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

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

The two given straight lines are at right angles.
Since \mathrm{AB}=\mathrm{AC}, the triangle is an isosceles right angled triangle.
The required equation is of the form \mathrm{y-2=m(x-1) \quad \quad \quad \dots (1)}
\begin{aligned} & \text { with }\mathrm{ \tan 45^{\circ}= \pm \frac{m+3 / 4}{I-3 m / 4}= \pm \frac{m-4 / 3}{I+4 m / 3}} \\ & \Rightarrow \mathrm{1= \pm \frac{m+3 / 4}{1-3 m / 4} \text { and } I= \pm \frac{m-4 / 3}{1+4 m / 3} \Rightarrow m=-7,\frac{1}{7} }\end{aligned}
Substitute the value of m in (1). We get the required equations.

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Ramraj Saini

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