Coordinate Geometry
Pair of Straight Lines
1
Coordinate Geometry
Pair of Straight Lines
#I § 1. Homogenous quadratic equation in two variables. Let us consider the general equation: ax2 + 2hxy + by 2 = 0 (1) Multiplying throughout by a, we write in the following form
⇒
(a2 x2 + 2ahxy + h2 y 2 ) − (h2 − ab)y 2 = 0 n on o √ √ (ax + hy) + y h2 − ab (ax + hy) − y h2 − ab = 0
This equation is saisfied by the coordinates of all points which make the first of these brackets zero, and also by the coordinates of those points that make the second bracket zero, i.e. by all the points which satisfy the following two equations: ´ ³ ´ ³ p p and ax + y h − h2 − ab = 0 ax + y h + h2 − ab = 0 But the above two equations represent straight lines passing through the origin. Hence, we conclude that Eq. (1) represents a pair of straight lines passing through the origin of the xy-coordinate plane whose equations are p ax + hy + y h2 − ab = 0 (2) p and ax + hy − y h2 − ab = 0 (3) These two straight lines are real and different if h2 > ab, real and coincident if h2 = ab, and imaginary if h2 − ab. In the case, when h2 < ab, the straight lines, though themselves imaginary, intersect in a real point: the origin. #I § 2. An equation such as (1), which is such that in each of the term the sum of the exponent of x and y is the same, is called a homogenous equation. Thus, the equation given by Eq. (1) is a homogenous equation of second degree, since in each term the sum of the exponents add up to 2. Similarly, the expression 3x3 + 4x2 y − 5xy 2 + 9y 3 is a homogenous expression expression of third degree. However, the expression 3x3 + 4x2 y − 5xy 2 + 9y 3 − 7xy is not homogenous (why??). Hence, from § 1 it follows that a homogenous equation of the second degree represents two straight lines, real and distinct, coincident, or imaginary.
Anant Kumar
• study circle for iitjee & aieee •
Mob. No. 9932347531
Coordinate Geometry
Pair of Straight Lines
2
#I § 3. Angle between the pair of lines. The axes being rectangular, to find the angle between the straight lines given by Eq. (1) i.e. the equation ax2 + 2hxy + by 2 = 0 Let the separate equations of the two lines be y − m1 x = 0
and
y − m2 x = 0
(4)
so that Eq. (1) must be equivalent to b(y − m1 x)(y − m2 x) = 0
(5)
Equating the coefficients of xy and x2 in Eq. (1) and Eq. (5), we have −b(m1 + m2 ) = 2h,
and
bm1 m2 = a,
and
m1 m2 =
so that
2h , b If θ be the angle between the lines Eq. (4), then m1 + m2 = −
⇒
m1 − m2 tan θ = = 1 + m1 m2 r 4h2 4a − b2 b = a 1+ b √ 2 h2 − ab tan θ = a+b
a b
p
(m1 + m2 )2 − 4m1 m2 1 + m1 m2
(6)
Thus, the required angle is found. #I § 4. From Eq. (6), the following two results follow: 1. If a + b = 0, then tan θ is undefined and hence θ = 90◦ ; the straight lines are therefore perpendicular. Hence, two straight lines, represented by one equation, are at right angles if the algebraic sum of the coefficients of x2 and y 2 be zero. For example, the equations x2 − y 2 = 0
and
6x2 + 11xy − 6y 2 = 0
both represent pair of straight lines at right angles. 2. If h2 = ab, the value of tan θ is zero and hence θ is zero. The angle between the straight line is zero and, since they both pass through the origin, they are therefore coincident. √ This may be directly seen from the origin. Indeed, if h2 = ab, i.e. h = ± ab, it may be written ³ √ √ √ ´2 ax2 ± 2 ab xy + by 2 = 0 ⇒ x a±y b =0 which is two coincident straight lines.
