MATH -127 (Coordinate Geometry)
Chapter-2
(Pair of Straight Lines)
General equation of second degree,
ax2+2hxy+by2+2hx+2fy+c=0
If we multiply two linear functions we get one non-linear function.
To remove the first-degree term from (1), we must shift the origin to
(α,β)
where,
α=bg−fhh2−ab=fh−bgab−h2β=af−ghh2−ab=gh−afab−h2
and to remove the term containing xy, we must rotate the axes through,
θ=12tan−12ha−b
Homogenous quadratic equation (where the total power of variables are same in every term):
ax2+2hxy+by2=0
we know, if m1 and m2 are two roots of an equation,
x2−(m1+m2)x+m1m2=0
Now, take b≠0 and divide throughout of (2) by bx2,
⟹ab+2hy6x+y2x2=0 is a quadratic equation in yx
we know, for every quadratic equation,
AX2+BX+C=0
Here,
A=1, B=2hb, C=ab, X=yx
If m1, m2 are the roots of (3), then we get,
m1+m2=−2hb and m1m2=ab
∴ From (2) we get,
ax2+2hxy+by2=(yx−m1)(yx−m2)=0⟹y−m1x=0 and y−m2x=0
∴ Equation (2) represents two straight lines that pass through the origin.
♦ Generalization:
an equation of the form,
aoyn+a1yn−1x1+a2yn−2x2+⋯+anxn=0
is a homogenous equation in xy of degree ′n′
Pn(x)=ao+a1x+⋯+anxn
if θ be the angle between
y−m1x=0 and y−m2x=0
then,
tanθ=m1−m21+m1m2=√(m1−m2)21+m1m2=√(m1+m2)2−4m1m21+m1m2=2√h2−aba+b∴ θ=tan−12√h2−aba+b
♦ Corollary:
1) if the lines
y−m1x=0 and y−m2x=0
are parallel:
θ=2√h2−aba+b=0∴ h2=ab
2)if the line represented by equation (2) are perpendicular:
tan90∘=2√h2−aba+b=10∴ a+b=0
•Under which condition equation (1) represents a pair of straight lines:
|ahghbfgfc|=0⟹abc+2fgh−af2−bg2−ch2=0
•angle between the lines represented by (1)
lx+my+n=0and l1x+m1y+n1=0
They will not pass through the origin. In case of homogenous, they will pass through the origin.
Question 1: Prove that the equation x2+6xy+9y2+12y−5=0 represents a pair of straight lines.
Solution: Given that
x2+6xy+9y2+12y−5=0
comparing equation (1.1) with the standard equation we get,
ax2+2hxy+by2+2gx+2fy+c=0
a=1, b=9, c=−5, h=3, g=2, f=6
the equation will represent a pair of straight lines if,abc+2fgh−af2−bg2−ch2=0So now, 1×9×(−5)+2×6.2×3−1×36−9×4+5×9=−45+72−36−36+45=0
∴ The condition is satisfied.
Question 2: Find the values of k, so that the equation 3x2+10xy+9y2−kx−26y+21=0 represents a pair of straight lines and also find the angle between them.
Solution: The given equation is,3x2+10xy+9y2−kx−26y+21=0
comparing equation (2.1) with the standard equation,
ax2+2hxy+by2+2gx+2fy+c=0we get,a=3, b=8, c=21, h=5, g=−k2, f=−13.
now, 3×8×21+2×(−13)×(−k2)×5−3×(−13)2−8(−k2)2−21×52=02k2−65k+528=0
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