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CBSE Class 12 Math 2020 Outside Delhi Set 3 Solved Paper

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Question : 11 of 11
Marks: +1, -0
Find the image of the point (−1,3,4)(-1,3,4) in the plane x−2y=0.x-2y=0.
Solution:     👈: Video Solution
We know that the image (x,y,z)(x, y, z) of point(x1,y1,z1)\text{point} (x_1, y_1, z_1)
In a plane ax+by+cz+d=0a x+b y+c z+d=0 is
  x−x1a=  y−y1b=  z−z1c=  −2(ax1+by1+cz1+d)a2+b2+c2\;\frac{x-x_1}{a}=\;\frac{y-y_1}{b}=\;\frac{z-z_1}{c}=\;\frac{-2(a x_1+b y_1+c z_1+d)}{a^2+b^2+c^2}
Thus, the image of point (−1,3,4)(-1,3,4) in a plane x−2y=0x-2y=0 is
  x+11=  y−3−2=  z−40=  −2[1(−1)+(−2)3+0(4)]1+0+4\;\frac{x+1}{1}=\;\frac{y-3}{-2}=\;\frac{z-4}{0}=\;\frac{-2[1(-1)+(-2)3+0(4)]}{1+0+4}
  x+11=  y−3−2=  z−40=  −2(−7)5\;\frac{x+1}{1}=\;\frac{y-3}{-2}=\;\frac{z-4}{0}=\;\frac{-2(-7)}{5}
x=  145−1=  95;y=  −285+3=  −135&z=4x=\;\frac{14}{5}-1=\;\frac{9}{5} ; y=\;\frac{-28}{5}+3=\;\frac{-13}{5} \& z=4
∴\therefore Image of (−1,3,4)(-1,3,4) is (  95,  135,4)\left(\;\frac{9}{5}, \;\frac{13}{5}, 4\right)
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