MATH -127 (Coordinate Geometry)
Chapter-2
(Pair of Straight Lines)
Question 3: Prove that 6x2+7xy−5y2−22x41y−8=p represents a pair of straight lines. Find their point of intersection and angle between them.
Hint: α=fh−bgab−h=bg−fhh2−ab, β=gh−afab−h2=af−ghh2−ab
Question 4: If the equation
ax2+3xy−2y2−5x+5y+c=0
represents a pair of straight lines perpendicular to eachother. Find the value of a and c
Hint:
abc+2fgh−af2−bg2−ch2=0 and a+b=0, a=2, c=−3
Condition for coincidence of two straight lines:
1.f2−bc=0
2.g2−ac=0
3.h2−ab=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,
(x−α)2−(y−β)2a−b=(x−α)(y−β)h
If the equation is homogenous quadratic equation
ax2+2hxy+by2=0
The Bisectors are,
x2−y2a−b=xyh
Solution: Let
(α,β)
be the point of intersection between the lines represented by the given equation (1).
Then, α=bg−fhh2−ab=3 and β=gh−afab−h2=−2
∴ The equation of the bisectors are,
(x−3)2−(y+2)26+5=(x−3)−(y+3)72⟹7(x2−6x+9−y2−4y−4)=22(xy+2x−3y−6)⟹7x2−42x+63−7y2−28y−28−22xy−44x+132=0⟹7x2−7y2−22x+38y−86x+167=0
Lines joining the origin to the points of intersection of a curve and a line.
(lx+my+n)=0⟹l−nx+m−ny=1⟹ax2+2hxy+by2+2(gx+fy)+c.12=0⟹ax2+2hxy+by+2(gx+fy)((l−nx+m−ny))+c(l−nx+m−ny)
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