Length of Perpendicular from a Point to a Line

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\(\textbf{Art 7 : } \qquad\boxed{{\text{Length of perpendicular}}}\)


Suppose that we are given the equation of a line L and we are required to find the length of the perpendicular dropped from an arbitrary point  \(P\left( {{x_1},{y_1}} \right)\,\,{\text{on}}\,\,L.\)

Suppose that the equation of L is in normal form, i.e,\(L \equiv x\cos \alpha  + y\sin \alpha  = p.\)

Based on the discussion in the figure above, the equation of the line \({L^{\,{\mathbf{'}}}}\,\,{\text{is}}\,\,x\cos \alpha  + y\sin \alpha  - {p_1} = 0.\) Since L1 passes through P, the co-ordinates of P must satisfy the equation of L1. Thus,

\[{x_1}\cos \alpha  + {y_1}\sin \alpha  - {p_1} = 0\]

Thus, we get \({p_1}\) as \(\left( {{x_1}\cos \alpha  + {y_1}\sin \alpha } \right).\) The length of perpendicular PQ is now simply \(\left| {{p_1} - p} \right| = \left| {{x_1}\cos \alpha  + {y_1}\sin \alpha  - p} \right|\). (Modulus sign is used since PQ is a length so it must be positive).

\[\boxed{PQ = \left| {{x_1}\cos \alpha  + {y_1}\sin \alpha  - p} \right|}\,\,\,\,:\,\,\,\,{\mathbf{Length}}{\text{ }}{\mathbf{of}}{\text{ }}{\mathbf{perpendicular}}\]

Let us now assume the case where L is given in the general form, i.e.\(L \equiv ax + by + c = 0.\)

We can easily adjust the equation of L so that c is negative. We do this so that we can convert L into the normal form:

\[\begin{align}& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\qquad ax + by + c = 0\qquad c < {\text{ }}0 \\&\qquad \Rightarrow  \quad ax + by =  - c  \\&\qquad\Rightarrow  \quad \left( {\frac{a}{{\sqrt {{a^2} + {b^2}} }}} \right)x + \left( {\frac{b}{{\sqrt {{a^2} + {b^2}} }}} \right)y = \left( {\frac{{ - c}}{{\sqrt {{a^2} + {b^2}} }}} \right) \\&\qquad
   \Rightarrow  \quad x\cos \alpha  + y\sin \alpha  = p\qquad \qquad \qquad \qquad \qquad \qquad \qquad ...{\text{ }}\left( 1 \right)  \\&  {\text{where}}\;\;\;\cos \alpha  = \frac{a}{{\sqrt {{a^2} + {b^2}} }},\,\,\sin \alpha  = \frac{b}{{\sqrt {{a^2} + {b^2}} }}\,\,{\text{and}}\,\,p = \frac{{ - c}}{{\sqrt {{a^2} + {b^2}} }}  \end{align} \]

The equation in (1) is in the normal form; we can now use the result obtained in the preceding discussion to obtain the length of the perpendicular PQ:

\[\begin{align}& PQ = \left| {{x_1}\cos \alpha  + {y_1}\sin \alpha  - p} \right|\,\,\,\,\left\{ \begin{gathered}
  {\text{Modulus sign is}}  \\  {\text{used since }}PQ  \\  {\text{must be  + ve}}  \end{gathered}  \right\}  \\
& \qquad  = \left| {\frac{{a{x_1}}}{{\sqrt {{a^2} + {b^2}} }} + \frac{{b{y_1}}}{{\sqrt {{a^2} + {b^2}} }} + \frac{c}{{\sqrt {{a^2} + {b^2}} }}} \right|  \\& \boxed{PQ = \frac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}}\qquad :\qquad {\mathbf{Length}}{\text{ }}{\mathbf{of}}{\text{ }}{\mathbf{perpendicular}}  \end{align} \]

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Straight Lines
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Straight Lines
grade 11 | Answers Set 2
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Straight Lines
grade 11 | Questions Set 2
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