# lines in perependicular position 3ax-8y+13=0(a+1)(x-2ay-21))=0 a=?

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### 1 Answer

You need to use the slope intercept form to write both equations such that:

`-8y = -3ax- 13 =gt y = 3ax/8 + 13/8`

You need to open the brackets of the second equation such that:

`ax + 2a^2y - 21a + x - 2ay - 21 = 0`

`y(2a^2 - 2a)+ x(a+1) - 21a - 21 = 0`

`y(2a^2 - 2a) = -x(a+1) + 21a+ 21`

`y = -x(a+1)/(2a^2 - 2a) + (21a+ 21)/(2a^2 - 2a)`

You need to remember that if you multiply the slopes of the lines yields -1 when they are perpendicular such that:

`(3a/8)*((a+1)/(2a^2 - 2a)) = -1`

`3a(a+1)/(16a(a-1)) = -1 =gt 3a(a+1) = -16a(a-1)`

You need to remember that you do not need to divide by a, to preserve all solutions to the equation.

`3a^2 + 3a + 16a^2 - 16a = 0`

`19a^2 - 13a = 0 =gt a(19a - 13) = 0`

If `a(19a - 13) = 0 =gt a = 0 ; 19a - 13 = 0 =gt a = 13/19`

You need to exclude a = 0 because this value cancel the denominator `2a^2 - 2a` of the slope of the second line.

**Hence, the value of a for the lines are orthogonal is `a = 13/19` .**