Problem 12 - Precalculus Essentials

Question

Use the given conditions to write an equation for each line in point-slope form and general form. Passing through (5,-9) and perpendicular to the line whose equation is \(x+7 y-12=0\)

Step-by-Step Solution

Answer

The equation of the line passing through (5,-9) and perpendicular to the line whose equation is \(x+7y-12=0\), in the general form is \(7x - y - 44 = 0\).

Step 1: Determine the slope of the given line.

First, we need to express the given line equation \(x+7y-12=0\) in the slope-intercept form \(y=mx+b\). By doing so, we get \(y=-\frac{1}{7}x+\frac{12}{7}\). Thus, the slope of the given line (m) is \(-\frac{1}{7}.\)

Step 2: Find the slope of the line that is perpendicular to the given line.

It's a known property of lines that the product of slopes of two perpendicular lines is \(-1\). So, if \(m_1\) is the slope of the given line and \(m_2\) is the slope of the line which is perpendicular to it, then \(m_1 * m_2 = -1\). From step 1, we have \(m_1 = -\frac{1}{7}\). Solving for \(m_2 = -\frac{1}{m_1}, we get \(m_2 = 7\).

Step 3: Write in point-slope form.

Now that we have a point (5,-9) and the slope (7), we can write the equation of the line in point-slope form. It is given by \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = (5,-9)\) is the given point. Substituting these values, we get \(y+9 = 7(x-5)\).

Step 4: Transform to General form.

The general form of a line is \(Ax + By = C\). To convert our point-slope form into general form, we need to distribute \(7\) in \(7(x-5)\) and bring other terms to the left. This gives \(7x - y - 44 = 0\) as the equation in general form.

Key Concepts

Perpendicular Lines in Geometry

When two lines intersect to form a right angle, they are called perpendicular lines. In a coordinate plane, the slopes of these lines have a special relationship: if one line has a slope of \(m\), the other line, which is perpendicular to it, will have a slope of \( -\frac{1}{m} \) if \(m\) is not zero. This means the slopes of perpendicular lines are negative reciprocals of each other. For instance, if a line has a slope of 3, the slope of a line perpendicular to it would be \( -\frac{1}{3} \), and vice versa. In our exercise, we used this principle to find that a line perpendicular to one with a slope of \( -\frac{1}{7} \) has a slope of 7.

Slope-Intercept Form

The slope-intercept form of a line's equation is one of the most commonly used forms to represent straight lines. It is expressed as \( y = mx + b \), where \(m\) represents the slope, and \(b\) represents the y-intercept, which is the point where the line crosses the y-axis. This form makes it very easy to graph a linear equation and to understand the rate of change of y with respect to x, which is the slope. To find the slope -intercept form from a general equation like \( Ax + By = C \), you simply solve for \( y \). In our example, converting \( x+7y-12=0 \) into slope -intercept form allowed us to isolate \( y \) and identify the slope, hence facilitating the process of finding a perpendicular line passing through a given point.

Writing Equations of Lines

When it comes to writing equations of lines, there are several forms to consider, each with different uses. Apart from the slope-intercept form, another common form is the point-slope form, which is ideal for when you have a point and the slope of the line. It looks like \( y - y_1 = m(x - x_1) \), with \( (x_1, y_1) \) being the given point and \(m\) the slope. After finding the slope of the perpendicular line in the previous steps, we use this point-slope form to write the new line's equation using the point _(5,-9)_. In our case, this delicate step involved plugging in the slope value of 7 and the coordinates of the given point to formulate the line's equation. The equation can then be rearranged into the general form \(Ax + By = C\), which often better suits certain analytical purposes.