Question 10 of 58intermediate🔧 ApplyMCQ1 mark

If area of triangle is 35 sq units with vertices (2, -6), (5, 4) and (k, 4). Then k is

Exercise: EXERCISE 4.2 | Q: 5 | (Page 8)
For More Understanding

Explanation

The question asks to find the value of k given the area of a triangle with vertices (2, -6), (5, 4), and (k, 4) is 35 sq. units. Using the determinant formula for area of a triangle from Section 4.3, we set up the equation and solve for k. Since area is given, we must consider both positive and negative values of the determinant as stated in the remarks. This yields two possible values for k: 12 and -2, which corresponds to option D.

Solution Steps

  1. Step 1: Using the formula for area of triangle with vertices (x1,y1),(x2,y2),(x3,y3)(x_1, y_1), (x_2, y_2), (x_3, y_3): Δ=12∣x1y11x2y21x3y31∣\Delta = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}

  2. Step 2: Substituting the given vertices (2, -6), (5, 4), (k, 4): 12∣2−61541k41∣=±35\frac{1}{2} \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} = \pm 35

  3. Step 3: Expanding the determinant: 12[2(4−4)+5(4+6)+k(−6−4)]=±35\frac{1}{2}[2(4-4) + 5(4+6) + k(-6-4)] = \pm 35

  4. Step 4: Simplifying: 12[0+50−10k]=±35\frac{1}{2}[0 + 50 - 10k] = \pm 35, which gives 50−10k=±7050 - 10k = \pm 70

  5. Step 5: Solving: When 50−10k=7050 - 10k = 70, we get k=−2k = -2. When 50−10k=−7050 - 10k = -70, we get k=12k = 12.

  6. Step 6: Therefore, k=12k = 12 or k=−2k = -2, which corresponds to option (D).

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