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Lecture 4 [24 Feb, 2020] MATH 127 (Prof Farid)

 MATH -127 (Coordinate Geometry)
Chapter-2 
(Pair of Straight Lines)



Question 3: Prove that  6x2+7xy5y222x41y8=p  represents a pair of straight lines. Find their point of intersection and angle between them.

Hint:  α=fhbgabh=bgfhh2ab,     β=ghafabh2=afghh2ab

Question 4: If the equation  ax2+3xy2y25x+5y+c=0  represents a pair of straight lines perpendicular to eachother. Find the value of a and c
Hint:  abc+2fghaf2bg2ch2=0  and  a+b=0, a=2, c=3



Condition for coincidence of two straight lines:
1.f2bc=0
2.g2ac=0
3.h2ab=0
4.gh=af
5.hf=bg
6.fg=ch

Bisectors of the angles between two straight lines:
  ax2+2hxy+by2+wgx+2fy+c=0     | (α,β) is the point of intersection  
Let (α, β) be the new origin,
(xα)2(yβ)2ab=(xα)(yβ)h
If the equation is homogenous quadratic equation  ax2+2hxy+by2=0  The Bisectors are,
x2y2ab=xyh


Question 5: Find the equation of the bisectors of the angles between the lines represented by 6x2+7xy5y222x41y8=0
Solution: Let (α,β) be the point of intersection between the lines represented by the given equation (1).
Then, α=bgfhh2ab=3   and   β=ghafabh2=2
The equation of the bisectors are,
(x3)2(y+2)26+5=(x3)(y+3)727(x26x+9y24y4)=22(xy+2x3y6)7x242x+637y228y2822xy44x+132=07x27y222x+38y86x+167=0
Lines joining the origin to the points of intersection of a curve and a line.
(lx+my+n)=0lnx+mny=1ax2+2hxy+by2+2(gx+fy)+c.12=0ax2+2hxy+by+2(gx+fy)((lnx+mny))+c(lnx+mny)



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