To find the equations of the straight lines passing through the point of intersection of the lines [tex]\(2x - 3y + 1 = 0\)[/tex] and [tex]\(x + y - 2 = 0\)[/tex] that make an angle of [tex]\(45^\circ\)[/tex] with the line [tex]\(x + 2y - 5 = 0\)[/tex], we can proceed step-by-step as follows:
### Step 1: Find the Point of Intersection
First, we need to find the point of intersection of the lines [tex]\(2x - 3y + 1 = 0\)[/tex] and [tex]\(x + y - 2 = 0\)[/tex].
1. Solve the system of equations:
[tex]\[ x + y - 2 = 0 \][/tex]
[tex]\[ y = 2 - x \][/tex]
2. Substitute [tex]\( y = 2 - x \)[/tex] into [tex]\(2x - 3y + 1 = 0\)[/tex]:
[tex]\[ 2x - 3(2 - x) + 1 = 0 \][/tex]
[tex]\[ 2x - 6 + 3x + 1 = 0 \][/tex]
[tex]\[ 5x - 5 = 0 \][/tex]
[tex]\[ x = 1 \][/tex]
[tex]\[ y = 2 - 1 = 1 \][/tex]
Therefore, the point of intersection is [tex]\((1, 1)\)[/tex].
### Step 2: Determine the Slope of the Given Line
We need the slope of the line [tex]\(x + 2y - 5 = 0\)[/tex].
1. Rewrite the equation in slope-intercept form [tex]\(y = mx + c\)[/tex]:
[tex]\[ x + 2y - 5 = 0 \][/tex]
[tex]\[ 2y = -x + 5 \][/tex]
[tex]\[ y = -\frac{1}{2}x + \frac{5}{2} \][/tex]
Thus, the slope ([tex]\(m_2\)[/tex]) of the line [tex]\(x + 2y - 5 = 0\)[/tex] is [tex]\(-\frac{1}{2}\)[/tex].
### Step 3: Determine the Slopes of the Desired Lines
To find the lines making a [tex]\(45^\circ\)[/tex] angle with the slope [tex]\(m_2 = -\frac{1}{2}\)[/tex], we use the tangent formula for angles between two lines:
[tex]\[ \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \tan(45^\circ) = 1 \][/tex]
where [tex]\(m_1\)[/tex] is the slope of the desired line.
Substitute [tex]\(m_2 = -\frac{1}{2}\)[/tex]:
[tex]\[ \left| \frac{m_1 + \frac{1}{2}}{1 - \frac{1}{2}m_1} \right| = 1 \][/tex]
This gives two cases:
[tex]\[ \frac{m_1 + \frac{1}{2}}{1 - \frac{1}{2}m_1} = 1 \][/tex]
[tex]\[ \frac{m_1 + \frac{1}{2}}{1 - \frac{1}{2}m_1} = -1 \][/tex]
Solve these equations separately:
1. First case ([tex]\( \frac{m_1 + \frac{1}{2}}{1 - \frac{1}{2}m_1} = 1 \)[/tex]):
[tex]\[ m_1 + \frac{1}{2} = 1 - \frac{1}{2}m_1 \][/tex]
[tex]\[ m_1 + \frac{1}{2}m_1 = 1 - \frac{1}{2} \][/tex]
[tex]\[ \frac{3}{2}m_1 = \frac{1}{2} \][/tex]
[tex]\[ m_1 = \frac{1}{3} \][/tex]
2. Second case ([tex]\( \frac{m_1 + \frac{1}{2}}{1 - \frac{1}{2}m_1} = -1 \)[/tex]):
[tex]\[ m_1 + \frac{1}{2} = - (1 - \frac{1}{2}m_1) \][/tex]
[tex]\[ m_1 + \frac{1}{2} = -1 + \frac{1}{2}m_1 \][/tex]
[tex]\[ m_1 - \frac{1}{2}m_1 = -1 - \frac{1}{2} \][/tex]
[tex]\[ \frac{1}{2}m_1 = -\frac{3}{2} \][/tex]
[tex]\[ m_1 = -3 \][/tex]
### Step 4: Equations of the Desired Lines
Using the point-slope form of the line equation, we can write the equations of the lines passing through [tex]\((1, 1)\)[/tex]:
1. For [tex]\(m_1 = \frac{1}{3}\)[/tex]:
[tex]\[ y - 1 = \frac{1}{3}(x - 1) \][/tex]
[tex]\[ y - 1 = \frac{1}{3}x - \frac{1}{3} \][/tex]
[tex]\[ y = \frac{1}{3}x + \frac{2}{3} \][/tex]
Multiplying by 3 to get the standard form:
[tex]\[ 3y = x + 2 \][/tex]
[tex]\[ x - 3y + 2 = 0 \][/tex]
2. For [tex]\(m_1 = -3\)[/tex]:
[tex]\[ y - 1 = -3(x - 1) \][/tex]
[tex]\[ y - 1 = -3x + 3 \][/tex]
[tex]\[ y = -3x + 4 \][/tex]
Multiplying by 1 to get the standard form:
[tex]\[ 3x + y - 4 = 0 \][/tex]
### Final Answer
The equations of the straight lines passing through the point of intersection of [tex]\(2x - 3y + 1 = 0\)[/tex] and [tex]\(x + y - 2 = 0\)[/tex] and making a [tex]\(45^\circ\)[/tex] angle with [tex]\(x + 2y - 5 = 0\)[/tex] are:
[tex]\[ x - 3y + 2 = 0 \][/tex]
[tex]\[ 3x + y - 4 = 0 \][/tex]
