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

Question. 10

In what ratio does the x–axis divide the segment joining \((-4,-6)\) and \((-1,7)\)? Find the coordinates of the point of division.

Answer:

Ratio \(6:7\) (internally), point \(\big(-\dfrac{34}{13},\,0\big)\).

Detailed Answer with Explanation:

Step 1: Let the required point on the x-axis be \(P(x,0)\) because every point on the x-axis has \(y = 0\).

Step 2: Suppose this point divides the line joining \(A(-4,-6)\) and \(B(-1,7)\) in the ratio \(m:n\).

Step 3: By 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 the coordinates are

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

Step 4: Apply this for \(A(-4,-6)\) and \(B(-1,7)\):

Coordinates of point = \( \Big( \dfrac{m(-1) + n(-4)}{m+n},\; \dfrac{m(7) + n(-6)}{m+n} \Big) \)

Step 5: Since the point lies on the x-axis, its \(y\)-coordinate must be 0.

So, \[ \dfrac{7m - 6n}{m+n} = 0 \]

Step 6: Multiply both sides by \((m+n)\):

\(7m - 6n = 0 \Rightarrow 7m = 6n\)

Therefore, \(m:n = 6:7\).

Step 7: Now find the \(x\)-coordinate:

\[ x = \dfrac{6(-1) + 7(-4)}{6+7} = \dfrac{-6 - 28}{13} = -\dfrac{34}{13} \]

Step 8: Since \(y = 0\), the point is \(\Big(-\dfrac{34}{13}, 0\Big)\).

Final Answer: Ratio is \(6:7\) and the point of division is \(\Big(-\dfrac{34}{13},0\Big)\).

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