Problem 31
Question
Find an equation for the line that passes through the point \(P(2,7)\) and is parallel to the line \(3 y-2 x+6=0\)
Step-by-Step Solution
Verified Answer
The equation of the line that passes through the point \(P(2,7)\) and is parallel to the line \(3 y-2 x+6=0\) is \(2x -3y = -5\).
1Step 1: Find the slope of the given line
Rearrange the given equation \(3y - 2x + 6 = 0\) into the slope-intercept form \(y = mx + c\). This results in \(y = 2/3x - 2\). Thus, the slope of the line, \(m\) is \(2/3\).
2Step 2: Use point-slope form to find the equation of the parallel line
The formula for the point-slope form is \(y - y1 = m(x - x1)\), where \(m\) is the slope and \((x1, y1)\) is a point on the line. With a slope of \(2/3\) and a given point of \(P(2,7)\), the equation becomes \(y - 7 = 2/3(x - 2)\) which simplify to \(y = 2/3x + 5/3\).
3Step 3: Convert to standard form
Equations of lines are often given in standard form: \(Ax + By = C\). To convert the equation from slope-intercept form to standard form, simply multiply every term by 3 to get rid of the fractions. This results in \(2x - 3y = -5\).
Key Concepts
Slope-Intercept FormPoint-Slope FormStandard Form
Slope-Intercept Form
The slope-intercept form of a linear equation is probably the most common and easily recognizable form. It is expressed as:
In the existing solution, the line equation \(3y - 2x + 6 = 0\) was rearranged into slope-intercept form \(y = \frac{2}{3}x - 2\), revealing a slope \(m = \frac{2}{3}\). Knowing this slope is key for finding equations of parallel lines since parallel lines share the same slope.
- \( y = mx + c \)
In the existing solution, the line equation \(3y - 2x + 6 = 0\) was rearranged into slope-intercept form \(y = \frac{2}{3}x - 2\), revealing a slope \(m = \frac{2}{3}\). Knowing this slope is key for finding equations of parallel lines since parallel lines share the same slope.
Point-Slope Form
Sometimes, when constructing the equation of a line, you already know a specific point on the line and the slope. The point-slope form of a linear equation is tailor-made for these scenarios, expressed as:
For the current exercise, since the line must be parallel to the given line, it uses the same slope \(m = \frac{2}{3}\). Given the point \(P(2,7)\), we plug these into the point-slope form to get \(y - 7 = \frac{2}{3}(x - 2)\). After simplifying, this can be rewritten in slope-intercept form as \(y = \frac{2}{3}x + \frac{5}{3}\).
- \( y - y_1 = m(x - x_1) \)
For the current exercise, since the line must be parallel to the given line, it uses the same slope \(m = \frac{2}{3}\). Given the point \(P(2,7)\), we plug these into the point-slope form to get \(y - 7 = \frac{2}{3}(x - 2)\). After simplifying, this can be rewritten in slope-intercept form as \(y = \frac{2}{3}x + \frac{5}{3}\).
Standard Form
The standard form of a line is another way of presenting a linear equation, generally noted as:
Converting from slope-intercept form to standard form becomes straightforward by eliminating any fractions through multiplying by the least common multiple. In the solution, the slope-intercept form \(y = \frac{2}{3}x + \frac{5}{3}\) was multiplied by 3 to clear the fractions, leading to the standard form: \(2x - 3y = -5\). This transformation gives a clear, fraction-free equation, commonly preferred for its simplicity and utility in different mathematical applications.
- \( Ax + By = C \)
Converting from slope-intercept form to standard form becomes straightforward by eliminating any fractions through multiplying by the least common multiple. In the solution, the slope-intercept form \(y = \frac{2}{3}x + \frac{5}{3}\) was multiplied by 3 to clear the fractions, leading to the standard form: \(2x - 3y = -5\). This transformation gives a clear, fraction-free equation, commonly preferred for its simplicity and utility in different mathematical applications.
Other exercises in this chapter
Problem 31
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Form the composition \(f \circ g \circ h\) and give the domain. $$f(x)=x-1, \quad g(x)=4 x, \quad h(x)=x^{2}$$
View solution Problem 32
Find an equation for the line tangent to the circle $$x^{2}+y^{2}+2 x-6 y-3=0$$ at the point (2,1)
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