Question

find k if one of the line given by 3x^2 - kxy + 5y^2 = 0 is perpendicular to line 5x + 3y = 0​

Answer 1

Step-by-step explanation:

given lines 3x²-kxy+5y²=0______(1)

and 5x+3y=0

given one line in pair of line 3x²-kxy+5y²=0 is perpendicular to the line 5x+3y=0 is 3x-5y=0

therefore x=5y/3_______(2)

substitute (2) in equation (1)

we get 3

$3(25 {y}^{2} \div 9) - ky(5y \div 3) + 5 {y}^{2} = 0$

(25/3) -(5k/3)+5=0

125-15k+27=0

15k=152

k=152/15

Answer 2

Answer:

The value of k is $\dfrac{152}{15}$  

Step-by-step explanation:

Given as :

The equation of one line , 3 x² + k x y + 5 y² = 0

This is in the form of auxiliary equation

i.e a x² + 2h x y + b y² = 0

i.e a = 3  , 2 h = k and b = 5

The auxiliary equation of the given line in the form of

b m² + 2 h m + a = 0

Or, 5 m² + k m + 3 = 0               ........A

Now, Equation of another line is

5 x + 3 y = 0

Or, y = $\dfrac{-5}{3}$ x

Since equation of line is

y = m x + c , where m is slope

So, comparing we get , Slope = m = $\dfrac{-5}{3}$  

Put the value of m in eq A

ie 5 ($\dfrac{-5}{3}$)² + k ($\dfrac{-5}{3}$ )+ 3 = 0

Or, 125 - 15 k + 27 = 0

Or, 15 k = 152

i.e k = $\dfrac{152}{15}$

Hence, The value of k is $\dfrac{152}{15}$  Answer