
The line segment joining the points \[P(3,3)\] and \[Q(6, - 6)\] is trisected by the points \[A\] and \[B\] such that \[A\] is nearer to \[P\]. If \[A\] also lies on the line given by \[2x\ + \ y\ + \ k\ = \ 0\] , find the value of \[k\].
Answer
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Hint: In this question, given that the line segment joining the point \[{PQ}\]. So it is clear that \[{PQ}\] is a line segment and also given that it is trisected by the points \[A\] and \[B\]. Also the point \[A\] is nearer to \[P\]. The point \[A\] lies on the line \[2x + y + k = 0\] . Here we need to find the value of \[k\]
Formula used :
Section formula to find the coordinate\[\ (x,\ y)\] ,
\[x = \dfrac{(mx{_2}+ nx{_1})}{\left(m+n\right)}\]
\[y = \dfrac{(my{_2}+ my{_1})}{\left(m+n\right)}\]
Diagram :
The point \[A\] lies on the line \[2x + y + k = 0\]
Complete step-by-step solution:
Let the coordinates of the point \[A\] be \[(x{_1} ,y{_1})\]
As \[P\] and \[Q\] are the point of trisection, we have \[PA = AB = QB\]
Hence we have
\[\dfrac{{AP}}{{PB}} = \dfrac{{AP}}{2AP} = \dfrac{1}{2}\]
Therefore \[A\] divides \[{PQ}\] in the ratio \[1:2\ \] and
Similarly \[B\] divides \[{PQ}\] in the ratio \[2:1\]
Now we need to find the coordinate of \[A\]
First we can find the \[x\] coordinate of \[A\]
\[x{_1} = \dfrac{1 \times 6 + 2 \times 3}{1 + 2}\]
By simplifying,
We get,
\[x{_1} = \dfrac{12}{3}\]
By dividing,
We get,
\[x{_1}= \ 4\]
Now we can find the \[y\] coordinate of \[A\]
\[y{_1}= \dfrac{\left( 1 \times \left( - 6 \right) + 2 \times 3 \right)}{1 + 2}\]
By simplifying,
We get,
\[y{_1}= \ 0\]
Now the coordinate of \[A\] is \[(4,\ 0)\ \]
Given that the point \[A\] lies on the line \[2x + y + k = 0\]
Now we can substitute the coordinate of \[A\] in the given line.
\[2(4)\ + 0 + k = 0\]
By multiplying,
We get,
\[8 + k = 0\]
Thus we get \[k = - 8\]
Final answer :
The value of \[k\] is \[\ - 8\]
Note: The concept used to find the value of \[k\] is co-ordinate geometry. In coordinate geometry, the section formula is used to find the ratio in which the line segment is divided by a point internally or externally. It is also used to find out the centroid of the triangle.
Formula used :
Section formula to find the coordinate\[\ (x,\ y)\] ,
\[x = \dfrac{(mx{_2}+ nx{_1})}{\left(m+n\right)}\]
\[y = \dfrac{(my{_2}+ my{_1})}{\left(m+n\right)}\]
Diagram :
The point \[A\] lies on the line \[2x + y + k = 0\]
Complete step-by-step solution:
Let the coordinates of the point \[A\] be \[(x{_1} ,y{_1})\]
As \[P\] and \[Q\] are the point of trisection, we have \[PA = AB = QB\]
Hence we have
\[\dfrac{{AP}}{{PB}} = \dfrac{{AP}}{2AP} = \dfrac{1}{2}\]
Therefore \[A\] divides \[{PQ}\] in the ratio \[1:2\ \] and
Similarly \[B\] divides \[{PQ}\] in the ratio \[2:1\]
Now we need to find the coordinate of \[A\]
First we can find the \[x\] coordinate of \[A\]
\[x{_1} = \dfrac{1 \times 6 + 2 \times 3}{1 + 2}\]
By simplifying,
We get,
\[x{_1} = \dfrac{12}{3}\]
By dividing,
We get,
\[x{_1}= \ 4\]
Now we can find the \[y\] coordinate of \[A\]
\[y{_1}= \dfrac{\left( 1 \times \left( - 6 \right) + 2 \times 3 \right)}{1 + 2}\]
By simplifying,
We get,
\[y{_1}= \ 0\]
Now the coordinate of \[A\] is \[(4,\ 0)\ \]
Given that the point \[A\] lies on the line \[2x + y + k = 0\]
Now we can substitute the coordinate of \[A\] in the given line.
\[2(4)\ + 0 + k = 0\]
By multiplying,
We get,
\[8 + k = 0\]
Thus we get \[k = - 8\]
Final answer :
The value of \[k\] is \[\ - 8\]
Note: The concept used to find the value of \[k\] is co-ordinate geometry. In coordinate geometry, the section formula is used to find the ratio in which the line segment is divided by a point internally or externally. It is also used to find out the centroid of the triangle.
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