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Combined Equations of Angle Bisectors of Lines

Combined equations of the angle bisectors of the lines represented by ax² + 2hxy +by² = 0

Let ax² + 2hxy + by² = 9 represent lines

y – m₁x = 0 and y – m₂x = 0

Let P(α, ß) be any point on one of bisectors.

⇒ (ß–m₁α)/√1+m₁² = + ß–m2α/√1+m₂²

⇒ (1 + m₂²) (ß – m₁α)² – (1 + m₁²) (ß – m₂α)² = 0

⇒ Locus of P(α, ß) is

x² – y² = 2hxy 1–m₁m₂/m₁+m₁

⇒ x₂–y₂/a–b = xy/h; is required equation of angle bisectors …… (1)

Note:

1. If a = b, then bisectors are x² – y² = 0 i.e. x – y = 0, x + y = 0

2. If h = 0, the bisectors are xy = 0 i.e. x = 0, y = 0

3. If in (i), coefficien
t of x² + coefficient of 2 = 0, then the two bisectors are always perpendicular to each other

Illustration:

Prove that the angle between one of the lines given by
ax² + 2hxy + by² = 0 and one of the lines given by ax² + 2hxy+ by² + K(x² + y²) = 0 is equal to the angle between the other two lines of the system.

Solution:

Let L₁ and L₂ be one pair and L₃ and L₄ be the other pair of lines.

If the angle between L₁ and L₃ is equal to the angle between L₂ and L₄ then pair of bisectors of L₁ and L₂ would be same as that of L₃ and L₄. Pair of bisectors of L₃ and L₄ is

x₂–y₂/(a+k)–(b+k) = xy/h

⇒ x₂–y₂/a–b = xy/h

Which is the same as the bisector pair of L₁ and L₂.

6. Angle bisectors of ax² + 2hxy + by² + 2gx + 2fy + c = 0 (i) are given by (x–x₀)²–(y–y₀)²/a–b = (x–x₀)(y–y₀)/h where (x₀, y₀) is the point of intersection of (i)

Enquiry: If we shift the origin of the coordinate system how can the coordinates of a point be known in the new system? What will happen if we rotate the axis?

(i) Linear Transformation:

CP = x, DP = y; AP = x, BP = y

If origin is shifted to (h, k) then the coordinates of a point
P(x, y) in the new system are

X = x – h

Y = y – k

You can check it by putting (h = 0, k = 0) that it gets reduced to the same original coordinate system.

(ii) Rotation of Axes:

If axes are rotated anticlockwise by angle q then the coordinates of a point P(x, y) changes to say P(x, y):

OL = x, PL = y

OM = X, PM = Y

OP = R (say)

cos (α + θ) = x/R …… (i)

sin (α + θ) = Y/R …… (ii)

cos α = x/R …… (ii)

sin α = Y/R …… (iv)

Eliminating α from (i), (ii), (iii) and (iv) we get

X = x cos θ + y sin θ

Y = –x sin θ + y cos θ

It can be checked that if θ = 0; then coordinates remain unchanged.

Illustration:

Find the equation of the line 2x + y = 7 when co-ordinates system is shifted to the point (3, 1)

Solution:

x = 3 + X and y = 1 + Y

Equation of line becomes

2(3 + X) + (1 + Y) = 7

or 2X + Y = 0

Note:

Slope of the line remains the same.

The general equation of second degree ax² + 2hxy + by² + 2gx + 2fy + c=0 represents a pair of straight lines if 1994_Combined equations 1.JPG

= 0.

⇒ abc + 2fgh – af² – bg² – ch² = 0 and h² > ab.

The homogeneous second degree equation ax² + 2hxy + by² = 0 represents a pair of straight lines through the origin.

If lines through the origin whose joint equation is ax² + 2hxy + by² = 0, are y = m₁x and y = m₂x, then y² – (m₁ + m₂)xy + m₁m₂x² = 0 and y² + 2h/b xy + a/b x² = 0 are identical. If θ is the angle between the two lines, then tanθ = + √(m₁+m₂)²–4m₁m₂/1+m₁m₂ = + 2√h²–ab/a+b.

The lines are perpendicular if a + b = 0 and coincident if h² = ab.

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.