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Maths-

General

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Question

The equation of perpendicular bisectors of sides AB and AC of a triangle ABC are $x minus y plus 5 equals 0 text and end text x plus 2 y equals 0$ respectively. If point A is (1, -2), then the equation of line BC is

$23 x + 14 y - 40 = 0$
$14 x - 23 y + 40 = 0$
$23 x - 14 y + 40 = 0$
$14 x + 23 y - 40 = 0$

The correct answer is: $14 x plus 23 y minus 40 equals 0$

Let the equation of perpendicular bisector FN of AB is
$x minus y plus 5 equals 0$

See Fig. 10.42. The midpoint F of AB is $open parentheses fraction numerator x subscript 1 end subscript plus 1 over denominator 2 end fraction comma fraction numerator y subscript 1 end subscript minus 2 over denominator 2 end fraction close parentheses$ that lies on the line given in Eq. (1).
Therefore,
$x subscript 1 end subscript minus y subscript 1 end subscript equals negative 13$
Also, AB is perpendicular to FN. So the product of their slopes is -1. That is,
$fraction numerator y subscript 1 end subscript plus 2 over denominator x subscript 1 end subscript minus 1 end fraction cross times 1 equals negative 1 text or end text x subscript 1 end subscript plus y subscript 1 end subscript equals negative 1$
On solving Eqs. (2) and (3), we get B(−7, 6). Similarly, we get $C open parentheses fraction numerator 11 over denominator 5 end fraction comma fraction numerator 2 over denominator 5 end fraction close parentheses$
Hence, the equation of BC is $14 x plus 23 y minus 40 equals 0$

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