(D) The slope of the given line $2x - 3y + 17 = 0$ is $m_1 = \frac{-2}{-3} = \frac{2}{3}$.Since the lines are perpendicular,the product of their slope

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(D) The slope of the given line $2x - 3y + 17 = 0$ is $m_1 = \frac{-2}{-3} = \frac{2}{3}$.Since the lines are perpendicular,the product of their slopes is $-1$. Let $m_2$ be the slope of the line passing through $(7, 17)$ and $(15, \beta)$.$m_2 = \frac{\beta - 17}{15 - 7} = \frac{\beta - 17}{8}$.Since $m_1 \times m_2 = -1$,we have $\frac{2}{3} \times \frac{\beta - 17}{8} = -1$.$\frac{\beta - 17}{12} = -1$.$\beta - 17 = -12$.$\beta = 17 - 12 = 5$.

Class 11 Mathematics Straight Line Angle between two straight lines

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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:

JEE MAIN 2019 All JEE MAIN

A
$\frac{35}{3}$
B
$-5$
C
$-\frac{35}{3}$
D
$5$

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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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