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  •  In a triangle , if <img alt="\mathrm{\mathrm{A}(2,-1)}" src="https://entrancecorner.oncodeco

 In a triangle \mathrm{ABC}, if \mathrm{\mathrm{A}(2,-1)} and \mathrm{7 \mathrm{x}-10 \mathrm{y}+1=0} and \mathrm{3 \mathrm{x}-2 \mathrm{y}+5=0} are equations of an altitude and an angle bisector respectively drawn from \mathrm{\mathrm{B}}, then equation of \mathrm{\mathrm{BC}} is 

Option: 1

\mathrm{x+y+1=0}


Option: 2

\mathrm{5 x+y+17=0}


Option: 3

\mathrm{4 x+9 y+30=0}


Option: 4

\mathrm{x-5 y-7=0}


Answers (1)

best_answer

\mathrm{BD} and \mathrm{BE} are intresect at \mathrm{B}

Coordinates of \mathrm{B} are \mathrm{(-3,-2)}

\mathrm{\mathrm{m}_{\mathrm{AB}}=1 / 5 }
\mathrm{\tan \theta=\left|\frac{\frac{3}{2}-\frac{1}{5}}{1+\frac{3}{10}}\right|=\left|\frac{\frac{3}{2}-\mathrm{m}}{1+\frac{3 \mathrm{~m}}{2}}\right| }
\mathrm{\Rightarrow \quad 1=\left|\frac{3-2 \mathrm{~m}}{2+3 \mathrm{~m}}\right| \quad \text { or } \quad \pm 1=\frac{3-2 \mathrm{~m}}{2+3 \mathrm{~m}}}

\mathrm{\Rightarrow \quad \mathrm{m}=1 / 5 \text { (rejected) or }-5 \quad }
\mathrm{\text{equation of BC}= y+2= -5\left ( x+3 \right )\Rightarrow 5 \mathrm{x}+\mathrm{y}+17=0 }
Alternatively : Take image of \mathrm{(2,-1)}  in the line \mathrm{\mathrm{BD}}  to get a point on \mathrm{\mathrm{BC}}

Posted by

Irshad Anwar

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