Statement (A): If the angle between the lines $2 x + 3 y + 5 = 0$ and $3 x - k y + 10 = 0$ is $\frac{π}{2}$ , then the value of $‘ k ’$ is 2.
Statement (B): If the angle between the lines $5 x + 7 y - 10 = 0$ and $a x - 14 y + 20 = 0$ is $0 °$ , then the value of $a$ is $‘ - 2 ’$

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Statement A is False, Statement B is True

Statement A is True, Statement B is True

Statement A is True, Statement B is False

Statement A is False, Statement B is False

answer is C.

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Detailed Solution

Statement (A): $2 x + 3 y + 5 = 0$ , $3 x - k y + 10 = 0$

$a_{1} x + b_{1} y + c_{1} = 0$ , $a_{2} x + b_{2} y + c_{2} = 0$

$\cos θ = | \frac{a_{1} a_{2} + b_{1} b_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2}}} |$

$\cos 90 ° = | \frac{2 \times 3 + 3 (- k)}{\sqrt{2^{2} + 3^{2}} \sqrt{3^{2} + {(- k)}^{2}}} |$

$0 = | \frac{6 - 3 k}{\sqrt{13} \sqrt{9 + k^{2}}} |$

$6 - 3 k = 0$

$k = 2$

Statement (B): $5 x + 7 y - 10 = 0$ , $a x - 14 y + 20 = 0$

If the angle between two lines is $0 °$

Then the lines are parallel

$5 x + 7 y - 10 = 0$ , $a x - 14 y + 20 = 0$

$a_{1} x + b_{1} y + c_{1} = 0$ , $a_{2} x + b_{2} y + c_{2} = 0$

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}$

$\frac{5}{7} = \frac{a}{- 14}$

$a = \frac{- 70}{7}$

$a = - 10$

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