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To find the equations of the straight lines passing through the point [tex]\((1, 3)\)[/tex] and making an angle of [tex]\(45^{\circ}\)[/tex] with the line [tex]\(x - 3y + 4 = 0\)[/tex], we will follow these steps:
### Step 1: Find the Slope of the Given Line
The given equation of the line is [tex]\(x - 3y + 4 = 0\)[/tex]. To find its slope, we can rewrite the equation in the slope-intercept form [tex]\(y = mx + c\)[/tex]:
[tex]\[ x - 3y + 4 = 0 \][/tex]
[tex]\[ -3y = -x - 4 \][/tex]
[tex]\[ y = \frac{1}{3}x + \frac{4}{3} \][/tex]
Therefore, the slope [tex]\(m\)[/tex] of the given line is [tex]\(\frac{1}{3}\)[/tex].
### Step 2: Determine the Slopes of the New Lines
Since the new lines make an angle of [tex]\(45^{\circ}\)[/tex] with the given line, their slopes can be found using the property of angles between lines:
[tex]\[ m_{\text{new}} = m_{\text{given}} \pm \tan(45^{\circ}) \][/tex]
Recall that [tex]\(\tan(45^{\circ}) = 1\)[/tex]. Hence, we have:
[tex]\[ m_{\text{new}} = \frac{1}{3} \pm 1 \][/tex]
This gives us two possible slopes:
[tex]\[ m_{\text{new}_1} = \frac{1}{3} + 1 = \frac{4}{3} \][/tex]
[tex]\[ m_{\text{new}_2} = \frac{1}{3} - 1 = -\frac{2}{3} \][/tex]
### Step 3: Write the Equations in Point-Slope Form
For the point [tex]\((1, 3)\)[/tex], we use the point-slope form of the equation of a line [tex]\(y - y_1 = m(x - x_1)\)[/tex]:
1. For the slope [tex]\(\frac{4}{3}\)[/tex]:
[tex]\[ y - 3 = \frac{4}{3}(x - 1) \][/tex]
2. For the slope [tex]\(-\frac{2}{3}\)[/tex]:
[tex]\[ y - 3 = -\frac{2}{3}(x - 1) \][/tex]
### Step 4: Simplify the Equations
1. Simplifying the first equation:
[tex]\[ y - 3 = \frac{4}{3}(x - 1) \][/tex]
[tex]\[ y - 3 = \frac{4}{3}x - \frac{4}{3} \][/tex]
[tex]\[ y = \frac{4}{3}x - \frac{4}{3} + 3 \][/tex]
[tex]\[ y = \frac{4}{3}x + \frac{5}{3} \][/tex]
Rewriting in standard form:
[tex]\[ 3y = 4x + 5 \][/tex]
[tex]\[ 4x - 3y = -5 \][/tex]
[tex]\[ \Rightarrow Eq(1.33333333333333x - y, -1.66666666666667) \][/tex]
2. Simplifying the second equation:
[tex]\[ y - 3 = -\frac{2}{3}(x - 1) \][/tex]
[tex]\[ y - 3 = -\frac{2}{3}x + \frac{2}{3} \][/tex]
[tex]\[ y = -\frac{2}{3}x + \frac{2}{3} + 3 \][/tex]
[tex]\[ y = -\frac{2}{3}x + \frac{11}{3} \][/tex]
Rewriting in standard form:
[tex]\[ 3y = -2x + 11 \][/tex]
[tex]\[ 2x + 3y = 11 \][/tex]
[tex]\[ \Rightarrow Eq(0.666666666666667x + y, 3.66666666666667) \][/tex]
### Final Equations
Thus, the equations of the straight lines passing through the point [tex]\((1, 3)\)[/tex] and making an angle of [tex]\(45^{\circ}\)[/tex] with [tex]\(x - 3y + 4 = 0\)[/tex] are:
[tex]\[ \boxed{1.33333333333333x - y = -1.66666666666667} \][/tex]
[tex]\[ \boxed{0.666666666666667x + y = 3.66666666666667} \][/tex]
### Step 1: Find the Slope of the Given Line
The given equation of the line is [tex]\(x - 3y + 4 = 0\)[/tex]. To find its slope, we can rewrite the equation in the slope-intercept form [tex]\(y = mx + c\)[/tex]:
[tex]\[ x - 3y + 4 = 0 \][/tex]
[tex]\[ -3y = -x - 4 \][/tex]
[tex]\[ y = \frac{1}{3}x + \frac{4}{3} \][/tex]
Therefore, the slope [tex]\(m\)[/tex] of the given line is [tex]\(\frac{1}{3}\)[/tex].
