Straight Line

Q: A straight line through P (–2 , –3) cuts the pair of straight lines x² + 3y² + 4xy – 8x – 6y – 9 = 0 in Q and R. Find the equation of the line if PQ . PR = 20.

Sol: Let line be $\displaystyle \frac{x+2}{cos\theta} = \frac{y+3}{sin\theta} = r $

x = r cosθ – 2, y = r sinθ – 3 …(i)

Now, x² + 3y² + 4xy – 8x – 6y – 9 = 0 …(ii)

Taking intersection of (i) with (ii) and considering terms of r₂ and constant (as we need PQ.PR = r₁.r₂ = product of the roots)

r²(cos² θ + 3 sin² θ + 4 sin θ cos θ) + (some terms)r + 80 = 0

$\displaystyle r_1 . r_2 = PQ.PR = \frac{80}{cos^2 \theta + 4 sin\theta cos\theta + 3 sin^2 \theta} $

⇒ cos² θ + 4 sin θ cos θ + 3 sin² θ = 4 (As PQ.PR = 20)

⇒ sin² θ – 4 sin θ cos θ + 3 cos² θ = 0

⇒ (sin θ – cos θ)(sin θ – 3 cos θ) = 0

⇒ tan θ = 1, tan θ = 3

hence equation of the line is y + 3 = 1(x + 2)

⇒ x – y = 1

and y + 3 = 3(x + 2)

⇒ 3x – y + 3 = 0.

Q: A ray of light generated from the source kept at (–3, 4) strikes the line 2x + y = 7 at R and then terminated at (0, 1). Find the point R so that ray travels through the shortest distance.

Sol. Slope of the line 2x + y = 7 is –2.

Let R be (a, 7 – 2a ), P (–3, 4) and Q (0, 1)

Slope of PR $\displaystyle = \frac{3-2a}{a+3}$ ,

Slope of RQ $\displaystyle = \frac{6-2a}{a}$

Now, ray travels through the shortest distance so, PR must be incident ray and RQ must be reflected ray.

$\displaystyle \frac{\frac{3-2a}{a+3} + 2}{1-2(\frac{3-2a}{a+3})} = \frac{-2 -\frac{6-2a}{a} }{1-2(\frac{6-2a}{a})}$

$\displaystyle a = \frac{42}{25} $

Hence R is $\displaystyle ( \frac{42}{25} , \frac{91}{25}) $

Q: 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.

Sol: A ≡ (a,0) , B ≡ (0 , b) , A’ ≡ (a’ ,0) , B’ ≡ (0 , b’)

Equation of A’B is $\displaystyle \frac{x}{a’} + \frac{y}{b} = 1$ ….(i)

and the equation of AB’ is $\displaystyle \frac{x}{a} + \frac{y}{b’} = 1$ ….(ii)

Subtracting (i) from (ii), we get,

$\displaystyle x(\frac{1}{a} – \frac{1}{a’}) + y(\frac{1}{b’} – \frac{1}{b}) = 0 $

$\displaystyle \frac{x(a’-a)}{a a’} + \frac{y(b-b’)}{b b’} = 0 $

Using a’ – a = b – b’

$\displaystyle \frac{x}{a a’} + \frac{y}{b b’} = 0 $

$\displaystyle b’ = \frac{a(a+b)y}{ay – bx} $ …(iii)

From (ii) b’x + ay = ab’

$\displaystyle b’ = \frac{a y}{a – x} $ …(iv)

Equating (iii) and (iv) we get x + y = a + b which is the required locus

Q: Find the equation of the straight lines passing through (-2 , -7) and having intercept of length 3 units between the straight lines 4x + 3y = 12 and 4x + 3y = 3.

