SATHEE: Straight Line and Pair of Straight Lines 1 Question 55

Straight Line and Pair of Straight Lines 1 Question 55

55. A line through $A (- 5, - 4)$ meets the line $x + 3 y + 2 = 0$ , $2 x + y + 4 = 0$ and $x - y - 5 = 0$ at the points $B, C$ and $D$ respectively. If $(15 / A B)^{2} + (10 / A C)^{2} = (6 / A D)^{2}$ , find the equation of the line.

(1993, 5M)

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Solution:

Let equation of line $A C$ is

$\frac{y + 4}{\sin θ} = \frac{x + 5}{\cos θ} = r$

Let line $A E$ make angle $θ$ with $X$ -axis and intersects $x + 3 y + 2 = 0$ at $B$ at a distance $r_{1}$ and line $2 x + y + 4 = 0$ at $C$ at a distance $r_{2}$ and line $x - y - 5 = 0$ at $D$ at a distance $r_{3}$ .

$\begin{aligned} ∴ A B = r_{1}, A C = r_{2}, A D = r_{3} . \\ r_{1} = - \frac{- 5 - 3 \times 4 + 2}{1 \cdot \cos θ + 3 \cdot \sin θ} ∵ r = - \frac{I}{(a \cos θ + b \sin θ)} \\ \Rightarrow r_{1} = \frac{15}{\cos θ + 3 \sin θ} \\ \Rightarrow r_{2} = \frac{10}{2 \cos θ + \sin θ} \\ and r_{3} = - \frac{- 5 \times 1 - 4 (- 1) - 5}{\cos θ - \sin θ} \\ \Rightarrow r_{3} = \frac{6}{\cos θ - \sin θ} \end{aligned}$

But it is given that,

$\begin{aligned} {\frac{15}{A B}}^{2} + {\frac{10}{A C}}^{2} = {\frac{6}{A D}}^{2} \\ \Rightarrow {\frac{15}{r_{1}}}^{2} + {\frac{10}{r_{2}}}^{2} = {\frac{6}{r_{3}}}^{2} \end{aligned}$

$\begin{array}{cc} \Rightarrow & (\cos θ + 3 \sin θ)^{2} + (2 \cos + \sin θ)^{2} = (\cos θ - \sin θ)^{2} \\ [from Eqs. (i), (ii) and (iii)] \\ \Rightarrow & \cos^{2} θ + 9 \sin^{2} θ + 6 \cos θ \sin θ + 4 \cos^{2} θ \\ + \sin^{2} θ + 4 \cos θ \sin θ = \cos^{2} θ + \sin^{2} θ - 2 \cos θ \sin θ \\ \Rightarrow & 4 \cos^{2} θ + 9 \sin^{2} θ + 12 \sin θ \cos θ = 0 \\ \Rightarrow & (2 \cos θ + 3 \sin θ)^{2} = 0 \\ \Rightarrow & 2 \cos θ + 3 \sin θ = 0 \\ \Rightarrow & \cos θ = - (3 / 2) \sin θ \end{array}$

On substituting this in equation of $A C$ , we get

$\begin{aligned} \frac{y + 4}{\sin θ} & = \frac{x + 5}{- \frac{3}{2} \sin θ} \\ \Rightarrow & - 3 (y + 4) & = 2 (x + 5) \\ \Rightarrow & - 3 y - 12 & = 2 x + 10 \\ \Rightarrow & 2 x + 3 y + 22 & = 0 \end{aligned}$

which is the equation of required straight line.

Straight Line and Pair of Straight Lines 1 Question 55

55. A line through $A (- 5, - 4)$ meets the line $x + 3 y + 2 = 0$ , $2 x + y + 4 = 0$ and $x - y - 5 = 0$ at the points $B, C$ and $D$ respectively. If $(15 / A B)^{2} + (10 / A C)^{2} = (6 / A D)^{2}$ , find the equation of the line.

Solution:

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Straight Line and Pair of Straight Lines 1 Question 55

55. A line through A(−5,−4) meets the line x+3y+2=0, 2x+y+4=0 and x−y−5=0 at the points B,C and D respectively. If (15/AB)2+(10/AC)2=(6/AD)2, find the equation of the line.

Solution:

Engineering Preparation

Medical Preparation

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Important Resources

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Syllabus

Test Platform

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NCERT Exercise Video Solution

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55. A line through $A (- 5, - 4)$ meets the line $x + 3 y + 2 = 0$ , $2 x + y + 4 = 0$ and $x - y - 5 = 0$ at the points $B, C$ and $D$ respectively. If $(15 / A B)^{2} + (10 / A C)^{2} = (6 / A D)^{2}$ , find the equation of the line.