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 given straight lines is farther from the origin, we will calculate the perpendicular distance from the origin (0, 0) to each line using the formula for the distance from a point to a line.

### Given Lines:
1. Line 1: \(2x - y + 3 = 0\)
2. Line 2: \(x - 4y - 7 = 0\)

### Step 1: Calculate the distance from the origin to Line 1

The formula for the distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by:

\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

For Line 1, we have:
- \(A = 2\)
- \(B = -1\)
- \(C = 3\)
- \((x_1, y_1) = (0, 0)\)

Substituting these values into the formula:

\[ d_1 = \frac{|2(0) - 1(0) + 3|}{\sqrt{2^2 + (-1)^2}} = \frac{|3|}{\sqrt{4 + 1}} = \frac{3}{\sqrt{5}} \]

### Step 2: Calculate the distance from the origin to Line 2

For Line 2, we have:
- \(A = 1\)
- \(B = -4\)
- \(C = -7\)
- \((x_1, y_1) = (0, 0)\)

Substituting these values into the formula:

\[ d_2 = \frac{|1(0) - 4(0) - 7|}{\sqrt{1^2 + (-4)^2}} = \frac{|-7|}{\sqrt{1 + 16}} = \frac{7}{\sqrt{17}} \]

### Step 3: Compare the distances

Now we have:
- Distance from origin to Line 1: \(d_1 = \frac{3}{\sqrt{5}}\)
- Distance from origin to Line 2: \(d_2 = \frac{7}{\sqrt{17}}\)

To compare \(d_1\) and \(d_2\), we can calculate their approximate values:

1. Calculate \(d_1\):
\[ d_1 \approx \frac{3}{2.236} \approx 1.34 \]

2. Calculate \(d_2\):
\[ d_2 \approx \frac{7}{4.123} \approx 1.69 \]

### Conclusion

Since \(d_2 > d_1\), the line \(x - 4y - 7 = 0\) is farther from the origin than the line \(2x - y + 3 = 0\).

### Final Answer
The line \(x - 4y - 7 = 0\) is farther from the origin.

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