
Let P and Q be the points of trisection of the line segment joining the points A (2, -2) and B (-7, 4) such that P is nearer to A. Find the coordinates of P and Q.
Answer
601.5k+ views
Hint: P and Q are the point of trisection, this means P and Q must be dividing the line segment AB as P is closer to A thus P will be dividing the line segment AB in the ratio 1: 2 and similar point Q will be dividing the line segment AB in ratio 2: 1. Apply this concept to the section formula to get the answer.
Complete step-by-step answer:
If point C divide the line AB in the ratio (m : n) internally as shown above then according to section formula the coordinates of C are:
$x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}},{\text{ y}} = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}$
Now it is given P and Q be the points of trisection of the line joining the points A (2, -2) and B (-7, 4) such that P is nearer to A.
Let A = (2, -2) $ \equiv \left( {{x_1},{y_1}} \right)$ and B = (-7, 4) $ \equiv \left( {{x_2},{y_2}} \right)$
So, P and Q are points of trisection (i.e. they divide the line into three equal parts) as shown in figure 1.
So P divides the line AB in the ratio (1 : 2) internally and Q divides the line AB in the ratio (2 : 1) internally.
Let the coordinates of P be $\left( {a,b} \right)$ and the coordinates of Q be $\left( {c,d} \right)$
Therefore according to section formula the coordinates of P are,
Here the value of m = 1 and the value of n = 2.
$ \Rightarrow \left( {a,b} \right) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},{\text{ }}\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right) = \left( {\dfrac{{1 \times \left( { - 7} \right) + 2 \times 2}}{{1 + 2}},\dfrac{{1 \times 4 + 2 \times \left( { - 2} \right)}}{{1 + 2}}} \right)$
Now simplify we have,
$ \Rightarrow \left( {a,b} \right) = \left( {\dfrac{{ - 3}}{3},\dfrac{0}{3}} \right) = \left( { - 1,0} \right)$
Now again according to section formula the coordinates of Q are,
Here the value of m = 2 and the value of n = 1.
$ \Rightarrow \left( {c,d} \right) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},{\text{ }}\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right) = \left( {\dfrac{{2 \times \left( { - 7} \right) + 1 \times 2}}{{2 + 1}},\dfrac{{2 \times 4 + 1 \times \left( { - 2} \right)}}{{2 + 1}}} \right)$
Now simplify we have,
$ \Rightarrow \left( {c,d} \right) = \left( {\dfrac{{ - 12}}{3},\dfrac{6}{3}} \right) = \left( { - 4,2} \right)$
So these are the required coordinates of P and Q respectively.
Note: Whenever we face such types of problems the key concept is to have a good gist of the section formula as it is applicable to all questions in which some specific points divide a line segment into some particular ratios, this helps finding the coordinates of these specific points.
Complete step-by-step answer:
If point C divide the line AB in the ratio (m : n) internally as shown above then according to section formula the coordinates of C are:
$x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}},{\text{ y}} = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}$
Now it is given P and Q be the points of trisection of the line joining the points A (2, -2) and B (-7, 4) such that P is nearer to A.
Let A = (2, -2) $ \equiv \left( {{x_1},{y_1}} \right)$ and B = (-7, 4) $ \equiv \left( {{x_2},{y_2}} \right)$
So, P and Q are points of trisection (i.e. they divide the line into three equal parts) as shown in figure 1.
So P divides the line AB in the ratio (1 : 2) internally and Q divides the line AB in the ratio (2 : 1) internally.
Let the coordinates of P be $\left( {a,b} \right)$ and the coordinates of Q be $\left( {c,d} \right)$
Therefore according to section formula the coordinates of P are,
Here the value of m = 1 and the value of n = 2.
$ \Rightarrow \left( {a,b} \right) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},{\text{ }}\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right) = \left( {\dfrac{{1 \times \left( { - 7} \right) + 2 \times 2}}{{1 + 2}},\dfrac{{1 \times 4 + 2 \times \left( { - 2} \right)}}{{1 + 2}}} \right)$
Now simplify we have,
$ \Rightarrow \left( {a,b} \right) = \left( {\dfrac{{ - 3}}{3},\dfrac{0}{3}} \right) = \left( { - 1,0} \right)$
Now again according to section formula the coordinates of Q are,
Here the value of m = 2 and the value of n = 1.
$ \Rightarrow \left( {c,d} \right) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},{\text{ }}\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right) = \left( {\dfrac{{2 \times \left( { - 7} \right) + 1 \times 2}}{{2 + 1}},\dfrac{{2 \times 4 + 1 \times \left( { - 2} \right)}}{{2 + 1}}} \right)$
Now simplify we have,
$ \Rightarrow \left( {c,d} \right) = \left( {\dfrac{{ - 12}}{3},\dfrac{6}{3}} \right) = \left( { - 4,2} \right)$
So these are the required coordinates of P and Q respectively.
Note: Whenever we face such types of problems the key concept is to have a good gist of the section formula as it is applicable to all questions in which some specific points divide a line segment into some particular ratios, this helps finding the coordinates of these specific points.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

