vipin verma sir from many days these ques r pending plz ans soon exams on my head u see earlier many times i reposted this q but no response

1- let u= ax+by+ a( b)1/3=0 , v= bx-ay +b (a)1/3=0, a,b belongs to r be 2 st lines eqn of bisectors of angle formed by k1u-k2v=o, and k1u + k2v=0 for non zero real k1 and k2 r ............

2- the parallelogram is bounded by lines y= ax+c, y= ax+d, y= bx+c, y=bx+d has area= 18 parallelogram bounded by line y= ax+c , y=ax-d,y= bx+C, And y= bx-d has area 72 find a, b,c,d

$$\begin{gathered} u = ax + by + a(b{)}^{1/3} , v = by -ay +b(a{)}^{1/3} \\ So with these two lines are formed \\ {k}_{1}u -k_{2}v =0 \left(1\right) \\ {k}_{1}u + k_{2}v = 0 \left(2\right) \\ So line \left(1\right) is (k_{1}a -k_{2}b)x +y(k_{1}b +k_{2}a) +k_{1}a(b{)}^{1/3} -k_{2}b(a{)}^{1/3} = 0 \\ Line \left(2\right) is (k_{1}a +k_{2}b)x +y(k_{1}b -k_{2}a) +k_{1}a(b{)}^{1/3} +k_{2}b(a{)}^{1/3} = 0 \end{gathered}$$

$$\begin{gathered} So equation of bisector of line \left(1\right) and \left(2\right) are \\ \frac{k_{1}u-k_{2}v}{\sqrt{(a{k}_1-b{k}_2{)}^2+(k_{1}b+a{k}_2{)}^2}} =\frac{\pm (k_{1}u+k_{2}v)}{\sqrt{(a{k}_1+b{k}_2{)}^2+(k_{1}b-a{k}_2{)}^2}} \end{gathered}$$      
$$\begin{gathered} So the denominators are same , for both sides. \\ Hence k_{1}u -k_{2}v = \pm (k_{1}u +k_{2}v) \\ Taking positive , we have \\ {k}_{1}u -k_{2}v = (k_{1}u +k_{2}v) \\ Or v = 0 \\ Taking negative , we have \\ {k}_{1}u -k_{2}v =- (k_{1}u +k_{2}v) \\ Or u = 0 \end{gathered}$$


2)

The plot of the parallelogram is like this
This parallelogram has area = 18
Area of ABD  = $\frac{1}{2}\times \frac{1}{2}\left|\right(c-d\left)\right|\times 2\times \left|\frac{c-d}{b-a}\right|$ = 9
Here mod is used because sides have positive values.
So |c-d|² /|b-a| = 18 

Or |c-d|² = 18|b-a| (1)

Similarly draw the second parallelogram ,which has area as 72 , you will get the relation as :
|c+d|² = 72|b-a| (2)

Subtracting (1) from (2) , we get
4cd = 54|b-a|
Or 2cd = 27|b-a|
As the left side is even , so |b-a| should be even . So that both sides should be even.  [ even X odd = even ]
And we can find the minimum value of  a+b , when | b-a| = 2 , when a = 1 and b = 3 or a = 3 and b = 1   [ as a ,b,c, d are positive integers]
So cd should be multiple of 27
So c +d can be minimum at c = 3 and d = 9 or c = 9 and d = 3

Hope you have got it.