Hi
K is 1

equation of the line we know
x = 3y – 7

3y=x+7
y=x/3+7/3----y=mx+c

when (a,b)= b=a/3+7/3
-->b=a+7/3

(a+3,b+k)=b+k=(a+3)/3+7/3
b+k=a+10/3
k=(a+10/3)-(a+7/3)
we get
k=(a+10-a-7)/3

where k=1

OA??

HTH
vishu

Since both the points are on the line so they satisfy the line equation X=3Y-7
Substitute both the points(a,b) and (a+3,b+k) in the equation to form two separate equations and solve them to find the value of k.

So,by substituting (a,b) we get
a=3b-7 -----------(1)
And by substituting (a+3,b+k) we get,
a+3=3(b+k)-7 => a=3b+3k-10 -------(2)

Now sub value of 'a' from (1) in (2)

3b-7=3b+3k-10
solving we get 3k=3 => k=1

So answer is (D)

Thanks ...OA is 1

simplest way would be :


Given eq. x=3y-7=>y=x/3+7/3
so slope is 1/3

applying slopes formula => 1/3=b+k-b/a+3-a=>k=1

(slope=diff. of y coordinates/diff. x coordinates)

Amit