Problem 63
Question
Find an equation for the line satisfying the given conditions. Through (-1,3) and perpendicular to the line through (0,1) and (2,3).
Step-by-Step Solution
Verified Answer
Question: Find the equation of the line that passes through the point (-1,3) and is perpendicular to the line passing through the points (0,1) and (2,3).
Answer: The equation of the line is x + y - 2 = 0.
1Step 1: Finding the slope of the line passing through the points (0,1) and (2,3)
To find the slope, use the slope formula, which is:
slope = (y2 - y1) / (x2 - x1)
Using the points (0,1) as (x1,y1) and (2,3) as (x2,y2):
slope = (3 - 1) / (2 - 0)
This simplifies to:
slope = 2 / 2 = 1
The slope of the line passing through (0,1) and (2,3) is 1.
2Step 2: Finding the perpendicular slope
Now that we have the slope of the line passing through (0,1) and (2,3), we need to find the slope of a line that is perpendicular to it. To do this, we'll use the fact that the product of the slopes of two perpendicular lines equal -1. So, if the slope of our original line is 1, the perpendicular slope is -1, because 1 * (-1) = -1.
The perpendicular slope is -1.
3Step 3: Finding the equation of the line passing through the point (-1,3) and with a perpendicular slope
We have the point (-1,3) which the required line passes through and the slope of the required line is -1. Now we'll use the point-slope form of the equation to find the equation of the line passing through this point. The point-slope form is:
y - y1 = slope * (x - x1)
Plugging in our given point (-1,3) as (x1, y1) and the perpendicular slope (-1) as the slope, we get:
y - 3 = -1 * (x - (-1))
Now, we simplify the equation:
y - 3 = -1 * (x + 1)
y - 3 = -x - 1
To write the equation in the standard form, add x to both sides of the equation:
x + y - 3 + 1 = 0
x + y - 2 = 0
The equation of the line passing through the point (-1,3) and perpendicular to the line through (0,1) and (2,3) is x + y - 2 = 0.
Key Concepts
Slope of a LinePoint-Slope FormPerpendicular SlopesEquation of a Line
Slope of a Line
Understanding the slope of a line is fundamental in Algebra. The slope is a measure of the steepness of a line, and it's usually represented by the letter 'm'. We calculate the slope by taking the difference in the y-coordinates of two points on the line (rise) and dividing it by the difference in the x-coordinates (run). The formula is as follows:\[\[\begin{align*}m = \frac{y_2-y_1}{x_2-x_1}\end{align*}\]\]For example, given two points on a line, say (0,1) and (2,3), the slope 'm' would be calculated by taking the change in y-values, which is \(3-1\), and dividing it by the change in x-values, \(2-0\), yielding a slope of 1. A positive slope indicates that the line rises as it moves from left to right, while a negative slope indicates that the line falls. A slope of zero means the line is horizontal and an undefined slope indicates a vertical line. Slopes are crucial when discussing lines, as they provide insight into the direction and angle of the line in relation to the axes.
Point-Slope Form
One method for writing the equation of a line is the point-slope form, particularly when you know a point on the line and its slope. The general form looks like this:\[\[\begin{align*}y - y_1 = m(x - x_1)\end{align*}\]\]In this equation, \(m\) represents the slope, and \(x_1, y_1\) are the coordinates of the known point through which the line passes. This form is exceptionally useful for creating an equation of a line when you do not have it in the y-intercept form. For instance, if you have a line with a slope of -1 that passes through the point (-1,3), the equation using the point-slope form would be \((y - 3) = -1(x + 1)\), which can be simplified and rearranged to fit different equation standards such as slope-intercept or standard form.
Perpendicular Slopes
When discussing perpendicular lines, we focus on their slopes and a unique relationship that exists between them. Two lines are perpendicular if and only if the product of their slopes is -1. This means that if one line has a slope of \(m\), the perpendicular line will have a slope of \(-\frac{1}{m}\). This relationship is critical in determining unknown slopes of lines that need to be perpendicular to existing ones.For example, given a line with a slope of 1, a line that is perpendicular to it must have a slope that multiplies with 1 to get -1. This leads us to find that the perpendicular slope is -1. This concept helps in drawing right angles and is widely used in geometry, trigonometry, and various applications such as construction, computer graphics, and more.
Equation of a Line
The equation of a line is an algebraic representation of the line on a coordinate plane. There are multiple forms to represent the equation of a line, including slope-intercept form \(y = mx + b\), standard form \(Ax + By = C\), and point-slope form, as previously discussed. Each form is useful depending on the given information.The process of finding the equation of a line involves identifying the slope and using a given point to determine the equation's specific constants. For instance, with a slope of -1 and passing through the point (-1,3), one can start with the point-slope form and manipulate the equation to reach the preferred form. When rearranged to the standard form, the equation \((x + y - 2 = 0\)) succinctly defines the line's behavior on the graph. The ability to write the equation of a line in multiple forms allows for flexibility in mathematical problem-solving and in various practical applications.
Other exercises in this chapter
Problem 62
Simplify the expression without using a calculator. Your answer should not have any radicals in it. $$\sqrt{12} \sqrt{3}$$
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Find all real solutions of the equation exactly. $$6 z^{4}-7 z^{2}+2=0$$
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Find the center and radius of the circle whose equation is given. $$x^{2}+y^{2}+8 x-6 y-15=0$$
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Simplify the expression without using a calculator. Your answer should not have any radicals in it. $$\sqrt{\frac{3}{5}} \sqrt{\frac{12}{5}}$$
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