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4.1 Q-7

Question Statement

Find the value of hh such that the points A(βˆ’1,h)A(-1, h), B(3,2)B(3, 2), and C(7,3)C(7, 3) are collinear.


Background and Explanation

To determine if three points are collinear (lie on the same straight line), we can use the concept of the area of the triangle formed by these points. If the area is zero, the points are collinear. This can be calculated using a determinant-based formula for the area of a triangle. If the area is zero, the points are collinear.

For three points A(x1,y1)A(x_1, y_1), B(x2,y2)B(x_2, y_2), and C(x3,y3)C(x_3, y_3), the area of the triangle formed by them is given by:

Area=12∣x1(y2βˆ’y3)+x2(y3βˆ’y1)+x3(y1βˆ’y2)∣\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|

For the points to be collinear, the area must be zero.


Solution

Step 1: Set up the Formula

We are given the points A(βˆ’1,h)A(-1, h), B(3,2)B(3, 2), and C(7,3)C(7, 3). Using the formula for the area of a triangle, we substitute the coordinates of the points:

Area=12βˆ£βˆ’1(2βˆ’3)+3(3βˆ’h)+7(hβˆ’2)∣\text{Area} = \frac{1}{2} \left| -1(2 - 3) + 3(3 - h) + 7(h - 2) \right|

For the points to be collinear, the area must be zero:

12βˆ£βˆ’1(2βˆ’3)+3(3βˆ’h)+7(hβˆ’2)∣=0\frac{1}{2} \left| -1(2 - 3) + 3(3 - h) + 7(h - 2) \right| = 0

This simplifies to:

βˆ£βˆ’1(2βˆ’3)+3(3βˆ’h)+7(hβˆ’2)∣=0\left| -1(2 - 3) + 3(3 - h) + 7(h - 2) \right| = 0

Step 2: Simplify the Expression

Now, expand the terms inside the absolute value:

βˆ£βˆ’1(βˆ’1)+3(3βˆ’h)+7(hβˆ’2)∣=0\left| -1(-1) + 3(3 - h) + 7(h - 2) \right| = 0

Simplify each term:

∣1+9βˆ’3h+7hβˆ’14∣=0\left| 1 + 9 - 3h + 7h - 14 \right| = 0

Combine like terms:

∣1+9βˆ’14+(7hβˆ’3h)∣=0\left| 1 + 9 - 14 + (7h - 3h) \right| = 0 βˆ£βˆ’4+4h∣=0\left| -4 + 4h \right| = 0

Step 3: Solve for hh

For the absolute value to be zero, the expression inside must also be zero:

βˆ’4+4h=0-4 + 4h = 0

Solving for hh:

4h=44h = 4 h=1h = 1

Thus, the value of hh is 1.


Key Formulas or Methods Used

  • Collinearity Condition: The points are collinear if the area of the triangle they form is zero. The area is given by the formula:
Area=12∣x1(y2βˆ’y3)+x2(y3βˆ’y1)+x3(y1βˆ’y2)∣ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|

Summary of Steps

  1. Use the formula for the area of a triangle formed by three points.
  2. Substitute the coordinates of the points A(βˆ’1,h)A(-1, h), B(3,2)B(3, 2), and C(7,3)C(7, 3).
  3. Simplify the resulting equation.
  4. Set the expression inside the absolute value to zero.
  5. Solve for hh, which gives h=1h = 1.