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Pair of Straight Lines The equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0. Represents a second degree equation where a, h, b doesn’t variables simultaneously. Let a ≠ 0. Now, the above equation becomes a2 x2 + 2ax (hy + g) = aby2 – 2afy – ac on completing the square on the left side, we get, a2 x2 + 2ax (hy + g) = y2 (h2 – ab) + 2y (gh – af) + g2 – ac. i.e. (ax + hy + g) = + √y2(h2–ab)+2y(gh–af)g2–ac We cannot obtain x in terms of y, involving only terms of the first degree, unless the quantity under the radical sign be a perfect square. The condition for this is, (gh – af)2 = (h2 – ab) (g2 – ac) i.e. g2h2 – 2afgh + a2f2 = g2h2 – abg2 – abg2 – ach2 + a2bc cancelling and diving by a, we have the required condition abc + 2fgh – af2 – af2 – bg2 – ch2 = 0 Illustration: What is the point of intersection of two straight lines given by general equation ax2+ 2hxy + by2 + 2gx + 2fy + c = 0? Solution: The general solution is ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 …… (1) Let (α, ß) be the point of intersection we consider line paralleled transformation. x = x’ + α, y = y’ + ß From (1) we have a(x’ + α)2 + 2h(x’ + α) (y’ + ß) + b(y’ + ß)2 + 2g(x’ + α) + 2f(y’ + ß) + c = 0 ⇒ ax’2 + 2hx’y’ + by’2 + a α2 + 2hαß + bß2 + 2gα + 2fß + 2x’(a α + hß + g) + 2y’ + 2y’ (hα + bß + f) = 0 ⇒ ax’2 + 2hx’y’ + by’2 + 2x’(aα + hß + g) + 2g’ + 2y’ (hα + bß + f) = 0 Which must be in the form ax'2 + 2hx’y’ + by’ = 0 This cannot be possible unless aα + hß + g = 0 hα + bß + f = 0 Solving α/hf–bg = ß/hg–af = 1/ab–h2 α = hf–bg/ab–b2, ß = hg–af/ab–h2 Illustration: Represent lines y = 2x and y = 3x by a homogeneous equation of second degree Solution: (y – 2x) (y – 3x) = 0 Or 6x2 – 5xy + y2 = 0 Illustration: Represent lines parallel to y = 2x and y = 3x by a second degree equation Solution: (y – 2x – c1) (y – 3x – c2) (where c1 and c2 are constants) = 6x2 – 5xy + y2 + (3c1 + 2c2) x + (– c1 – c2) y + c1 c2 = 0 Note: 1. Homogeneous part is same as for the equation of above illustration. Therefore, the homogeneous part of a general second degree equation determines the slope of the lines i.e. lines parallel to ax2 + 2hxy + by2 + c = 0 and through the origin are represented by the equation ax2 + 2hxy + by2 = 0 2. The equation ax2 + 2hxy + by2 + 2fy + c = 0 represents a pair of parallel straight lines if h/a = b/h = f/g or bg2 = af2 The distance between them is given by 2√g2–ac/a(a+b) or √f2–bc/b(a+b) Illustration: Does the second degree equation x2 + 3xy + 2y2 – x – 4y – 6 = 0 represents a pair of lines. If yes, find their point of intersection. Solution: We observe that a = 1, h = 3/2, b = 2, g = –1/2, f = 2, c = – 6 abc + 2fgh – af2 – bg2 – ch2 = – 12 + 3 – 4 – 1/2 + 27/2 = 0 Therefore the given second-degree equation represents a pair of lines, x2 + 3xy + 2y2 – x – 4y – 6 = (x + 2y + 2) (x + y – 3). Consider the equations formed by first two rows of . i.e. ax + hy + g = 0 and hx + by + f = 0 i.e. x + 3/2 y – 1/2 and 3/2 x + 2y – 2 = 0 Solving these, we get the required point of intersection. i.e. 2x + 3y – 1 = 0 3x – 4y – 4 = 0 Solving the above equation, we get x = 8, y = –5. Note: (2x + 3y – 1)(3x + 4y – 4) ≠ x2 + 3xy + 2y2 – x – 4y – 6. To read more, Buy study materials of Straight Lines comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
The equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0. Represents a second degree equation where a, h, b doesn’t variables simultaneously.
