Question

If one of the lines represents by
3x2-4xy+y2=0 is perpendicular to one of the
line 2x2-5xy+ky?=0 then k =
1)-3, 1379 2) 3, -13/9 3) -3, -13/94) -7, -33

please fast help me ​

Answer 1

Answer:

Option (4) -7, -33

Step-by-step explanation:

Ist pair of lines

$3x^2-4xy+y^2=0$

$\implies 3x^2-3xy-xy+y^2=0$

$\implies 3x(x-y)-y(x-y)=0$

$\implies(x-y)(3x-y)=0$

Thus the two lines are

$y=x$ & $y=3x$

slope of these lines are 1 and 3

Slope of the lines perpendicular to these lines will be -1 and -1/3

Since the constant term and the coefficients of x and y in the second pair of lines are zero

therefore these two lines should also be passing through the origin

These lines will be

$y=-x$ and $y=-\frac{1}{3}x$

Putting these values in $2x^2-5xy+ky^2=0$

$2x^2-5x(-x)+k(-x)^2=0$

$x^2(k+7)=0$

$\implies k=-7$

Also

$2x^2-5x(-x/3)+k(-x/3)^2=0$

$2x^2+5x^2/3+kx^2/9=0$

$\implies 18x^2+15x^2+kx^2=0$

$\implies x^2(33+k)=0$

$\implies k = -33$

Therefore,

$\boxed{k=-7,-33}$

Hope it helps.