If the straight line 2x - 3y + 17 = 0 is perp | vedclass

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(B) The slope of the line passing through $(7, 17)$ and $(15, \beta)$ is given by $m_1 = \frac{\beta - 17}{15 - 7} = \frac{\beta - 17}{8}$.The given l

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$5$

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

If 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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