A line $4 x + y = 1$ through the point $A (2, - 7)$ meets the line $B C$ whose equation is $3 x - 4 y + 1 = 0$ at the point $B$ . If $A B = A C$ , then the equation of the line $A C$ is

52 x - 89 y + 519 = 0

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52 x + 89 y - 519 = 0

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52 x + 89 y + 519 = 0

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None of these

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Solution

The correct option is C $52 x + 89 y + 519 = 0$
Since, $A B = A C$
$∴ ∠ A B C = ∠ A C B = θ$ (say)

Now, slope of $B C = \frac{3}{4}$ and slope of $A B = - 4$
Let slope of $A C$ be $m$ .
Equating the two values of $tan θ$ , we get
$∣ ∣ ∣ ∣ ∣ ∣ \frac{- 4 - \frac{3}{4}}{1 - 4. \frac{3}{4}} ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ \frac{\frac{3}{4} - m}{1 + \frac{3}{4} m} ∣ ∣ ∣ ∣ ∣ ∣ \Rightarrow \pm \frac{19}{8} = \frac{3 - 4 m}{4 + 3 m}$
$\Rightarrow 76 + 57 m = 24 - 32 m \Rightarrow m = - \frac{52}{89}$
or $76 + 57 m = 32 m - 24 \Rightarrow m = - 4 (not possible as it is slope of A B$
$∴$ Equation of $A C$ is $(y + 7) = - \frac{52}{89} (x - 2)$
$\Rightarrow 52 x + 89 y + 519 = 0$

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Similar questions

4 x + y = 1

A (2, - 7)

B C

3 x - 4 y + 1 = 0

B

A B = A C

A C

Find the equation of the line passing through the points (2, 3) and the point of intersection of the lines $4 x - 3 y = 7$ and $3 x + 4 y + 1 = 0$ .

{(\frac{15}{A B})}^{2} + {(\frac{10}{A C})}^{2} = {(\frac{6}{A D})}^{2}

x - y = 7

x + 4 y = 2

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