Let a ≠ 0.
Now, the above equation becomes
a2 x2 + 2ax (hy + g) = aby2 – 2afy – ac
on completing the square on the left side, we get,
a2 x2 + 2ax (hy + g) = y2 (h2 – ab) + 2y (gh – af) + g2 – ac.
i.e. (ax + hy + g) = + √y2(h2–ab)+2y(gh–af)g2–ac
We cannot obtain x in terms of y, involving only terms of the first degree, unless the quantity under the radical sign be a perfect square. The condition for this is,
(gh – af)2 = (h2 – ab) (g2 – ac)
i.e. g2h2 – 2afgh + a2f2 = g2h2 – abg2 – abg2 – ach2 + a2bc
cancelling and diving by a, we have the required condition
abc + 2fgh – af2 – af2 – bg2 – ch2 = 0
What is the point of intersection of two straight lines given by general equation ax2+ 2hxy + by2 + 2gx + 2fy + c = 0?
The general solution is
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 …… (1)
Let (α, ß) be the point of intersection we consider line paralleled transformation.
x = x’ + α, y = y’ + ß
From (1) we have
a(x’ + α)2 + 2h(x’ + α) (y’ + ß) + b(y’ + ß)2 + 2g(x’ + α) + 2f(y’ + ß) + c = 0
⇒ ax’2 + 2hx’y’ + by’2 + a α2 + 2hαß + bß2 + 2gα + 2fß + 2x’(a α + hß + g) + 2y’ + 2y’ (hα + bß + f) = 0
⇒ ax’2 + 2hx’y’ + by’2 + 2x’(aα + hß + g) + 2g’ + 2y’ (hα + bß + f) = 0
Which must be in the form
ax'2 + 2hx’y’ + by’ = 0
This cannot be possible unless
aα + hß + g = 0
hα + bß + f = 0
Solving
α/hf–bg = ß/hg–af = 1/ab–h2
α = hf–bg/ab–b2, ß = hg–af/ab–h2
Represent lines y = 2x and y = 3x by a homogeneous equation of second degree
(y – 2x) (y – 3x) = 0
Or 6x2 – 5xy + y2 = 0
Illustration:
Represent lines parallel to y = 2x and y = 3x by a second degree equation
Solution:
(y – 2x – c1) (y – 3x – c2) (where c1 and c2 are constants)
= 6x2 – 5xy + y2 + (3c1 + 2c2) x + (– c1 – c2) y + c1 c2 = 0
1. Homogeneous part is same as for the equation of above illustration. Therefore, the homogeneous part of a general second degree equation determines the slope of the lines i.e. lines parallel to ax2 + 2hxy + by2 + c = 0 and through the origin are represented by the equation ax2 + 2hxy + by2 = 0
2. The equation ax2 + 2hxy + by2 + 2fy + c = 0 represents a pair of parallel straight lines if h/a = b/h = f/g or bg2 = af2
The distance between them is given by 2√g2–ac/a(a+b) or √f2–bc/b(a+b)
Does the second degree equation x2 + 3xy + 2y2 – x – 4y – 6 = 0 represents a pair of lines. If yes, find their point of intersection.
We observe that
a = 1, h = 3/2, b = 2, g = –1/2, f = 2, c = – 6
abc + 2fgh – af2 – bg2 – ch2 = – 12 + 3 – 4 – 1/2 + 27/2 = 0
Therefore the given second-degree equation represents a pair of lines, x2 + 3xy + 2y2 – x – 4y – 6 = (x + 2y + 2) (x + y – 3).
Consider the equations formed by first two rows of .
i.e. ax + hy + g = 0 and hx + by + f = 0
i.e. x + 3/2 y – 1/2 and 3/2 x + 2y – 2 = 0
Solving these, we get the required point of intersection.
i.e. 2x + 3y – 1 = 0
3x – 4y – 4 = 0
Solving the above equation, we get x = 8, y = –5.
(2x + 3y – 1)(3x + 4y – 4) ≠ x2 + 3xy + 2y2 – x – 4y – 6.
To read more, Buy study materials of Straight Lines comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
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