A) \[\frac{x-1}{5}=\frac{y-4}{7}=\frac{z-0}{1}\]
B) \[\frac{x+1}{5}=\frac{y+4}{7}=\frac{z-0}{1}\]
C) \[\frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}\]
D) \[\frac{x-1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}\]
Correct Answer: C
Solution :
Let a, b, c be the d.r.'s of required line \[\therefore \] \[3a+2b+c=0\] and \[a+b-2c=0\] \[\frac{a}{-4-1}=\frac{b}{1+6}=\frac{c}{3-2}\] or \[\frac{a}{-5}=\frac{b}{7}=\frac{c}{1}\] In order to find a point on the required line we put \[z=0\] in the two given equation to obtain, \[3x+2y=5\] and \[x+y=3\]. Solving these two equations, we obtain \[x=-1,\,y=4\]. \[\therefore \] Co-ordinates of point on required line are \[(-1,\,4,\,0)\]. Hence required line is\[\frac{x+1}{-5}=\frac{y-4}{7}=\frac{z-0}{1}\].
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