Sol: Distance between the two given parallel lines

$\displaystyle =\frac{c_1 – c_2}{\sqrt{a^2 + b^2}} = \frac{12 – 3}{\sqrt{16 + 9}} = \frac{9}{5} $

$\displaystyle tan\theta = \frac{9}{12} = \frac{3}{4}$

Slope of the parallel lines $\displaystyle =- \frac{4}{3} = m_2 $

Also , $\displaystyle tan\theta = \pm \frac{m_1 – m_2}{1 + m_1 m_2}$

$\displaystyle \frac{3}{4} = \pm \frac{m_1 + \frac{4}{3}}{1 – \frac{4}{3} m_1}$

$\displaystyle m_1 + \frac{4}{3} = \frac{3}{4} (1 – \frac{4}{3} m_1)$

and , $\displaystyle m_1 + \frac{4}{3} = -\frac{3}{4} (1 – \frac{4}{3} m_1)$

The slopes are m₁ = -7/24 , m₁ = ∞ (the line is parallel to the y – axis)

The required equations of the lines are 7x + 24y + 182 = 0 and x+2 =0.

Q: Let P (sinθ , cosθ) (0 ≤ θ ≤ 2π) be a point and let OAB be a triangle with vertices (0, 0), (√(3/2),0 ) and (0 ,√(3/2) ) . Find θ if P lies inside the ΔOAB.

Sol. Equations of lines along OA, OB and AB are y = 0, x = 0, and $ x + y = \sqrt{\frac{3}{2}}$ respectively.
Now P and B will lie on the same side of y = 0 if cosθ > 0.

Similarly P and A will lie on the same side of x = 0 if sin θ > 0 and P and O will lie on the same side of $ x + y = \sqrt{\frac{3}{2}}$ if $ sin\theta + cos\theta < \sqrt{\frac{3}{2}} $ .

Hence P will lie inside the ΔABC, if sinθ > 0, cosθ > 0 and $ sin\theta + cos\theta < \sqrt{\frac{3}{2}} $

Now $ sin\theta + cos\theta < \sqrt{\frac{3}{2}} $

$\displaystyle sin(\theta + \pi/4) \sqrt{\frac{3}{2}} $

i.e. 0 < θ + π/4 < π/3

or, 2π/3 < θ + π/4 < π

Since sinθ >0 and cos θ > 0 , so

$\displaystyle 0 < \theta < \frac{\pi}{12}$

$\displaystyle \frac{5 \pi}{12} < \theta < \frac{\pi}{2}$

Q: Two consecutive sides of a parallelogram are 4x + 5y = 0 and 7x + 2y = 0. If the equation of one of the diagonals is 11x + 7y = 9, find the equation of the other diagonal.

Sol. The sides are 4x + 5y = 0 …(i)

and 7x + 2y = 0 …(ii)

Equation of the diagonal,

11x + 7y = 9 …(iii)

Solving (i) and (iii) co-ordinates of A are (5/3 , -4/3)

Solving (ii) and (iii) co-ordinates of B are (-2/3 , 7/3)

Mid-point of AB is M(1/2 , 1/2)

equation of the diagonal OC is y = x.

Q: A variable line is drawn through O, to cut two fixed straight lines L₁ and L₂ in A₁ and A₂, respectively. A point A is taken on the variable line such that $\displaystyle \frac{m+n}{OA} = \frac{m}{OA_1} + \frac{n}{OA_2} $ . Show that the locus of A is a straight line passing through the point of intersection of L₁ and L₂ , where O is being the origin.

Sol. Let the variable line passing through the origin is

$\displaystyle \frac{x}{cos\theta} = \frac{y}{sin\theta} = r_i $ ….(i)

Let the equation of the line L₁ is p₁ x + q₁ y = 1 …. (ii)

Equation of the line L₂ is p₂ x + q₂ y = 1 …. (iii)

Let the variable line intersects the line (ii) at A₁ and (iii) at A₂

Let OA₁ = r₁

Then A₁ = (r₁ cosθ , r₁ sinθ)

A₁ lies on L₁

$\displaystyle r_1 = OA_1 = \frac{1}{p_1 cos\theta + q_1 sin\theta}$

$\displaystyle r_2 = OA_2 = \frac{1}{p_2 cos\theta + q_2 sin\theta}$

Let OA = r

Let co-ordinate of A are (h, k)

