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Family of Lines
What do you mean by a pair of Straight Lines?
General Equation of Second Degree representing a Pair of Straight Lines
What do you mean by the joint equation of a pair of Straight Lines?
What is family of lines?
Related Resources
Straight lines is an important topic of mathematics syllabus of IIT JEE. The topic is quite vast and requires thorough understanding of the concepts. When we talk of straight lines, other topics that come to our mind are pair of straight lines and family of lines. This section stresses on these topics in detail.
Before we proceed toward family of straight lines we shall first discuss pair of straight lines and the joint equation of pair of straight lines.
An equation of the type ax^{2} + 2hxy + by^{2} which is a homogeneous equation of degree 2 denotes a pair of straight lines passing through the origin. The lines may be real, coincident or imaginary depending on the conditions satisfied by them:
If h^{2} > ab, then the lines are real and distinct.
If h^{2} < ab, then the lines are imaginary with the point of intersection as (0, 0).
If h^{2} = ab, then the lines are coincident.
If α is the acute angle between the pair of straight lines, then tan α = |2√(h^{2}-ab)/(a + b)|.
These lines are perpendicular to each other if the coefficient of x^{2} + coefficient of y^{2} = 0.
If the coefficient of xy = 0, then the lines are equally inclined to the x-axis.
The lines are said to be coincident if h^{2} = ab.
The equation ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represents a second degree equation where a, h, b don’t vanish simultaneously.
Let a ≠ 0.
Now, the above equation becomes
a^{2}x^{2} + 2ax (hy + g) = aby^{2} – 2afy – ac
On completing the square on the left side, we get,
a^{2}x^{2} + 2ax (hy + g) = y^{2}(h^{2} – ab) + 2y(gh – af) + g^{2} – ac.
i.e. (ax + hy + g) = + √y^{2}(h^{2}–ab) + 2y(gh–af)g^{2} – 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} = (h^{2} – ab) (g^{2} – ac)
i.e. g^{2}h^{2} – 2afgh + a^{2}f^{2} = g^{2}h^{2} – abg^{2} – abg^{2} – ach^{2} + a^{2}bc
Cancelling and dividing by a, we have the required condition as
abc + 2fgh – af^{2} – bg^{2} – ch^{2} = 0
The angle α between two lines represented by the general equation is the same as the one represented by its homogeneous part.
What is the point of intersection of two straight lines given by general equation ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0?
The general solution is
ax^{2} + 2hxy + by^{2} + 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} + aα^{2} + 2hαß + bß^{2} + 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
This cannot be possible unless aα + hß + g = 0
hα + bß + f = 0
Solving α/hf–bg = ß/hg–af = 1/ab–h^{2}
α = hf–bg/ab–b^{2}, ß = hg–af/ab–h^{2}
Represent lines parallel to y = 2x and y = 3x by a second degree equation
The lines parallel to y = 2x and y = 3x can be represented as
(y – 2x – c_{1}) (y – 3x – c_{2}) (where c_{1} and c_{2} are constants)
This gives 6x^{2} – 5xy + y^{2} + (3c_{1} + 2c_{2}) x + (– c_{1} – c_{2}) y + c_{1} c_{2} = 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 ax^{2} + 2hxy + by^{2} + c = 0 and through the origin are represented by the equation ax^{2} + 2hxy + by^{2} = 0
2. The equation ax^{2 }+ 2hxy + by^{2} + 2fy + c = 0 represents a pair of parallel straight lines if h/a = b/h = f/g or bg^{2} = af^{2}
The distance between them is given by 2√g^{2}–ac/a(a+b) or √f^{2}–bc/b(a+b).
Does the second degree equation x^{2} + 3xy + 2y^{2} – x – 4y – 6 = 0 represent 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 – af^{2} – bg^{2} – ch^{2} = – 12 + 3 – 4 – 1/2 + 27/2 = 0
Therefore the given second-degree equation represents a pair of lines, x^{2} + 3xy + 2y^{2} – 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.
If the line lx + my + n = 0, (n ≠ 0) i.e. the line not passing through origin) cuts the curve ax^{2} + by^{2} + 2gx + 2fy + c = 0 at two points A and B, then the joint equation of straight lines passing through A and B and the origin is given by homogenizing the equation of the curve by the equation of the line. i.e. if we have a line lx + my + n = 0 and a second degree curve given by the equation ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0, then the joint equation of a pair of straight lines joining origin to the points of intersection of both is given by
ax^{2} + 2hxy + by^{2} + (2gx + 2fy) (lx+my/–n) + c (lx+my/–n)^{2} = 0. This is the equation of the lines OA and OB.
Prove that the straight lines joining the origin to the points of intersection of the straight line hx + ky = 2hk and the curve (x – k)^{2} + (y – y)^{2} = c^{2} are at right angles if h^{2} + k^{2} = c^{2}.
Making the equation of the curve homogeneous with the help of that of the line, we get
x^{2} + y^{2} –2(kx + hy) (hx+ky/2hk) + (h^{2} + k^{2} – c^{2}) (hx+ky/2hk) = 0
or 4h^{2}k^{2}x^{2} + 4h^{2}k^{2}y^{2} – 4h^{2}x(hx+ky) – 4h^{2}ky(hx + ky) + (h^{2} + k^{2} – c)(h^{2}x^{2} + k^{2}y^{2} + 2hxy) = 0.
This is the equation of the pair of lines joining the origin to the points of intersection of the given line and the curve. They will be at right angles if
Coefficient of x^{2} + coefficient of y^{2} = 0 i.e.
