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  • Find the equation to the straight line which passes through the intersection of the straight lines <img alt="\mathrm{2x-3y+4=0,3x+4y-5=0,}" src="https://entrancecorner.oncodecogs.com/gif.latex

Find the equation to the straight line which passes through the intersection of the straight lines \mathrm{2x-3y+4=0,3x+4y-5=0,} and is perpendicular to the straight line \mathrm{6x-7y+8=0.}

Option: 1

\mathrm{79 x+51 y=125}


Option: 2

\mathrm{119 x+102 y=125}


Option: 3

\mathrm{119 x+51 y=25}


Option: 4

\mathrm{79 x+102 y=25}


Answers (1)

best_answer

Solving the equations \mathrm{\left ( 1 \right ),} the coordinates \mathrm{x_{1},y_{1}} of their common point are given by 
\mathrm{\frac{x_1}{(-3)(-5)-4 \times 4}=\frac{y_1}{4 \times 3-2 \times(-5)}=\frac{1}{2 \times 4-3 \times(-3)}=\frac{1}{17},}
so that \mathrm{x_1=-1 / 17 \text { and } y_1=22 / 17}
The equation of any straight line through this common point is therefore \mathrm{y-22 / 17=m(x+1 / 17)}
This straight line is, by Art. \mathrm{69}, perpendicular to \mathrm{\left ( 2 \right )} if \mathrm{\mathrm{m} \times 6 / 7=-1 \text {, i.e. if } \mathrm{m}=-7 / 6}
The required equation is therefore \mathrm{y-\frac{22}{17}=\frac{-7}{6}\left(x+\frac{1}{17}\right), \text { i.e. } 119 x+102 y=125}
Alter: Any straight line through the intersection of the straight lines (1) is
\mathrm{\begin{aligned} & 2 x-3 y+4+\lambda(3 x+4 y-5)=0 \\ & \text { i.e. }(2+3 \lambda) x+y(4 \lambda-3)+4-5 \lambda=0 \ldots \ldots \ldots \left ( 3 \right ) \end{aligned}}
This straight line is perpendicular to \mathrm{ (2),} if \mathrm{ 6(2+3 \lambda)-7(4 \lambda-3)=0}
\mathrm{ \text { i.e. if } \lambda=33 / 10}
The equation \mathrm{\left ( 3 \right )} is therefore \mathrm{x(2+99 / 10)+y(132 / 10-3)+4-165 / 10=0}
\mathrm{\text { i.e. } 119 \mathrm{x}+102 \mathrm{y}-125=0 \text {. }}

 

Posted by

Ritika Kankaria

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