Equation of perpendicular bisectors of sides AB and AC of a triangle ABC are x

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9.Straight Line

hard

The equation of perpendicular bisectors of the sides $AB$ and $AC$ of a triangle $ABC$ are $x - y + 5 = 0$ and $x + 2y = 0$ respectively. If the point $A$ is $(1,\; - \;2)$, then the equation of line $BC$ is

$23x + 14y - 40 = 0$

$14x - 23y + 40 = 0$

$23x - 14y + 40 = 0$

$14x + 23y - 40 = 0$

(IIT-1986)

Solution

(d) Let the equation of perpendicular bisector $FN$ of $AB$ is $x – y + 5 = 0$ ……$(i)$

The middle point $F$ of $AB$ is $\left( {\frac{{{x_1} + 1}}{2},\frac{{{y_1} – 2}}{2}} \right)$ lies on line $(i)$. Therefore ${x_1} – {y_1} = – 13$…..$(ii)$

Also $AB$ is perpendicular to $FN$. So the product of their slopes is $-1$.

i.e. $\frac{{{y_1} + 2}}{{{x_1} – 1}} \times 1 = – 1$or ${x_1} + {y_1} = – 1$……$(iii)$

On solving $(ii)$ and $(iii)$, we get $B( – 7,6)$.

Similarly $C{\rm{ }}\left( {\frac{{11}}{5},\frac{2}{5}} \right)$.

Hence the equation of $BC$ is $14x + 23y – 40 = 0$.

Standard 11

Mathematics

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(JEE MAIN-2014)

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The area of the parallelogram formed by the lines $y = mx,\,y = mx + 1,\,y = nx$ and $y = nx + 1$ equals

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(IIT-2001)

The equation of perpendicular bisectors of the sides $AB$ and $AC$ of a triangle $ABC$ are $x - y + 5 = 0$ and $x + 2y = 0$ respectively. If the point $A$ is $(1,\; - \;2)$, then the equation of line $BC$ is

Solution

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