Question

Find the equation of a locus of a point which form a triangle of area is 2 with the points A(1,1)andB(-2,3)
​

Answer 1

Step-by-step explanation:

above images are the answer

Answer 2

Answer:

$\sf \: Hence, \: equations \: of \: locus \: of \: point \: are \: 2x + 3y + 9 = 0 </p><p> \sf \\and \: 2x + 3y + 1 = 0$

Step-by-step explanation:

Let P = (x y) is the locus of point which forms a triangle of area 2 sq unit with points A (1,1) and B(-2,3).

We know, area of triangle formed by points

$\sf( \:^{}x_{1}, \:^{}y_{1}),( \:^{}x_{2}, \:^{}y_{2}) and ( \:^{}x_{3}, \:^{}y_{3}). \\ \sf \: Then, \: \: area \: \: of \: \: triangle \: \: = \\ \sf \implies \frac{1}{2} [ \:^{}x_{1}( \:^{}y_{2} - \:^{}y_{3}) + \:^{}x_{2}( \:^{}y_{3} - \:^{}y_{1})$

$\sf( \:^{}x_{1}, \:^{}y_{1}),( \:^{}x_{2}, \:^{}y_{2}) and ( \:^{}x_{3}, \:^{}y_{3}). \\ \sf \: Then, \: \: area \: \: of \: \: triangle \: \: = \\ \sf \implies \frac{1}{2} [ \:^{}x_{1}( \:^{}y_{2} - \:^{}y_{3}) + \:^{}x_{2}( \:^{}y_{3} - \:^{}y_{1})+ \:^{}x_{3}( \:^{}y_{1} - \:^{}y_{2})]$

$\sf \: So, \: \: area \: \: of \: triangle \: Formed \: by \: P(x,y), A(1,1) and B \: (-2,3)$

$\sf2 = 1/2 x(1 - 3) + 1(3 - y) -2(y - 1)|$

$\sf4 = |-2x + 3 - y -2y + 21|$

$\sf \bold {breaking \: mod}</p><p>$

$\sf+4 = (-2x -3y+ 5)$

$\sf 4 = |-2x -3y + 51 \\ \\ </p><p> \sf-2x - 3y + 5 = +4$

$\implies \sf-2x - 3y = 5 + 4$

$\sf \: Hence, \: equations \: of \: locus \: of \: point \: are \: 2x + 3y + 9 = 0 \\ </p><p> \sf \: and \: 2x + 3y + 1 = 0$