NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 7: Coordinate Geometry - Exercise 7.3
Question 18

Question. 18

Find the coordinates of the point \(R\) on the segment joining \(P(-1,3)\) and \(Q(2,5)\) such that \(PR=\dfrac{3}{5}PQ\).

Answer:

\(R\big(\dfrac{4}{5},\dfrac{21}{5}\big)\)

Detailed Answer with Explanation:

Step 1: Recall the idea.

The point \(R\) lies on the line segment from \(P(-1,3)\) to \(Q(2,5)\). We are told that the length \(PR\) is \(\dfrac{3}{5}\) of the total length \(PQ\). This means \(R\) divides the line from \(P\) to \(Q\) in the ratio \(3:2\).

Step 2: Write the coordinates of \(P\) and \(Q\).

\(P(x_1,y_1) = (-1,3), \; Q(x_2,y_2) = (2,5)\)

Step 3: Use the section formula.

If a point divides the line joining \((x_1,y_1)\) and \((x_2,y_2)\) in the ratio \(m:n\), then its coordinates are:

\[ \Bigg(\dfrac{mx_2+nx_1}{m+n}, \; \dfrac{my_2+ny_1}{m+n}\Bigg) \]

Step 4: Identify the ratio.

Here, \(PR:PQ = \tfrac{3}{5}\). So, the ratio of \(PR:RQ = 3:2\). That means \(m=3\) and \(n=2\).

Step 5: Substitute the values.

\[ x = \dfrac{3(2)+2(-1)}{3+2} = \dfrac{6-2}{5} = \dfrac{4}{5} \]

\[ y = \dfrac{3(5)+2(3)}{3+2} = \dfrac{15+6}{5} = \dfrac{21}{5} \]

Step 6: Write the final coordinates.

So, the point \(R\) is:

\(R\big(\tfrac{4}{5},\tfrac{21}{5}\big)\)

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 7: Coordinate Geometry – Exercise 7.3 | Detailed Answers