Question

If the adjacent sides of a parallelogram are
2x2 – 5xy + 3y2 = 0 and one diagonal is
x + y + 2 = 0 then the other diagonal is
(A) 9.2 – 1ly=0
(B) 9x + 1ly=0
(C) 11. – 9y = 0
(D) 11x + 9y = 0​

Answer 1

Step-by-step explanation:

Let y AB = 4 x +5 y = 0 or y = -4/ 5x

let Ad = 7x + 2y = 0 or y = -7/2 x

so the point A = (0,0) as these two intersect at origin

let BD = 11 x + 7 Y = 9 , we know it does not pass through origin so, C can not be on that line

intersection of AB and BD : 11 x - 28 28 x /5 = 9

x = 5/ 3 and so y = -4/3

so B( 5/ 3 , -4 / 3 )

intersection of AD and BD : 11 x -49 x / 2 = 9

x = -2/3 and so y = 7/3

so, D(-2/3, 7/3)

now midpoint of BD = O + (1/2 , 1/2)

line OA is the diagonal AC , SO its equations is y = x as its slope is (1/ 2 ) / (1/2) = 1 and it passes through origin

O is midpoint of AC hence C = (1, 1)

equation of BC : parallel to AD 7x + 2 y = K

As (1, 1) lines on it, so satisfying

7x +2y = 9 as K = 9

equation of CD : it is parallrl to AB . hence it is 4 x + 5 y = K

As (1, 1) lies on it , so satisfying

hence K= 9

so CD : 4x +5y = 9

Answer 2

The correct answer is 9x - 11y = 0.

Given: Equation of diagonal = x + y + 2 = 0.

Adjacent sides of a parallelogram are $2x^{2} -5xy+3y^{2} =0$.

To Find: Equation of other diagonal.

Solution:

$2x^{2} -5xy+3y^{2} =0$

$2x^{2} -2xy-3xy+3y^{2} =0$

= 2x(x - y) -3y(x-y)

= (2x - 3y)(x - y)

So

x - y = 0  (equation 1)

2x - 3y = 0   (equation 2)

x + y + 2 = 0   (equation 3)

Solve 1 and 3

x - y = 0

x = y

x + y + 2 =0

2x = -2

x = -1  and y = -1

Coordinates of A (-1, -1)

Solve 2 and 3

2x - 3y = 0

x = $\frac{3y}{2}$

x + y + 2 =0

$\frac{3y}{2}+y+2=0$

$\frac{5y}{2}=-2$

y = $\frac{-4}{5}$

x = $\frac{-6}{5}$

Coordinates of C ( $\frac{-6}{5}$,  $\frac{-4}{5}$ ).

Mid point of OD = Mid point of AC

$(\frac{a}{2} ,\frac{b}{2} )=(\frac{-1-\frac{6}{5} }{2} ,\frac{-1-\frac{4}{5} }{2} )$

$(\frac{a}{2} ,\frac{b}{2} )=(\frac{-11}{10} ,\frac{-9}{10} )$

$a=\frac{-11}{5} ,b=\frac{-9}{5}$

Coordinates of D ($\frac{-11}{5},\frac{-9}{5}$)

Hence equation of line is $\frac{y-y_{1} }{x-x_{1} }=\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$

$\frac{y-0}{x-0}=\frac{\frac{-9}{5}-0 }{\frac{-11}{5} -0}$

11y = 9x

9x - 11y = 0

Hence, the equation of diagonal is 9x - 11y = 0.

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