If the straight line `y=mx+c` passes through the points (2,4) and `(-3,6)`, then the value of m and c are
(i) `m=-(2)/(5),c=(24)/(5)`
(ii) `m=(2)/(5),c=(24)/(5)`
(iii) `m=-(2)/(5),c=-(24)/(5)`
(iv) `m=(2)/(5),c=-(24)/(5)`

Options

A. `m=-(2)/(5),c=(24)/(5)`
B. `m=(2)/(5),c=(24)/(5)`
C. `m=-(2)/(5),c=-(24)/(5)`
D. `m=(2)/(5),c=-(24)/(5)`

Answer

To find the values of \( m \) and \( c \) for the straight line equation \( y = mx + c \) that passes through the points (2, 4) and (-3, 6), we can follow these steps:

### Step 1: Set up the equations using the given points
The line passes through the point (2, 4). Substituting \( x = 2 \) and \( y = 4 \) into the equation \( y = mx + c \):
\[ 4 = m(2) + c \]
This simplifies to:
\[ 2m + c = 4 \quad \text{(Equation 1)} \]

Next, the line also passes through the point (-3, 6). Substituting \( x = -3 \) and \( y = 6 \):
\[ 6 = m(-3) + c \]
This simplifies to:
\[ -3m + c = 6 \quad \text{(Equation 2)} \]

### Step 2: Solve the system of equations
Now we have a system of two equations:
1. \( 2m + c = 4 \)
2. \( -3m + c = 6 \)

To eliminate \( c \), we can subtract Equation 2 from Equation 1:
\[ (2m + c) - (-3m + c) = 4 - 6 \]
This simplifies to:
\[ 2m + c + 3m - c = -2 \]
\[ 5m = -2 \]
Now, solving for \( m \):
\[ m = -\frac{2}{5} \]

### Step 3: Substitute \( m \) back to find \( c \)
Now that we have \( m \), we can substitute it back into Equation 1 to find \( c \):
\[ 2\left(-\frac{2}{5}\right) + c = 4 \]
This simplifies to:
\[ -\frac{4}{5} + c = 4 \]
Adding \( \frac{4}{5} \) to both sides:
\[ c = 4 + \frac{4}{5} \]
Converting 4 to a fraction:
\[ c = \frac{20}{5} + \frac{4}{5} = \frac{24}{5} \]

### Final Result
Thus, the values of \( m \) and \( c \) are:
\[ m = -\frac{2}{5}, \quad c = \frac{24}{5} \]

### Conclusion
The correct option is:
(i) \( m = -\frac{2}{5}, c = \frac{24}{5} \)