NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 7: Coordinate Geometry - Exercise 7.2
Question 2

Question. 2

Point \(P(-4,2)\) lies on the line segment joining \(A(-4,6)\) and \(B(-4,-6)\).

Answer:

True.

Detailed Answer with Explanation:

Step 1: If a point lies on the line segment joining two points, then the three points are collinear. In that case, the area of the triangle formed by them is zero.

Step 2: Use the area of triangle formula for points \(A(x_1,y_1)\), \(B(x_2,y_2)\), \(P(x_3,y_3)\):

\(\text{Area} = \dfrac{1}{2}\left|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)\right|\).

Step 3: Substitute \(A(-4,6)\), \(B(-4,-6)\), \(P(-4,2)\).

  • \(x_1 = -4,\ y_1 = 6\)
  • \(x_2 = -4,\ y_2 = -6\)
  • \(x_3 = -4,\ y_3 = 2\)

Step 4: Compute inside the absolute value:

\(x_1(y_2-y_3) = (-4)\big((-6)-2\big) = (-4)(-8) = 32\)

\(x_2(y_3-y_1) = (-4)\big(2-6\big) = (-4)(-4) = 16\)

\(x_3(y_1-y_2) = (-4)\big(6-(-6)\big) = (-4)(12) = -48\)

Step 5: Add them:

\(32 + 16 - 48 = 0\)

Step 6: Area:

\(\text{Area} = \dfrac{1}{2}|0| = 0\)

Final Step: Since the area of \(\triangle ABP\) is zero, points A, B, and P are collinear. Therefore, \(P(-4,2)\) lies on the line segment joining A and B (its y-value 2 is between -6 and 6).

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 7: Coordinate Geometry – Exercise 7.2 | Detailed Answers