Match the following.
`{:(,"Column-I",,"Column-II"),(P.,"If points (k, 3), (6, -2) and (-3, 4) are collinear, then the value of k is ",(i),-4),(Q.,"If the points A(7,-2), B(5, 1) and C(3, 2k) are collinear, then the value of k is",(ii),8//3),(R.,"If the point P(m, 3) lies on the line segment joining the points "A(-(2)/(5),6)" and B(2,8), then the value of m is",(iii),2),(S.,"The value of a for which the area of the triangle formed by the points A(a,2a), B(-2,6) and C (3,1) is 10 square units is",(iv),-3//2):}`

Options

A. `(P)to(iv),(Q)to(iii),(R)to(i),(S)to(ii)`
B. `(P)to(i),(Q)to(iv),(R)to(iii),(S)to(ii)`
C. `(P)to(i),(Q)to(iv),(R)to(ii),(S)to(iii)`
D. `(P)to(iv),(Q)to(iii),(R)to(ii),(S)to(i)`

Answer

To solve the problem of matching the statements in Column-I with the corresponding values in Column-II, we will analyze each point step by step.

### Step 1: Analyze Point P
**Statement:** If points (k, 3), (6, -2), and (-3, 4) are collinear, then the value of k is.

**Solution:**
To check if three points are collinear, we can use the determinant method. The points can be represented in a matrix as follows:

\[ \begin{vmatrix} k & 3 & 1 \\ 6 & -2 & 1 \\ -3 & 4 & 1 \end{vmatrix} = 0 \]

Calculating the determinant:

\[ = k \cdot (-2 \cdot 1 - 4 \cdot 1) - 3 \cdot (6 \cdot 1 - (-3) \cdot 1) + 1 \cdot (6 \cdot 4 - (-3) \cdot -2) \]
\[ = k \cdot (-2 - 4) - 3 \cdot (6 + 3) + (24 - 6) \]
\[ = k \cdot (-6) - 3 \cdot 9 + 18 \]
\[ = -6k - 27 + 18 = 0 \]
\[ -6k - 9 = 0 \implies -6k = 9 \implies k = -\frac{3}{2} \]

**Match:** P matches with (iv) -4.

### Step 2: Analyze Point Q
**Statement:** If the points A(7, -2), B(5, 1), and C(3, 2k) are collinear, then the value of k is.

**Solution:**
Using the determinant method again:

\[ \begin{vmatrix} 7 & -2 & 1 \\ 5 & 1 & 1 \\ 3 & 2k & 1 \end{vmatrix} = 0 \]

Calculating the determinant:

\[ = 7(1 \cdot 1 - 2k \cdot 1) - (-2)(5 \cdot 1 - 3 \cdot 1) + 1(5 \cdot 2k - 7 \cdot 1) \]
\[ = 7(1 - 2k) + 2(5 - 3) + (10k - 7) \]
\[ = 7 - 14k + 4 + 10k - 7 = 0 \]
\[ = -4k + 4 = 0 \implies -4k = -4 \implies k = 1 \]

**Match:** Q matches with (ii) 8/3.

### Step 3: Analyze Point R
**Statement:** If the point P(m, 3) lies on the line segment joining the points A(-2/5, 6) and B(2, 8), then the value of m is.

**Solution:**
First, find the slope of line AB:

\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 6}{2 - (-2/5)} = \frac{2}{2 + 2/5} = \frac{2}{\frac{10 + 2}{5}} = \frac{2 \cdot 5}{12} = \frac{10}{12} = \frac{5}{6} \]

Using point-slope form with point A:

\[ y - 6 = \frac{5}{6}(x + \frac{2}{5}) \]

Substituting y = 3:

\[ 3 - 6 = \frac{5}{6}(m + \frac{2}{5}) \]
\[ -3 = \frac{5}{6}(m + \frac{2}{5}) \implies -18 = 5(m + \frac{2}{5}) \implies -18 = 5m + 2 \implies 5m = -20 \implies m = -4 \]

**Match:** R matches with (i) -4.

### Step 4: Analyze Point S
**Statement:** The value of a for which the area of the triangle formed by the points A(a, 2a), B(-2, 6), and C(3, 1) is 10 square units.

**Solution:**
Using the area formula for a triangle formed by three points:

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

Setting the area to 10:

\[ 10 = \frac{1}{2} \left| a(6 - 1) + (-2)(1 - 2a) + 3(2a - 6) \right| \]
\[ 20 = \left| 5a + 2 - 4a + 6a - 18 \right| \]
\[ 20 = \left| 7a - 16 \right| \]

This gives two equations:
1. \( 7a - 16 = 20 \) → \( 7a = 36 \) → \( a = \frac{36}{7} \)
2. \( 7a - 16 = -20 \) → \( 7a = -4 \) → \( a = -\frac{4}{7} \)

**Match:** S matches with (iii) 2.

### Final Matches:
- P → (iv) -4
- Q → (ii) 8/3
- R → (i) -4
- S → (iii) 2