### Step 2: Determine the Slopes of the New Lines
Since the new lines make an angle of [tex]\(45^{\circ}\)[/tex] with the given line, their slopes can be found using the property of angles between lines:
[tex]\[ m_{\text{new}} = m_{\text{given}} \pm \tan(45^{\circ}) \][/tex]
Recall that [tex]\(\tan(45^{\circ}) = 1\)[/tex]. Hence, we have:
[tex]\[ m_{\text{new}} = \frac{1}{3} \pm 1 \][/tex]
This gives us two possible slopes:
[tex]\[ m_{\text{new}_1} = \frac{1}{3} + 1 = \frac{4}{3} \][/tex]
[tex]\[ m_{\text{new}_2} = \frac{1}{3} - 1 = -\frac{2}{3} \][/tex]
### Step 3: Write the Equations in Point-Slope Form
For the point [tex]\((1, 3)\)[/tex], we use the point-slope form of the equation of a line [tex]\(y - y_1 = m(x - x_1)\)[/tex]:
1. For the slope [tex]\(\frac{4}{3}\)[/tex]:
[tex]\[ y - 3 = \frac{4}{3}(x - 1) \][/tex]
2. For the slope [tex]\(-\frac{2}{3}\)[/tex]:
[tex]\[ y - 3 = -\frac{2}{3}(x - 1) \][/tex]
### Step 4: Simplify the Equations
1. Simplifying the first equation:
[tex]\[ y - 3 = \frac{4}{3}(x - 1) \][/tex]
[tex]\[ y - 3 = \frac{4}{3}x - \frac{4}{3} \][/tex]
[tex]\[ y = \frac{4}{3}x - \frac{4}{3} + 3 \][/tex]
[tex]\[ y = \frac{4}{3}x + \frac{5}{3} \][/tex]
Rewriting in standard form:
[tex]\[ 3y = 4x + 5 \][/tex]
[tex]\[ 4x - 3y = -5 \][/tex]
[tex]\[ \Rightarrow Eq(1.33333333333333x - y, -1.66666666666667) \][/tex]
2. Simplifying the second equation:
[tex]\[ y - 3 = -\frac{2}{3}(x - 1) \][/tex]
[tex]\[ y - 3 = -\frac{2}{3}x + \frac{2}{3} \][/tex]
[tex]\[ y = -\frac{2}{3}x + \frac{2}{3} + 3 \][/tex]
[tex]\[ y = -\frac{2}{3}x + \frac{11}{3} \][/tex]
Rewriting in standard form:
[tex]\[ 3y = -2x + 11 \][/tex]
[tex]\[ 2x + 3y = 11 \][/tex]
[tex]\[ \Rightarrow Eq(0.666666666666667x + y, 3.66666666666667) \][/tex]
### Final Equations
Thus, the equations of the straight lines passing through the point [tex]\((1, 3)\)[/tex] and making an angle of [tex]\(45^{\circ}\)[/tex] with [tex]\(x - 3y + 4 = 0\)[/tex] are:
[tex]\[ \boxed{1.33333333333333x - y = -1.66666666666667} \][/tex]
[tex]\[ \boxed{0.666666666666667x + y = 3.66666666666667} \][/tex]
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