Question

Find the value of k so that the equations 3y−(2/3)x=5 and ky+2x+8=0 are perpendicular.

Answer 1

Answer:

PLZ MARK AS BRAINLIEST

Answer 2

we know that ,.

Condition for perpendicular lines is that product of their slopes is (-1)

=> If slope of L1 is m1 and that of L2 is m2 then ,

for L1 and L2 to be perpendicular ,

m1 × m2 = -1

=> Also , slope = - (co efficient of x) / ( co efficient of y )

=> Slope of 3y - (2/3)x -5 = 0 is , - (-2/3) / 3 = 2/9

AND , slope of ky + 2x + 8 = 0 is , -2/k

=> -2/k × 2/9 = -1

=> k = 4/9

Answer is 4/9.