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If the straight lines ax + by + p = 0 and x cosa + y sina = p are inclined at an angle p/4 and concurrent with the straight line x sina – y cosa = 0, then the value of a2 +b2 is
- a)0
- b)1
- c)2
- d)None of these .
Correct answer is option 'C'. Can you explain this answer?
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If the straight lines ax + by + p = 0 and x cosa+ y sina= p are inclin...
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If the straight lines ax + by + p = 0 and x cosa+ y sina= p are inclin...
ax+by+p=0 ...(1)
⇒by=−ax+p
⇒y=−a/bx+p/b
So, Slope,m1 = −a/b
and xcosθ+ysinθ−p=0 ..(2)
⇒ysinθ=−xcosθ+p
⇒y=−cosθ/sinθx+p/sinθ
So, Slope,m2=−cosθ/sinθ
Now, straight lines (1) and (2) include an angle π/4
i,e tanπ/4 = ∣∣(m1−m2)/(1+m1m2)∣∣
⇒ 1 = ∣∣(m1−m2)/(1+m1m2)∣∣
⇒1+m1m2|=|m1−m2|
⇒∣∣1+(−a/b×(−cosθ/sinθ))∣∣
=∣∣−a/b−(−cosθ/sinθ)∣∣
⇒∣∣(bsinθ+acosθ)/bsinθ∣∣
=∣∣(−asinθ+bcosθ)/bsinθ∣∣
⇒|bsinθ+acosθ|=|−asinθ+bcosθ|
Squaring both sides, we get
b2sin2θ+a2cos2θ+2absinθ cosθ=a2sin2θ+b2cos2θ−2absinθ cosθ ...(3)
Given three lines are concurrent
Now, we will find the point of intersection of
xcosθ+ysinθ=p and xsinθ−ycosθ=0
which is x=pcosθ and y=psinθNow, point of intersection will satisfy (1)
i.e apcosθ+bpsinθ+p=0
⇒apcosθ+bpsinθ=−p
⇒acosθ+bsinθ=−1
Squaring both sides, we get
⇒a2cos2θ+b2sin2θ+2absinθ cosθ=1 ...(4)
Put the value of (4) in (3), we get
a2sin2θ+b2cos2θ−2absinθ cosθ =1 ....(5)
Adding (4) and (5), we get
a2cos2θ+b2sin2θ+2absinθ cosθ+a2sin2θ+b2cos2θ−2absinθ cosθ=1+1
⇒a2(sin2θ+cos2θ)+b2(sin2θ+cos2θ)=2
⇒a2+b2=2
⇒by=−ax+p
⇒y=−a/bx+p/b
So, Slope,m1 = −a/b
and xcosθ+ysinθ−p=0 ..(2)
⇒ysinθ=−xcosθ+p
⇒y=−cosθ/sinθx+p/sinθ
So, Slope,m2=−cosθ/sinθ
Now, straight lines (1) and (2) include an angle π/4
i,e tanπ/4 = ∣∣(m1−m2)/(1+m1m2)∣∣
⇒ 1 = ∣∣(m1−m2)/(1+m1m2)∣∣
⇒1+m1m2|=|m1−m2|
⇒∣∣1+(−a/b×(−cosθ/sinθ))∣∣
=∣∣−a/b−(−cosθ/sinθ)∣∣
⇒∣∣(bsinθ+acosθ)/bsinθ∣∣
=∣∣(−asinθ+bcosθ)/bsinθ∣∣
⇒|bsinθ+acosθ|=|−asinθ+bcosθ|
Squaring both sides, we get
b2sin2θ+a2cos2θ+2absinθ cosθ=a2sin2θ+b2cos2θ−2absinθ cosθ ...(3)
Given three lines are concurrent
Now, we will find the point of intersection of
xcosθ+ysinθ=p and xsinθ−ycosθ=0
which is x=pcosθ and y=psinθNow, point of intersection will satisfy (1)
i.e apcosθ+bpsinθ+p=0
⇒apcosθ+bpsinθ=−p
⇒acosθ+bsinθ=−1
Squaring both sides, we get
⇒a2cos2θ+b2sin2θ+2absinθ cosθ=1 ...(4)
Put the value of (4) in (3), we get
a2sin2θ+b2cos2θ−2absinθ cosθ =1 ....(5)
Adding (4) and (5), we get
a2cos2θ+b2sin2θ+2absinθ cosθ+a2sin2θ+b2cos2θ−2absinθ cosθ=1+1
⇒a2(sin2θ+cos2θ)+b2(sin2θ+cos2θ)=2
⇒a2+b2=2
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If the straight lines ax + by + p = 0 and x cosa+ y sina= p are inclined at an anglep/4 and concurrent with the straight line x sina– y cosa= 0, then the value of a2+b2isa)0b)1c)2d)None of these .Correct answer is option 'C'. Can you explain this answer?
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