The straight line 2x - 3y + 17 = 0 is perpend | vedclass

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(A) The equation of the given line is $2x - 3y + 17 = 0$,which can be written as $3y = 2x + 17$ or $y = \frac{2}{3}x + \frac{17}{3}$.Thus,the slope of

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$5$

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(A) The equation of the given line is $2x - 3y + 17 = 0$,which can be written as $3y = 2x + 17$ or $y = \frac{2}{3}x + \frac{17}{3}$.Thus,the slope of this line is $m_1 = \frac{2}{3}$.The slope of the line passing through $(7, 17)$ and $(15, \beta)$ is $m_2 = \frac{\beta - 17}{15 - 7} = \frac{\beta - 17}{8}$.Since the two lines are perpendicular,the product of their slopes must be $-1$,i.e.,$m_1 	imes m_2 = -1$.$\frac{2}{3} 	imes \frac{\beta - 17}{8} = -1$.$\frac{\beta - 17}{12} = -1$.$\beta - 17 = -12$.$\beta = 17 - 12 = 5$.

The straight line $2x - 3y + 17 = 0$ is perpendicular to the line passing through the points $(7, 17)$ and $(15, \beta)$. Then,$\beta$ equals:

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