Question

P(x, y) moves such that the area of the triangle with vertices at P(x, y), (1, –2), (–1, 3) is equal to the area of the triangle with vertices at P(x, y), (2, –1), (3, 1). The locus of P is the pair of lines-

• A. 3x + 3y + 4 = 0 = 7x + y – 6
• B. x + y – 2 = 0 = 7x + y + 4
• C. x + y – 6 = 0 = x + 7y + 4
• D. 7x + 3y + 4 = 0 = 3x + y – 6

Answer 1

Answer: (A)

3x + 3y + 4 = 0 = 7x + y – 6

Answer 2

Locus of P is given by $\frac { 1 } { 2 }$ $\left| \begin{matrix} 1 & x & y \\ 1 & 1 & –2 \\ 1 & –1 & 3 \end{matrix} \right|$ = ± $\frac { 1 } { 2 }$ $\left| \begin{matrix} 1 & x & y \\ 1 & 2 & –1 \\ 1 & 3 & 1 \end{matrix} \right|$i.e.,

1 – 5x – 2y = ± (5 – 2x + y) giving the two lines

3x + 3y + 4 = 0 and 7x + y – 6 = 0