Question

two straight lines 3x-2y=5 and 2x+ky+7=0 are perpendicular to each other. find the value of 'k'?

Answer 1

since the lines are perpendicular to each other,

product of their slopes is -1

$( \frac{3}{2} )( \frac{ - 2}{k} ) = - 1$
$- 6 = - 2k$
$k = 3$

Answer 2

Given:

Given equations are 3x-2y=5 and 2x+ky+7=0

Lines are perpendicular.

To Find:

Find the value of k.

Solution:

convert the lines into slope forms and find their slopes.

1.    3x-2y=5

      $y=\frac{3}{2} x-\frac{5}{2}$

      comparing it with y=mx +c

so, m₁ = 3/2

2. 2x+ky+7=0

  $y=-\frac{2}{k} -\frac{7}{k}$

  m₂ = -2/k

for the lines to be perpendicular,

    m₁×m₂ = -1

    (3/2)×(-2/k)=-1

     k=3

Hence the value of k is 3.