(C) The given equations of the lines are: $2x + 3y - 3 = 0$ --- $(1)$ $x + ky + 7 = 0$ --- $(2)$ The slope of line $(1)$ is $m_1 = -\frac{2}{3}$. Th

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(C) The given equations of the lines are: $2x + 3y - 3 = 0$ --- $(1)$ $x + ky + 7 = 0$ --- $(2)$ The slope of line $(1)$ is $m_1 = -\frac{2}{3}$. The slope of line $(2)$ is $m_2 = -\frac{1}{k}$. Since the lines are perpendicular,the product of their slopes must be $-1$: $m_1 \times m_2 = -1$ $(-\frac{2}{3}) \times (-\frac{1}{k}) = -1$ $\frac{2}{3k} = -1$ $2 = -3k$ $k = -\frac{2}{3}$

Class 11 Mathematics Straight Line Angle between two straight lines

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If the straight lines $2x + 3y - 3 = 0$ and $x + ky + 7 = 0$ are perpendicular,then the value of $k$ is:

KCET 2016 All KCET

A
$2/3$
B
$3/2$
C
$-2/3$
D
$-3/2$

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Straight Line

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Angle between two straight lines

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