Position of two points with respect to a given line

Let the line be ax + by + c = 0 and P(x₁, y₁), Q(x₂, y₂) be two points.

Case 1:

If P(x₁, y₁) and Q(x₂, y₂) are on the opposite sides of the line 
ax + by + c = 0, then the point R on the line ax + by + c = 0 divides the line PQ internally in the ratio m₁ : m₂, where m₁/m₂ must be positive.

Co-ordinates of R

are (m₁x₂+m₂x₁/m₁+m₂, m₁y₂+m₂y₁/m₁+m₂).

Point R lies on the line ax + by + c = 0.

⇒ m₁/m₂ = ax₁+by₁+c/ax₂+by₂+c > 0

So that ax₁ + by₁ + c and ax₂ + by₂ + c should have opposite signs.

Case 2:

If ax₁ + by₁ + c and ax₂ + by₂ + c have the same signs then m₁/m₂ = –ve, so that the point R on the line ax + by + c = 0 will divide the line PQ externally in the ratio m₁ : m₂ and the points P(x₁, y₁) and Q(x₂, y₂) are on the same side of the line ax + by + c = 0.
 

Illustration:

Find the range of θ in the interval (0, π) such that the points (3, 5) and (sinθ, cosθ) lie on the same side of the line x + y – 1 = 0.

Solution:

3 + 5 – 1 =7 > 0 ⇒ sinθ + cosθ – 1 > 0

⇒ sin(π/4 + θ) > 1/√2

⇒ π/4 < π/4 + θ < 3π/4

⇒ 0 < θ < π/2.

Illustration:

Find a, if (α, α²) lies inside the triangle having sides along the lines

2x + 3y = 1, x + 2y – 3 = 0, 6y = 5x – 1.

Solution:

Let A, B, C be vertices of the triangle.

A ≡ (–7, 5), B ≡ (5/4, 7/8),

C ≡ (1/3, 1/9).

Sign of A w.r.t. BC is –ve.

If p lies in-side the ¦ABC, then sign of P will be the same as sign of a w.r.t. the line BC

⇒     5α – 6α² – 1 < 0.                         …… (1)

Similarly    2α + 3α² – 1 > 0.                                …… (2)

And,          α + 2α² – 3 < 0.                          …… (3)

Solving, (1), (2) and (3) for α and then taking intersection,

We get      α ? (1/2, 1) ∪ (–3/2, –1).
 

Illustration:

The equations of the perpendicular bisectors of the sides AB and AC of a triangle ABC are respectively x – y + 5 = 0 and x + 2y = 0. If the co-ordinates of A are (1, –2), find the equation of BC.

Solution:

From the figure,

E ≡ (x₁+1/2, y₁–2/2),

F ≡ (x₂+1/2, y₂–2/2).

Alt text : Equations of the perpendicular bisectors of sides of triangle

Since E and F lie on OE and OF respectively,

x₁ – y₁ + 13 = 0                                  … (1)

and   x₂ + 2y₂ – 3 = 0                                  … (2)

Also, slope of AB = –1 and slope of AC is 2, so that

x₁ + y₁ + 1 = 0.                                  … (3)

And   2x₂ – y₂ – 4 = 0                                  … (4)

Solving these equations, we get the co-ordinates of B and C as

B ≡ (–7, 6) and C ≡ (11/5, 2/5)

⇒ Equation of BC is 14x + 23y – 40 = 0.
 

Illustration:

Two fixed points A and B are taken on the co-ordinate axes such that OA = a and OB = b. Two variable points A’ and B’ are taken on the same axes such that OA’ + OB’ = OA + OB. Find the locus of the point of intersection of AB’ and A’B.

Solution:

Let A ≡ (a, 0), B (0, b), A’ ≡ (a’, 0), B’ ≡ (0, b’).

Equation of A’B is x/a' + y/b' = 1.                                          …. (1)

and equation of AB’ is x/a + y/b'  = 1.                                    …. (2)

Subtracting (1) from (2), we get, x (1/a – 1/a') + y(1/b' – 1/b) = 0.

⇒ x(a'–a)/aa' + y(b–b')/bb' = 0.                             [Using a’ – a = b – b’]

⇒ x/a(b–b'+a) + y/bb', 0 ⇒ b’ = a(a+b)y/ay–bx.                              ….. (3)

From (2) b’x + ay = (4) we get x + y = a + b

which is the required locus.

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