In the xy-coordinate system, if (a, b) and (a + 3, b + k) are two points on the line defined by the equation x = 3y – 7, then k =
(A) 9
(B) 3
(C) 7/3
(D) 1
(E) 1/3
Coordinate geometry
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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
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
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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)
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)