Anant Kumar
• study circle for iitjee & aieee •
Mob. No. 9932347531
Coordinate Geometry
Pair of Straight Lines
3
y B
L2 A θ2 θ
L1 x
O
M1
θ1 A0 M2 B0
Figure 1: Angle between a pair of straight lines. #I § 5. Angle bisectors. To find the equation of the pair of straight lines bisecting the angle between the pair of lines given by Eq. (1) i.e. the equation ax2 + 2hxy + by 2 = 0 Let the Eq. (1) represent the two straight lines L1 OM1 and L2 OM2 inclined at angles θ1 and θ2 to the x axis, so that Eq. (1) is equivalent to b(y − x tan θ1 )(y − x tan θ2 ) = 0 Hence,
2h and b Let OA and OB be the required bisectors. Since tan θ1 + tan θ2 = −
tan θ1 tan θ2 =
a b
∠AOL1 = ∠L2 OA, ∴
∠AOX − θ1 = θ2 − ∠AOX. ∴
2∠AOX = θ1 + θ2
Also, ∠BOX = 90◦ + ∠AOX. Therefore, 2∠BOX = 180◦ + θ1 + θ2 . Hence, if θ stand for either of the angles AOX or BOX, we have tan 2θ = tan(θ1 + θ2 ) =
tan θ1 + tan θ2 2h =− 1 − tan θ1 tan θ2 b−a
y But, if (x, y) be the coordinates of any point on either of the lines OA or OB, we have tan θ = . x Therefore, we have 2 tan θ 2h = tan 2θ = = − b−a 1 − tan2 θ
y x = 2xy x2 − y 2 y2 1− 2 x 2
Thus, the equation of the angle bisectors become x2 − y 2 xy = a−b h Anant Kumar
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(7) Mob. No. 9932347531
Coordinate Geometry
Pair of Straight Lines
4
general equation of the second degree #I § 6. The most general expression which contains terms involving x and y in a degree not exceeding two, must contain terms involving x2 , xy, y 2 , x, y, and a constant. The notation which is in general use for this expression is given by ax2 + 2hxy + by 2 + 2gx + 2f y + c
(8)
This expression is known as the general expression of the second degree, and when equated to zero is called the general equation of the second degree. #I § 7. In general, the general equation represents a curve locus. However, if a certain condition holds between the coefficients of its terms it will represent a pair of straight lines. This condition is derived next. #I § 8. To find the condition that the general equation of the second degree ax2 + 2hxy + by 2 + 2gx + 2f y + c = 0
(9)
may represent two straight lines. If we can break the left-hand side of Eq. (9) in to two factors, each of the first degree, then it will represent two straight lines. Writing Eq. (9) in descending powers of x and equating it to zero: ax2 + 2x(hy + g) + by 2 + 2f y + c = 0 Solving this quadratic in x, we have: x=
−(hy + g) ±
p
(hy + g)2 − a(by 2 + 2f y + c) a
p ax + hy + g = ± y 2 (h2 − ab) + 2y(hg − af ) + (g 2 − ac)
or
(10)
From Eq. (10) we cannot obtain x in terms of y, involving only terms of the first degree, unless the quantity under the radical sign must be a perfect square. The condition for this is (gh − af )2 = (h2 − ab)(g 2 − ac)
⇒
g 2 h2 − 2af gh + a2 f 2 = g 2 h2 − abg 2 − ach2 + a2 bc
Transposing and diving by a, we obtain abc + 2f gh = af 2 + bg 2 + ch2
(11)
which is the required condition. #I § 9. To prove that a homogenous equation of the n-th degree represents n straight lines, real or imaginary, which all pass through the origin. Let the equation be y n + a1 xy n−1 + a2 x2 y n−2 + a3 x3 y n−3 + . . . + an xn = 0 On division by xn , it may be written ³ y ´n x Anant Kumar
+ a1
³ y ´n−1 x
+ a2
³ y ´n−2 x
+ . . . + an = 0
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(12) Mob. No. 9932347531
Coordinate Geometry
Pair of Straight Lines
5
y , and hence must have n roots. Let these roots be m1 , x m2 , m3 , . . . , mn . Then Eq. (12) must be equivalent to the equation ³y ´³y ´³y ´ ³y ´ − m1 − m2 − m3 · · · − mn = 0 (13) x x x x This is an equation of the n-th degree in
The Eq. (13) is satisfied by all points which satisfy the separate equations y y y y − m1 = 0, − m2 = 0, − m3 = 0, . . . − mn = 0 x x x x i.e. by all points that lie on the straight lines y − m1 x = 0, y − m2 x = 0, y − m3 x = 0, . . . , y − mn x = 0, all of which pass through the origin. Conversely, the coordinates of all the points which satisfy any one of these n equations satisfy Eq. (9). #I § 10. To find the equation of the two straight lines joining the points in which the straight line lx + my = n
(14)
ax2 + 2hxy + by 2 + 2gx + 2f y + c = 0
(15)
meets the locus whose equation is
The equation (14) may be written
lx + my =1 (16) n The coordinates of the points in which the straight line meets the locus satisfy both equation (15) and (16) and hence satisfy the equation µ 2
2
ax + 2hxy + by + 2(gx + f y)
lx + my n
¶
µ +c
lx + my n
¶2 =0
(17)
Hence, Eq. (17) represents some locus which pass through the intersections of (15) and (16). But since the Eq. (17) is homogenous of second degree, it represents two straight lines passing through the origin. It must, therefore, represent the two straight lines joining the origin to the intersection of (15) and (16). #I § 11. We know that in general an equation of the second degree represents a curve locus, including as a particular case a pair of straight lines. However, in some cases it will represent only isolated points. For example consider the equation (x − y + c)2 + (x + y − c)2 = 0
(18)
We know that the sum of the squares of two quantities cannot be zero unless each of the squares is individually zero. The only real points that satisfy Eq. (18) therefore satisfy both the equations x−y+c=0
and x + y − c = 0.
But the only solution of these two equations is x=0
and y = c
The only real points represented by Eq. (18) is therefore (0, c). Anant Kumar
• study circle for iitjee & aieee •
Mob. No. 9932347531
