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Chapter 0: Problem 95

Find the equation of the line that passes through the given point and also satisfies the additional piece of information. Express your answer in slope- intercept form, if possible. (-2,-7)\(;\) parallel to the line \(\frac{1}{2} x-\frac{1}{3} y=5\)

Short Answer

Expert verified
The equation of the line is \(y = \frac{3}{2}x - 4\).

Step by step solution

01

Identify the slope of the given line

The original line is given as \(\frac{1}{2}x - \frac{1}{3}y = 5\). To find its slope, convert it to the slope-intercept form \(y = mx + b\), where \(m\) is the slope. Rearrange the equation to isolate \(y\): \(-\frac{1}{3}y = -\frac{1}{2}x + 5\). Multiply the whole equation by -3 to solve for \(y\): \(y = \frac{3}{2}x - 15\). Thus, the slope \(m\) of the given line is \(\frac{3}{2}\).
02

Use the slope of the parallel line

Lines that are parallel to each other have the same slope. Therefore, the slope \(m\) of our desired line is also \(\frac{3}{2}\).
03

Use the point-slope form

With the slope \(\frac{3}{2}\) and a point it passes through, \((-2,-7)\), use the point-slope form equation: \(y - y_1 = m(x - x_1)\). Here \((x_1, y_1) = (-2, -7)\) and \(m = \frac{3}{2}\), so: \(y + 7 = \frac{3}{2}(x + 2)\).
04

Expand and simplify

Distribute \(\frac{3}{2}\): \(y + 7 = \frac{3}{2}x + 3\). Subtract 7 from both sides to express the equation in slope-intercept form: \(y = \frac{3}{2}x - 4\).
05

Verify the equation

Verify by checking that the point \((-2, -7)\) lies on the line \(y = \frac{3}{2}x - 4\). Substitute \(-2\) for \(x\) and check: \(y = \frac{3}{2}(-2) - 4 = -3 - 4 = -7\). The point satisfies the equation, confirming that \(y = \frac{3}{2}x - 4\) is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Slope-Intercept Form
Understanding the slope-intercept form of a linear equation is crucial. It is written as \( y = mx + b \).
  • \( y \) is the dependent variable, commonly representing a value you wish to find.
  • \( m \) is the slope of the line, representing how steep the line is.
  • \( x \) is the independent variable, usually the input or starting point.
  • \( b \) is the y-intercept, the point where the line crosses the y-axis.
Moving a line equation into slope-intercept form can make graphing easier. The slope \( m \) shows how much \( y \) will increase per unit of \( x \).
For example, with a slope \( m = \frac{3}{2} \), it means for every additional unit in \( x \), \( y \) increases by 1.5. This form is extremely helpful for quickly understanding and graphing linear relationships.
Parallel Lines
Parallel lines are lines in a plane that never intersect. This means they have the same slope. Understanding this concept is essential because it helps in predicting and verifying linear relationships.
For example, if a line has the equation \( y = \frac{3}{2}x - 4 \), any line parallel to it will also have a slope of \( \frac{3}{2} \).
Key points to remember about parallel lines:
  • Same slope \( m \) means the lines are parallel.
  • Different y-intercepts \( b \) mean they are distinct lines.
Always check that the slopes match to confirm two lines are parallel. This rule is applied when finding equations of lines parallel to a given line through a specific point.
Point-Slope Form
The point-slope form is a useful tool in linear equations. It鈥檚 useful when you know a line鈥檚 slope and a single point on the line. It is written as \( y - y_1 = m(x - x_1) \).
  • \( (x_1, y_1) \) is the given point.
  • \( m \) is the slope.
  • \( x \) and \( y \) are variables representing any point on the line.
Using point-slope form can quickly lead to the slope-intercept form. Simply substitute the known values and rearrange.
This form helps in quickly establishing a linear equation when some details are already known. After applying the point-slope form, it鈥檚 often straightforward to convert to the slope-intercept form for easier interpretation.

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Most popular questions from this chapter

Determine whether the lines are parallel, perpendicular, or neither, and then graph both lines in the same viewing screen using a graphing utility to confirm your answer. $$\begin{aligned} &y_{1}=0,16 x+2.7\\\ &y_{2}=6.25 x-1.4 \end{aligned}$$

Solve for the indicated variable in terms of other variables. Solve \(s=\frac{1}{2} g t^{2}\) for \(t\)

Determine whether the lines are parallel, perpendicular, or neither, and then graph both lines in the same viewing screen using a graphing utility to confirm your answer. $$\begin{aligned} &y_{1}=17 x+22\\\ &y_{2}=-\frac{1}{17} x-13 \end{aligned}$$

Solve for the indicated variable in terms of other variables. Solve \(a^{2}+b^{2}=c^{2}\) for \(c\)

Solve the equation by factoring: \(x^{4}-4 x^{2}=0\).
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