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Chapter 1: Problem 9

In Exercises 5 through 14, find an equation of the line satisfying the given conditions. $$ \text { The } x \text { intercept is }-3, \text { and the } y \text { intercept is } 4 \text {. } $$

Short Answer

Expert verified
The equation is \( y = \frac{4}{3} x + 4 \).

Step by step solution

01

Define the intercepts

The given intercepts are the x-intercept at \(x = -3\) and the y-intercept at \(y = 4\). This means the line passes through the points \((-3, 0)\) and \((0, 4)\).
02

Use the two-point form of a line's equation

The two-point form of a line's equation is given by \[y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1) \] where \((x_1, y_1)\) and \((x_2, y_2)\) are the given points.
03

Substitute the given points

Substitute \((-3, 0)\) and \((0, 4)\) into the formula:\[ y - 0 = \frac{4 - 0}{0 + 3} (x + 3) \]Simplify the fraction:\[ y = \frac{4}{3} (x + 3) \]
04

Distribute and simplify

Distribute \( \frac{4}{3} \):\[ y = \frac{4}{3} x + 4 \]This is the equation of the line in slope-intercept form \( y = mx + b \).

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Key Concepts

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

x-intercept
The x-intercept of a line is where the line crosses the x-axis. This happens when the value of y is zero. In this exercise, the x-intercept is \-3. This means the line touches the x-axis at the point \((-3, 0)\). Understanding the x-intercept helps in constructing the line's equation and visualizing how it behaves on a graph. Knowing the x-intercept is also crucial in breaking down more complex algebra problems into simpler parts.
y-intercept
The y-intercept is the point where the line crosses the y-axis. At this point, the value of x is zero. In this problem, the y-intercept is 4, meaning the line touches the y-axis at the point \((0, 4)\). The y-intercept gives us a starting point to graph the line and is a basic component in the equation of a line, helping further in understanding its slope.
two-point form
The two-point form of a line's equation resolves finding an equation when you know two specific points on the line. The formula is \[: y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1) \]. Here, \(x_1, y_1\) and \(x_2, y_2\) are the coordinates of the given points. In this problem, \((-3, 0)\) and \((0, 4)\) are used. Substituting these into the equation: \[: y - 0 = \frac{4 - 0}{0 + 3} (x + 3) \]. This simplifies to \[: y = \frac{4}{3} (x + 3) \]. This form is handy in cases where neither the slope nor specific forms are initially given.
slope-intercept form
The slope-intercept form of a line's equation is easily recognizable and tremendously useful for graphing. It is given by \[: y = mx + b \]. Here, \(m\) represents the slope of the line, and \(b\) is the y-intercept. In general, the slope \(m\) shows how steep the line is and in what direction it goes. In our problem, the equation was simplified to \[: y = \frac{4}{3} x + 4 \], where the slope \frac{4}{3}\ and the y-intercept is \4 \. This form becomes quite intuitive and straightforward for plotting and interpreting linear equations.

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Most popular questions from this chapter

In Exercises 15 through 20 , determine whether the graph is a circle, a point- circle, or the empty set. $$ x^{2}+y^{2}+2 x-4 y+5=0 $$

In Exercises 11 through 34, the function is the set of all ordered pairs \((x, y)\) satisfying the given equation. Find the domain and range of the function, and draw a sketch of the graph of the function. $$ F: y=\frac{x^{4}+x^{3}-9 x^{2}-3 x+18}{x^{2}+x-6} $$

Find an equation of the common chord of the two circles \(x^{2}+y^{2}+4 x-6 y-12=0\) and \(x^{2}+y^{2}+8 x-2 y+8=0\). (HINT: If the coordinates of a point satisfy two different equations, then the coordinates also satisfy the difference of the two equations.)

In Exercises 5 through 10, find an equation of the circle satisfying the given conditions. Tangent to the line \(3 x+4 y-16=0\) at \((4,1)\) and with a radius of 5 . (Two possible circles.)

In Exercises 1 through 4, find an equation of the circle with center at \(C\) and radius \(r\). Write the equation in both the centerradius form and the general form. $$ C(-5,-12), r=3 $$
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