A) -3
B) -3/8
C) -3/2
D) -3/16
Correct Answer: D
Solution :
Given line \[3x+4y=12\] can be rewritten as \[\frac{3x}{12}+\frac{4y}{12}=1\Rightarrow \frac{x}{4}+\frac{y}{3}=1\] \[\Rightarrow \]\[x-\]intercept = 4 and \[y-\]intercept = 3 Let the required line be \[L:\frac{x}{a}+\frac{y}{b}=1\]where \[a=x-\]mtercept and \[b=y-\]intercept According to the question \[a=4\times 2=8\]and \[b=3/2\] \[\therefore \] Required line is \[\frac{x}{8}+\frac{2y}{3}=1\] \[\Rightarrow \] \[3x+16y=24\] \[\Rightarrow \] \[y=\frac{-3}{16}x+\frac{24}{16}\] Hence, required slope \[=\frac{-3}{16}.\]You need to login to perform this action.
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