The triangle formed by the lines 3x + y + 4 = | vedclass

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(D) Let the lines be $L_1: 3x + y + 4 = 0$,$L_2: 3x + 4y - 15 = 0$,and $L_3: 24x - 7y - 3 = 0$. The slopes are $m_1 = -3$,$m_2 = -\frac{3}{4}$,and $m_

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scalene triangle

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(D) Let the lines be $L_1: 3x + y + 4 = 0$,$L_2: 3x + 4y - 15 = 0$,and $L_3: 24x - 7y - 3 = 0$. The slopes are $m_1 = -3$,$m_2 = -\frac{3}{4}$,and $m_3 = \frac{24}{7}$. The angle between $L_1$ and $L_2$ is $	an 	heta_1 = |\frac{-3 - (-3/4)}{1 + (-3)(-3/4)}| = |\frac{-9/4}{7/4}| = \frac{9}{7}$. The angle between $L_2$ and $L_3$ is $	an 	heta_2 = |\frac{-3/4 - 24/7}{1 + (-3/4)(24/7)}| = |\frac{-21/28 - 96/28}{1 - 18/7}| = |\frac{-117/28}{-11/7}| = \frac{117}{44}$. The angle between $L_3$ and $L_1$ is $	an 	heta_3 = |\frac{24/7 - (-3)}{1 + (24/7)(-3)}| = |\frac{45/7}{1 - 72/7}| = |\frac{45/7}{-65/7}| = \frac{9}{13}$. Since all slopes are distinct and the angles are not equal,the triangle is a scalene triangle.

The triangle formed by the lines $3x + y + 4 = 0$,$3x + 4y - 15 = 0$,and $24x - 7y = 3$ is a/an:

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