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Mathematics Question on Coordinate Geometry

The equations of two sides AB and AC of a triangle ABC are 4x+y=144x + y = 14 and 3x−2y=53x - 2y = 5, respectively. The point (2,−43)\left(2, -\frac{4}{3}\right) divides the third side BC internally in the ratio 2 : 1. The equation of the side BC is:

A

x – 6y – 10 = 0

B

x – 3y – 6 = 0

C

x + 3y + 2 = 0

D

x + 6y + 6 = 0

Answer

x + 3y + 2 = 0

Explanation

Solution

Given:
The given equations are:
AB:4x+y=14,AC:3x−2y=5.AB : 4x + y = 14, \quad AC : 3x - 2y = 5.
Coordinates of BB and CC
The coordinates of BB on ABAB:
B(x1,y1)where y1=14−4x1.B(x_1, y_1) \quad \text{where } y_1 = 14 - 4x_1.
The coordinates of CC on ACAC:
C(x2,y2)where y2=3x2−52.C(x_2, y_2) \quad \text{where } y_2 = \frac{3x_2 - 5}{2}.
Use Section Formula
The point P(2,−43)P(2, -\frac{4}{3}) divides BCBC in the ratio 2:12 : 1. Using the section formula:
xp=2x2+x13,yp=2y2+y13.x_p = \frac{2x_2 + x_1}{3}, \quad y_p = \frac{2y_2 + y_1}{3}.
Substitute xp=2x_p = 2 and yp=−43y_p = -\frac{4}{3} into these equations.
Solve for x1x_1 and x2x_2
For xp=2x_p = 2:
2=2x2+x13.2 = \frac{2x_2 + x_1}{3}.
Rearrange:
6=2x2+x1  ⟹  x1=6−2x2.6 = 2x_2 + x_1 \quad \implies \quad x_1 = 6 - 2x_2.
For yp=−43y_p = -\frac{4}{3}:
−43=2y2+y13.-\frac{4}{3} = \frac{2y_2 + y_1}{3}.
Substitute y1=14−4x1y_1 = 14 - 4x_1 and y2=3x2−52y_2 = \frac{3x_2 - 5}{2}:
−43=2(3x2−52)+(14−4x1)3.-\frac{4}{3} = \frac{2\left(\frac{3x_2 - 5}{2}\right) + (14 - 4x_1)}{3}.
Simplify:
−43=3x2−5+14−4x13.-\frac{4}{3} = \frac{3x_2 - 5 + 14 - 4x_1}{3}.
Multiply through by 3:
−4=3x2−5+14−4x1.-4 = 3x_2 - 5 + 14 - 4x_1.
Combine terms:
−4=3x2+9−4x1  ⟹  3x2−4x1=−13.-4 = 3x_2 + 9 - 4x_1 \quad \implies \quad 3x_2 - 4x_1 = -13.
Substitute x1=6−2x2x_1 = 6 - 2x_2 into 3x2−4x1=−133x_2 - 4x_1 = -13:
3x2−4(6−2x2)=−13.3x_2 - 4(6 - 2x_2) = -13.
Simplify:
3x2−24+8x2=−13  ⟹  11x2=11  ⟹  x2=1.3x_2 - 24 + 8x_2 = -13 \quad \implies \quad 11x_2 = 11 \quad \implies \quad x_2 = 1.
Substitute x2=1x_2 = 1 into x1=6−2x2x_1 = 6 - 2x_2:
x1=6−2(1)=4.x_1 = 6 - 2(1) = 4.
--- Step 4: Solve for y1y_1 and y2y_2
Substitute x1=4x_1 = 4 into y1=14−4x1y_1 = 14 - 4x_1:
y1=14−4(4)=−2.y_1 = 14 - 4(4) = -2.
Substitute x2=1x_2 = 1 into y2=3x2−52y_2 = \frac{3x_2 - 5}{2}:
y2=3(1)−52=−22=−1.y_2 = \frac{3(1) - 5}{2} = \frac{-2}{2} = -1.
Thus, B(4,−2)B(4, -2) and C(1,−1)C(1, -1).
Slope of BCBC
The slope of BCBC is:
m=y2−y1x2−x1=−1−(−2)1−4=1−3=−13.m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - (-2)}{1 - 4} = \frac{1}{-3} = -\frac{1}{3}.
Equation of BCBC
Using the point-slope form at B(4,−2)B(4, -2):
y−(−2)=−13(x−4).y - (-2) = -\frac{1}{3}(x - 4).
Simplify:
y+2=−13x+43.y + 2 = -\frac{1}{3}x + \frac{4}{3}.
Multiply through by 3:
3y+6=−x+4  ⟹  x+3y+2=0.3y + 6 = -x + 4 \quad \implies \quad x + 3y + 2 = 0.
Final Answer:
x+3y+2=0.\boxed{x + 3y + 2 = 0.}