = 1$. $C = (7/3, 1)$. $x_B = 2(7/3) - 2 = 14/3 - 6/3 = 8/3$. $y_B = 2(1) + 4/3 = 10/3$. $B = (8/3, 10/3)$. The slope of BC is $\frac{1-10/3}{7/3-8/3} = \frac{-7/3}{-1/3} = 7$. $y - 1 = 7(x - 7/3)$ $y - 1 = 7x - 49/3$ $3y - 3 = 21x - 49$ $21x - 3y - 46 = 0$. This is not the answer.

Let's try to construct the line $x+6y+6=0$ and see if (2, -4/3) divides it in 2:1 ratio with AB and AC: The intersection of $4x+y=14$ and $x+6y+6=0$ gives B. $x = -6y - 6$. $4(-6y-6) + y = 14$ $-24y - 24 + y = 14$ $-23y = 38$ $y = -38/23$. $x = -6(-38/23) - 6 = 228/23 - 138/23 = 90/23$. So $B = (90/23, -38/23)$.

The intersection of $3x-2y=5$ and $x+6y+6=0$ gives C. $x = -6y - 6$. $3(-6y-6) - 2y = 5$ $-18y - 18 - 2y = 5$ $-20y = 23$ $y = -23/20$ $x = -6(-23/20) - 6 = 69/10 - 60/10 = 9/10$. So $C = (9/10, -23/20)$. If (2,-4/3) divides BC in the ratio 2:1, then $2 = \frac{90/23 + 2(9/10)}{3}$, so $6 = 90/23 + 9/5 = \frac{450 + 207}{115} = 657/115 \neq 6$.

Common Mistakes & Tips

• Ratio Order: Ensure the section formula is applied with the correct order of the ratio and corresponding points. The ratio $m:n$ means $BP:PC = m:n$.
• Algebraic Manipulation: Systems of equations require careful manipulation. Double-check all calculations, especially with fractions.
• Check the answer: After finding a solution, substitute back into the original equations to verify.

Summary

The problem requires finding the equation of side BC of a triangle given the equations of sides AB and AC and a point that divides BC in a 2:1 ratio. By using the section formula and the fact that the vertices lie on their respective lines, we can set up a system of equations to find the coordinates of B and C. Then, using the two-point form of a line, we can find the equation of BC. Despite the initial calculations and double checks, the correct answer is not matching the option (C), but option (A) is stated as correct. Let's assume there is an error in the options, so the calculation leads us to the side BC as $x+3y+2=0$

The final answer is \boxed{x+3y+2=0}.