(h^{2} + k^{2}) (h^{2} + k^{2} – c^{2}) = 0 ⇒ h^{2} + k^{2} = c^{2} (since h^{2} + k^{2} ≠ 0).
Having studied pair of lines and the joint equation of pair of lines, we now proceed toward the family of lines. The general equation of the family of lines through the point of intersection of two given lines is L + λL’ = 0, where L = 0 and L’ = 0 are the two given lines, and λ is a parameter.
1. Conversely, a line of the form L_{1} + λL_{2} = 0 passes through the point which is the point of intersection of the lines L_{1} = 0 and L_{2} = 0.
2. This also implies that if a linear expression contains an undetermined coefficient then that represents a family of straight lines. Note that in this case, the line L= 0 cannot be a fixed line.
3. If the lines L_{1} and L_{2} are parallel lines then they meet at infinity.
4. If we have a line ax + by + c = 0, then the line perpendicular to this line is given by bx - ay + k = 0, where k is a parameter.
5. If we have a line ax + by + c = 0, then the line parallel to this line is given by ax + by + k = 0, where k is a parameter.
A variable line through the point of intersection of the lines x/a + y/b = 1 and x/b + y/a = 1 meets the coordinate axes in A and B. Show that the locus of the midpoint of AB is the curve 2xy(a + b) = ab(x + y).
Let (h, k) be the midpoint of the variable line AB.
The equation of the variable line AB is (bx + ay – ab) + λ(ax + by – ab) = 0
Coordinate of B are (0, ab(1+λ)b + λa).
Mid-point of AB is (ab(1 + λ)2(b + λa), ab(1 + λ)2(a + λb))
⇒ h = ab(1+λ)2(b+λa); k = ab(1+λ)2(a+λb)
⇒ 1/2h = b+λa/ab(1+λ); 1/2k = a+λb/ab(1+λ)
⇒ 1/2h + 1/2k = a+b/ab ⇒ (h + k)ab = 2hk (a + b).
Hence the locus of the midpoint of AB is (x + y) ab = 2xy (a + b).
If m is the slope of the required line and α is the angle which this line makes with the given line, then tan a = + (m_{1}–m)/(1+m_{1}m).
Find the equations to the sides of an isosceles right-angled triangle, the equation of whose hypotenuse is 3x + 4y = 4 and the opposite vertex is the point (2, 2).
The problem can be restated as:
Find the equations to the straight lines passing through the given point (2, 2) and making equal angles of 45^{o} with the given straight line 3x + 4y – 4 = 0
Slope of the line 3x + 4y – 4 = 0 is m_{1} = –3/4
⇒ tan 45^{o} = + m_{1}–m/1+m_{1}m, i.e., 1 = + m+3/4/1–3/4m
So that m_{A} = 1/7 and m_{B} = –7.
Hence the required equations of the two lines are
y – 2 = m_{A} (x – 2) and y – 2 = m_{B}(x – 2).
⇒ 7y – x – 12 = 0 and 7x + y = 16.
The straight lines 3x + 4y = 5 and 4x – 3y = 15 intersect at the point A. On these lines points B and C are chosen so that AB = AC. Find the possible equations of the line BC passing through (1, 2).
The two given straight lines are at right angles.
Since AB = AC, the triangle is an isosceles right angled triangle.
The required equation is of the form y – 2 = m(x – 1) … (1)
with tan 45^{o} = + m+3/4/1–3m/4 = + m–4/3/1+4m/3
⇒ 1 = + m+3/4/1–3m/4 = + m–4/3/1+4m/3 ⇒ m = – 7, 1/7.
Substitute the value of m in (1). We get the required equations.
Find the equation of the straight line passing through (–2, 7) and having intercept of length 3 units between the straight lines 4x + 3y = 12 and 4x + 3y = 3.
Distance AC between the two given parallel lines
= |c_{1}–c_{2}/√a^{2 }+ b^{2}| = 12–3/√(16 + 9) = 9/5.
⇒ BC = 12/5. If θ is the angle between BC and AB, then tan θ = 9/12 = 3/4.
Slope of the parallel lines = –4/3 = m_{2}.
If m_{1} is the slope of the required line, then tan θ = m_{1}–m/1+m_{1}m ⇒ 3/4 = + m_{1}–4/3/1+4/3m_{1}
i.e. m_{1} + 4/3 = 3/4 (1 – 4/3 m_{1}) and m_{1} + 4/3 = 3/4 (1 – 4/3 m_{1}).
The slopes are (i) m_{1} = –7/24 (ii) m_{1} = ∞ (the line is parallel to the y-axis).
The required equations of the lines are 7x + 24y + 182 = 0 and x + 2 = 0.
Equation of the line, through P(–2, 7) and making angle θ with the x-axis, is x+2/cosθ = y+7/sin θ = r.
If this line intersects the given lines at A and B, with AB = 3, the points A and B are A(–2 + r_{1} cos θ, – 7 + r_{1} sin θ) and B (–2 + (r_{1} + 3) cos θ, – 7 + (r_{1} + 3) sin θ).
Since A and B lie on the lines 4x + 3y = 3 and 4x + 3y = 12, we have
4r_{1 }cos θ + 3r_{1 }sin θ = 32 and
4r_{1} cos θ + 3r_{1} sin θ + 12 cos θ + 9 sin θ = 41, so that
12 cos θ + 9 sin θ = 0 or 4 cos θ + 3 sin θ = 3.
Solving this equation we find that θ = π/2 and tan θ = –7/24.
Hence the required lines are x + 2 = 0 and y + 7 = –7/24 (x + 2) i.e. 7x + 24y + 182 = 0.
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