(h, k) = (r cosθ , r sinθ)

$\displaystyle \frac{m+n}{OA} = \frac{m}{OA_1} + \frac{n}{OA_2} $

$\displaystyle \frac{m+n}{r} = \frac{m}{r_1} + \frac{n}{r_2} $

m + n = m (p₁ rcosθ + q₁ rsinθ ) + n(p₂ rcosθ + q₂ rsinθ)

(p₁ h + q₁ k – 1) + (n/m)(p₂ h + q₂ k – 1) = 0

Therefore, locus of A is (p₁ x + q₁ y -1) + (n/m)(p₂ x + q₂ y -1) = 0

L₁ + λ L₂ = 0 where λ = n/m

This is the equation of the line passing through the intersection of L₁ and L₂.

Q: If orthocentre of the triangle formed by ax^2 + 2hxy + by^2 = 0 and px + qy = 1 is (r, s) then prove that $\displaystyle \frac{r}{p} = \frac{s}{q} = \frac{a+b}{aq^2 + bp^2-2hpq} $

Sol: Let ax² + 2hxy + by² = 0 be OA and OB.

Let OD and BE be perpendiculars on AB and OA respectively.

Equation of OD is = r₁ …(i)

H ≡ (pr₁ , qr₁) also lies on BE.

$\displaystyle B= (\frac{1}{p + m_2 q} , \frac{m_2}{p + m_2 q}) $

Equation of BE is

$\displaystyle y – \frac{m_2}{p + m_2 q} = -\frac{1}{m_1}(x-\frac{1}{p + m_2 q} ) $

But (pr₁ , qr₁) also lies on BE

$\displaystyle r_1(p+qm_2) = \frac{1 + m_1 m_2}{p + m_1 q}$

$\displaystyle r_1 = \frac{1 + m_1 m_2}{(p + m_1 q)(p+qm_2)}$

$\displaystyle m_1 + m_2 = -\frac{2h}{b}$

$\displaystyle m_1 . m_2 = \frac{a}{b}$

$\displaystyle r_1 = \frac{a+b}{a q^2 -2hpq + bp^2}$

$\displaystyle \frac{r}{p} = \frac{s}{q} = \frac{a+b}{aq^2 + bp^2-2hpq} $

Q: Two points A and B move on the +ve direction of x-axis and y-axis respectively, such that OA + OB = α . Show that the locus of the foot of the perpendicular from the origin O on the line AB is (x + y)(x² + y²) = αxy.

Sol: Let the equation of AB be

$\displaystyle \frac{x}{a} + \frac{y}{b} = 1$ ….(i)

given, a + b = α (ii)

now, m_AB × m_OM =- 1

⇒ ah = bk ….(iii)

from (ii) and (iii),

$\displaystyle a = \frac{k \alpha}{h + k} $

$\displaystyle b = \frac{h \alpha}{h + k} $

from (i)

$\displaystyle \frac{x(h+k)}{k \alpha} + \frac{y(h+k)}{h \alpha} = 1$

as it passes through (h, k)

$\displaystyle \frac{h(h+k)}{k \alpha} + \frac{k(h+k)}{h \alpha} = 1$

⇒ (h + k) (h² + k² ) = α h k

locus of (h, k) is (x + y) (x² + y²) = α x y

Problem:   A(2, 0), B(0, 2) and C are the vertices of a triangle inscribed in the circle $\large x^2 + y^2 = 4 $ . The bisector of angle C belongs to either of two families of concurrent lines, whose points of concurrency are

(A) (√2 , √2 )

(B) (-√2 , -√2 )

(C) (√2 , -√2 )

(D) (-√2 , √2 )

Ans:    (A), (B).

Sol: The points of concurrency lie on the intersection of perpendicular bisectors of AB with the circle x² + y² = 4.

Hence the points are (√2 , √2 )   or (-√2 , -√2 )