SATHEE: Straight Line Question 34

Straight Line Question 34

Question: Equation of angle bisector between the lines $ 3x+4y-7=0 $ and $ 12x+5y+17=0 $ are[RPET 1995]

Options:

A) $ \frac{3x+4y-7}{\sqrt{25}}=\pm \frac{12x+5y+17}{\sqrt{169}} $

B) $ \frac{3x+4y+7}{\sqrt{25}}=\frac{12x+5y+17}{\sqrt{169}} $

C) $ \frac{3x+4y+7}{\sqrt{25}}=\pm \frac{12x+5y+17}{\sqrt{169}} $

D) None of these

Show Answer

Answer:

Correct Answer: A

Solution:

By direct formulae $ \frac{a_1x+b_1y+c_1}{\sqrt{a_1^{2}+b_1^{2}}}=\pm \frac{a_2x+b_2y+c_2}{\sqrt{a_2^{2}+b_2^{2}}} $$ \frac{3x+4y-7}{\sqrt{3^{2}+4^{2}}}=\pm \frac{12x+5y+17}{\sqrt{{{(12)}^{2}}+{{(5)}^{2}}}} $ $ \frac{3x+4y-7}{5}=\pm \frac{12x+5y+17}{13} $ .

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Straight Line Question 34

Question: Equation of angle bisector between the lines $ 3x+4y-7=0 $ and $ 12x+5y+17=0 $ are[RPET 1995]

Options:

Answer:

Solution:

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