Which of the st.lines `2x-y+3=0` and `x-4y-7=0` is farther from the origin ?

Answer

To determine which of the two straight lines is farther from the origin, we can use the formula for the distance from a point to a line. The distance \( d \) from a point \( (x_0, y_0) \) to the line given by the equation \( Ax + By + C = 0 \) is given by:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

In this case, the point is the origin \( (0, 0) \).

### Step 1: Identify the coefficients for the first line \( 2x - y + 3 = 0 \)

Rearranging the equation, we have:
- \( A = 2 \)
- \( B = -1 \)
- \( C = 3 \)

### Step 2: Calculate the distance from the origin to the first line

Using the distance formula:

\[ d_1 = \frac{|2(0) + (-1)(0) + 3|}{\sqrt{2^2 + (-1)^2}} \]

Calculating the numerator:

\[ |2(0) + (-1)(0) + 3| = |3| = 3 \]

Calculating the denominator:

\[ \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

Thus, the distance from the origin to the first line is:

\[ d_1 = \frac{3}{\sqrt{5}} \]

### Step 3: Identify the coefficients for the second line \( x - 4y - 7 = 0 \)

Rearranging the equation, we have:
- \( A = 1 \)
- \( B = -4 \)
- \( C = -7 \)

### Step 4: Calculate the distance from the origin to the second line

Using the distance formula:

\[ d_2 = \frac{|1(0) + (-4)(0) - 7|}{\sqrt{1^2 + (-4)^2}} \]

Calculating the numerator:

\[ |1(0) + (-4)(0) - 7| = |-7| = 7 \]

Calculating the denominator:

\[ \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17} \]

Thus, the distance from the origin to the second line is:

\[ d_2 = \frac{7}{\sqrt{17}} \]

### Step 5: Compare the distances \( d_1 \) and \( d_2 \)

To determine which distance is greater, we can compare \( \frac{3}{\sqrt{5}} \) and \( \frac{7}{\sqrt{17}} \).

To compare these fractions, we can cross-multiply:

\[ 3 \cdot \sqrt{17} \quad \text{and} \quad 7 \cdot \sqrt{5} \]

Calculating \( 3 \cdot \sqrt{17} \) and \( 7 \cdot \sqrt{5} \):

- \( 3 \cdot \sqrt{17} \approx 3 \cdot 4.123 = 12.369 \)
- \( 7 \cdot \sqrt{5} \approx 7 \cdot 2.236 = 15.652 \)

Since \( 12.369 < 15.652 \), we have:

\[ d_1 < d_2 \]

### Conclusion

The line \( x - 4y - 7 = 0 \) is farther from the origin than the line \( 2x - y + 3 = 0 \).