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  {\text{Find the equation of the line passing through the point }}\left( { - 4, - 3} \right){\text{ and perpendicular to the straight }} \\
  {\text{line joining }}\left( {1,3} \right){\text{ and }}\left( {2,7} \right). \\
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  {\text{Now, we have to find the equation of line passing through point }}P\left( { - 4, - 3} \right){\text{ and perpendicular}} \\
  {\text{to the line joining the given points}}{\text{.}} \\
  {\text{First we find slope of line formed after joining two points Q}}\left( {1,3} \right){\text{ and }}R\left( {2,7} \right){\text{.}} \\
  {\text{Let slope of line joining points Q}}\left( {1,3} \right){\text{ and }}R\left( {2,7} \right){\text{ be }}{m_1}{\text{.}} \\
  {\text{Them }}{m_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \dfrac{{7 - 3}}{{2 - 1}} = 4 \\
  {\text{Let the slope of required line be }}{m_2} \\
  {\text{As, it is give that the required line is perpendicular to the line joining points Q}}\left( {1,3} \right){\text{ and }}R\left( {2,7} \right). \\
  {\text{So, }}{m_1}*{m_2} = - 1 \\
  {\text{Putting value of }}{m_1}{\text{, we get }}{m_2} = \dfrac{{ - 1}}{4} \\
  {\text{Now, finding the equation of line passing through point P and having slope }}{m_2}{\text{ using point slope form,}} \\
  \left( {y - \left( { - 3} \right)} \right) = \dfrac{{ - 1}}{4}\left( {x - \left( { - 4} \right)} \right){\text{ }}\left( 1 \right) \\
  {\text{Cross - multiplying equation 1 we get,}} \\
  4y + 12 = - x - 4 \\
  {\text{Solving above equation we get }}4y + x + 16 = 0 \\
  {\text{Hence, equation of line passing through point P and perpendicular to line joining points Q and R}} \\
  {\text{is }}4y + x + 16 = 0 \\
  {\text{NOTE: - Whenever you come up with a type of problem the first find the slope of lines}}{\text{. As for}} \\
  {\text{perpendicular lines }}{m_1}*{m_2} = - 1{\text{ and for parallel line }}{m_1} = {m_2}{\text{. Then use point - slope form to }} \\
  {\text{find equation of line}}{\text{.}} \\
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