A) \[2x+5y\pm 16=0\]
B) \[4x+3y\pm 25=0\]
C) \[4x+3y\pm 5=0\]
D) \[2x+5y\pm 4=0\]
Correct Answer: B
Solution :
Let \[y=mx+c\]be the equation of straight line \[\Rightarrow \] \[mx-y+c=0\] ??(i) Now, by condition Perpendicular distance from origin to line (i) = 5 \[\Rightarrow \] \[\frac{|0-0+c|}{\sqrt{{{m}^{2}}+1}}=5\] \[\Rightarrow \] \[c=\pm 5\sqrt{{{m}^{2}}\pm 1}\] ?..(ii) Also, given that Slope of perpendicular to line (i) from the origin \[=\frac{3}{4}\] \[\Rightarrow \] \[-\frac{1}{m}=\frac{3}{4}\] \[\Rightarrow \] \[m=\frac{-4}{3}\] From Eq. (ii), \[c=\pm 5\sqrt{{{m}^{2}}+1}\]\[c=\pm 5\sqrt{1+\frac{16}{9}}=\pm 5\times \frac{5}{3}=\pm \frac{25}{3}\] On putting the values of m and c in Eq. (i), we get \[-\frac{4x}{3}-y\pm \frac{25}{3}=0\] \[\Rightarrow \] \[4x+3y\pm 25=0\] which is required equation of straight line.
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