A) \[(-1,-4)\]
B) \[(-3\,,\,\,-8)\]
C) \[(1,-4)\]
D) \[(3,\,8)\]
Correct Answer: A
Solution :
Equation of the line passing through (3, 8) and perpendicular to \[x+3y-7=0\]is \[3x-y-1=0\]. The intersection point of both the lines is (1, 2). Now let the image of \[A(3,8)\]be \[{A}'({{x}_{1}},{{y}_{1}}),\]then point (1, 2) will be the midpoint of \[A{A}'\]. Þ \[\frac{{{x}_{1}}+3}{2}=1\Rightarrow {{x}_{1}}=-1\] and\[\frac{{{y}_{1}}+8}{2}=2\]Þ \[{{y}_{1}}=-4\]. Hence the image is (?1, ?4).
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