The equation of the straight line passing through the intersection of `x+2y-19=0,x-2y-3=0` and at a distance of 5 from (-2,4) are…………

Answer

To solve the problem of finding the equations of the straight line passing through the intersection of the lines \(x + 2y - 19 = 0\) and \(x - 2y - 3 = 0\) that is at a distance of 5 from the point \((-2, 4)\), we can follow these steps:

### Step 1: Find the point of intersection of the two lines.
To find the intersection, we solve the equations simultaneously:

1. From the first equation:
\[ x + 2y - 19 = 0 \implies x = 19 - 2y \]

2. Substitute \(x\) in the second equation:
\[ (19 - 2y) - 2y - 3 = 0 \implies 19 - 4y - 3 = 0 \implies 16 = 4y \implies y = 4 \]

3. Substitute \(y = 4\) back into the first equation to find \(x\):
\[ x + 2(4) - 19 = 0 \implies x + 8 - 19 = 0 \implies x = 11 \]

Thus, the point of intersection \(P\) is \((11, 4)\).

### Step 2: Write the equation of the line in point-slope form.
Using the point-slope form of the equation of a line, we have:
\[ y - 4 = m(x - 11) \]
where \(m\) is the slope of the line.

### Step 3: Find the perpendicular distance from the point \((-2, 4)\) to the line.
The formula for the perpendicular distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

First, we need to express the line in the form \(Ax + By + C = 0\):
\[ y - 4 - m(x - 11) = 0 \implies mx - y + (4 - 11m) = 0 \]
Here, \(A = m\), \(B = -1\), and \(C = 4 - 11m\).

### Step 4: Set the distance equal to 5.
Now, we set the distance equal to 5:
\[ \frac{|m(-2) - 1(4) + (4 - 11m)|}{\sqrt{m^2 + 1}} = 5 \]

Simplifying the numerator:
\[ |-2m - 4 + 4 - 11m| = |-13m| = 13|m| \]
Thus, we have:
\[ \frac{13|m|}{\sqrt{m^2 + 1}} = 5 \]

### Step 5: Solve for \(m\).
Squaring both sides:
\[ \frac{169m^2}{m^2 + 1} = 25 \]
Cross-multiplying gives:
\[ 169m^2 = 25(m^2 + 1) \implies 169m^2 = 25m^2 + 25 \implies 144m^2 = 25 \implies m^2 = \frac{25}{144} \implies m = \pm \frac{5}{12} \]

### Step 6: Write the equations of the lines.
Substituting \(m = \frac{5}{12}\) and \(m = -\frac{5}{12}\) back into the point-slope form:
1. For \(m = \frac{5}{12}\):
\[ y - 4 = \frac{5}{12}(x - 11) \implies 12(y - 4) = 5(x - 11) \implies 5x - 12y - 7 = 0 \]

2. For \(m = -\frac{5}{12}\):
\[ y - 4 = -\frac{5}{12}(x - 11) \implies 12(y - 4) = -5(x - 11) \implies 5x + 12y - 67 = 0 \]

Thus, the equations of the lines are:
1. \(5x - 12y - 7 = 0\)
2. \(5x + 12y - 67 = 0\)