A) one
B) two
C) infinite
D) zero
Correct Answer: D
Solution :
Let the slope be m of the straight line which passes through \[(-2,2),\] then equation \[(y-2)=m(x+2)\] \[mx-y+(2m+2)=0\] ?(i) \[\because \] Perpendicular distance of line (i) from point (given)\[(3,-1)=6\] \[\Rightarrow \] \[\frac{|m(3)-(-1)+(12m+2)|}{\sqrt{{{m}^{2}}+{{(-1)}^{2}}}}=6\] \[|3m+1+2m+2|=6\sqrt{{{m}^{2}}+1}\] Squaring on both sides, \[\Rightarrow \] \[{{(5m+3)}^{2}}=36({{m}^{2}}+1)\] \[\Rightarrow \] \[25{{m}^{2}}+9+30m=36{{m}^{2}}+36\] \[\Rightarrow \] \[11{{m}^{2}}-30m+27=0\] ?..(ii) Now, \[\Delta ={{B}^{2}}-4AC=900-4(11)\,(27)<0\] \[\because \] Discriminate of Eq. (ii) is negative. i.e., slope of the given line is imaginary. So, no line drawn through the point \[(\_2,2)\].
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