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ત્રિકોણ $ABC$ ની બાજુઓ $AB$ અને $AC$ ના લંબદ્વિભાજકના સમીકરણો અનુક્રમે $x - y + 5 = 0$ અને $x + 2y = 0$ છે.જો બિંદુ $A$ એ $(1,\; - \;2)$ આપેલ હોય તો રેખા $BC$ નું સમીકરણ મેળવો.
$23x + 14y - 40 = 0$
$14x - 23y + 40 = 0$
${\tan ^{ - 1}}(2)$
$14x + 23y - 40 = 0$
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 $14x + 23y – 40 = 0$.
