Question

Find the coordinate of the point on $ y- $ axis which is nearest point $ \left( -2,5 \right) $ .\[\]

Answer 1

Hint: We use the fact that the nearest point of any point $ P\left( {{x}_{1}},{{y}_{1}} \right) $ from a line $ L:ax+by+c=0 $ is the foot of the perpendicular say $ Q\left( {{x}_{2}},{{y}_{2}} \right) $ drawn from $ P $ on the line $ L $ . We find the co-ordinates of the foot $ Q\left( {{x}_{2}},{{y}_{2}} \right) $ from the relation $ \dfrac{{{x}_{2}}-{{x}_{1}}}{a}=\dfrac{{{y}_{2}}-{{y}_{1}}}{b}=\dfrac{-\left( a{{x}_{1}}+b{{y}_{1}}+c \right)}{{{a}^{2}}+{{b}^{2}}} $ . \[\]

Complete step by step answer:
We know that the nearest point of any point $ P\left( {{x}_{1}},{{y}_{1}} \right) $ from a line $ L:ax+by+c=0 $ is the foot of the perpendicular drawn from $ P $ on the line $ L $ . The co-ordinates of the foot say $ Q\left( {{x}_{2}},{{y}_{2}} \right) $ can be obtained from the following relation
\[\dfrac{{{x}_{2}}-{{x}_{1}}}{a}=\dfrac{{{y}_{2}}-{{y}_{1}}}{b}=\dfrac{-\left( a{{x}_{1}}+b{{y}_{1}}+c \right)}{{{a}^{2}}+{{b}^{2}}}\]

We denoted the given point as $ P\left( -2,5 \right) $ . We drop a perpendicular from the point $ P $ on $ y- $ axis and we denote the foot of the perpendicular as $ Q\left( {{x}_{2}},{{y}_{2}} \right) $ \[\]

We know that equation of the $ y- $ axis is $ x=0 $ as all the $ x- $ coordinates of all the points in $ y- $ axis are zero. We denote $ L:x=0 $ which we compare with the line $ L:ax+by+c=0 $ and get $ a=1,b=0,c=0 $ . We are given in the question $ P\left( {{x}_{1}},{{y}_{1}} \right)=P\left( -2,5 \right) $ . Let use the formula for the foot of the perpendicular and get $ Q\left( {{x}_{2}},{{y}_{2}} \right) $ . We have
 $ \Rightarrow $ \[\dfrac{{{x}_{2}}-\left( -2 \right)}{1}=\dfrac{{{y}_{2}}-5}{0}=\dfrac{-\left( 1\times \left( -2 \right)+0\times 5+0 \right)}{{{0}^{2}}+{{1}^{2}}}=2\]
We first find the $ x- $ co-ordinate of $ Q $
\[\begin{align}
  & \dfrac{{{x}_{2}}-\left( -2 \right)}{1}=2 \\
 & \Rightarrow {{x}_{2}}+2=2 \\
 & \Rightarrow {{x}_{2}}=0 \\
\end{align}\]
We now find the $ y- $ co-ordinate of $ Q $
\[\begin{align}
  & \dfrac{{{y}_{2}}-5}{0}=2 \\
 & \Rightarrow {{y}_{2}}-5=0 \\
 & \Rightarrow {{y}_{2}}=5 \\
\end{align}\]
 $ \Rightarrow $ So the co-ordinates of $ Q $ are given by $ Q\left( {{x}_{2}},{{y}_{2}} \right)=Q\left( 0,5 \right) $ \[\]
 $ \Rightarrow $ We can alternatively solve by finding with the knowledge that the modulus of the abscissa of the co-ordinates of $ P\left( -2,5 \right) $ is the perpendicular distance from $ y- $ axis and modulus of the ordinate is the perpendicular distance from $ x- $ axis . We have PR as the perpendicular on $ x- $ axis. So we have $ PR=\left| 5 \right|=5 $ . In the rectangle PROQ $ PR=OQ $ . The point Q lies on $ y- $ axis so its abscissa is 0 and now we have obtained its ordinate $ OQ=5 $ . So the co-ordinates of $ Q $ are given by $ Q\left( 0,5 \right) $ .\[\]

Note:
 We note that the distance is always a positive quantity and we have to reject negative values. If we are given the points $ P\left( {{x}_{1}},{{y}_{1}} \right),Q\left( {{x}_{2}},{{y}_{2}} \right) $ where PQ is perpendicular on the line L we can find the equation of line L by first by finding the equation of PQ with two point $ y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right) $ form then slope of line L as $ {{m}_{2}}=\dfrac{-1}{{{m}_{1}}} $ where $ {{m}_{1}} $ is the slope of PQ and finally equation of L by slope- point form $ y-{{y}_{2}}={{m}_{2}}\left( x-{{x}_{2}} \right) $ .