The line 4x−3y+2=0 is rotated through an angle of π4 in clockwise direction about the point (1,2). The equation of the line in its new position is
A
x−7y+13=0
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B
y−7x+5=0
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C
x+7y−15=0
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D
y+7x−15=0
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Solution
The correct option is Ax−7y+13=0
(1,2) lies on the line 4x−3y+2=0 The slope of the given line is tanθ=43 Let the slope of the new line be m. Angle between these two lines is 45∘ ⇒tan45∘=∣∣
∣
∣
∣∣43−m1+43m∣∣
∣
∣
∣∣ ⇒4−3m3+4m=±1 ⇒4−3m=±(3+4m) ⇒m=17(m=−7rejected)
Required equation of line is y−2=17(x−1) ⇒x−7y+13=0