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  • The straight lines <img alt="\mathrm{3 x+4 y=5 \: and \: 4 x-3 y=15}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B3%20x&plus;4%20y%3D5%20%5C%3A%20and%20%5C%3A%204%20x-3%20y

The straight lines \mathrm{3 x+4 y=5 \: and \: 4 x-3 y=15} intersect at the point A. On these lines the points \mathrm{\mathrm{B}\: and\: \mathrm{C}} are chosen so that \mathrm{\mathrm{AB}=\mathrm{AC}}. Find the possible equation of the line \mathrm{BC} passing through (1,2)
 

Option: 1

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


Option: 2

\mathrm{7x+y-9=0}


Option: 3

\mathrm{7x+3y-2=0}


Option: 4

\mathrm{x-7y+9=0}


Answers (1)

best_answer

The two given straight lines are at right angles.

Since \mathrm{A B=A C}, the triangle is an isosceles right angled triangle.

The required equation is of the form \mathrm{y-2=m(x-1)} (1)

\mathrm{\text { with } \tan 45^{\circ}= \pm \frac{m+3 / 4}{1-3 m / 4}= \pm \frac{m-4 / 3}{1+4 m / 3} }

\mathrm{\Rightarrow 1= \pm \frac{m+3 / 4}{1-3 m / 4} \text { and } 1= \pm \frac{m-4 / 3}{1+4 m / 3} \Rightarrow m=-7,1 / 7 }

Substitute the value of \mathrm{\mathrm{m}} in (1). We get the required equations.

Hence option 2 is correct.

Posted by

shivangi.bhatnagar

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