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.