1.

show the straight line x-2y+3=0 6x+3y+8=0 are perpendicular

Answer»

Given straight lines are

x - 2y + 3 = 0

⇒ 2y = x+3

⇒ y = x/2 + 3/2 .......(1)

And 6x+3y+8 = 0

⇒ 3y = -6x - 8

⇒ y = -2x - 8/3 ........(2)

The slope of line (1) is m= 1/2 (By comparing equation y = mx + c)

And slope of line (2) is m2 = -2

We obsened that m1 m2 = 1/2 x -2 = -1

which is condition that both lines are pependicular.

Hence, given lines are pependicular to each other.



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