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Chapter 3: Problem 55

Find the \(x\) - and \(y\) -intercepts. Then graph each equation. $$ -\frac{2}{3} y=x $$

Short Answer

Expert verified
The x- and y-intercepts are both at (0, 0). Graph the line through (0,0) and (2,-3).

Step by step solution

01

- Find the y-intercept

To find the y-intercept, set \(x = 0\) and solve for \(y\). \[-\frac{2}{3} y = 0\]The equation simplifies to \(y = 0\). Hence, the y-intercept is the point \( (0, 0) \).
02

- Find the x-intercept

To find the x-intercept, set \(y = 0\) and solve for \(x\). \[-\frac{2}{3} (0) = x\]Simplifying, we get \(x = 0\). Hence, the x-intercept is the point \( (0, 0) \).
03

- Graph the equation

To graph the equation, plot the intercept point \((0, 0)\). Since this is the only intercept and the line is linear, it passes through the origin with a slope determined by the coefficient of \(y\).Rewrite the equation in slope-intercept form \(y = mx + b\):\[y = -\frac{3}{2}x\]The slope is \(-\frac{3}{2}\). Starting from the point \((0,0)\), use the slope to identify another point. For example, from \((0,0)\), go down 3 units and right 2 units to find the point \((2, -3)\).
04

- Draw the line

Draw a straight line through the points \((0, 0)\) and \((2, -3)\) to graph the entire equation.

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Key Concepts

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

x-intercept
Let's start by understanding the concept of the x-intercept. The x-intercept of a line is the point where the graph of the equation crosses the x-axis. This means that at the x-intercept, the value of y is always 0. To find the x-intercept, we plug y = 0 into the equation and solve for x. In our example, we set y to 0 in the equation \(-\frac{2}{3} y = x\), simplifying it to \(0 = x\). Therefore, the x-intercept is (0, 0). If y wasn't 0, we would repeat the process accordingly.
y-intercept
Next, let's tackle the y-intercept. The y-intercept is where the line crosses the y-axis. Here, the value of x is always 0. To find the y-intercept of a linear equation, we set x to 0 and solve for y. In our given equation \(-\frac{2}{3} y = x\), substituting x with 0 gives us \(-\frac{2}{3} y = 0\). Solving this equation, we see that y equals 0. Thus, the y-intercept is also the point (0, 0). Once you have the y-intercept, you'll have a clearer picture of where your line starts on the graph.
slope-intercept form
Let's explore the slope-intercept form of a linear equation, which is written as \ (y = mx + c) \ where m is the slope and c is the y-intercept. This form is incredibly useful for graphing linear equations. In our case, the initial equation \(-\frac{2}{3} y = x\) can be rewritten as \ (y = -\frac{3}{2} x)\, after rearranging. Here, m = -\frac{3}{2} indicates the slope. The slope tells us how steep the line is and the direction it moves. A negative slope like \ (-\frac{3}{2})\ tells us the line will go down as we move from left to right.
plotting points
Now we move on to plotting points. This is a practical method used to graph the line on a coordinate plane. Starting with our intercept points, at (0, 0), which is both the x- and y-intercept. To plot another point, use the slope \(-\frac{3}{2}\). Start from (0, 0), and follow the slope: go down 3 units and then 2 units to the right. This gives you the point (2, -3). By plotting these points, (0, 0) and (2, -3), on a graph and connecting them with a straight line, you effectively graph the equation. Repeat the process if necessary for more accuracy.

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Most popular questions from this chapter

Solve each inequality. $$ -x+4>3 $$

Find the slope of the line through each pair of points.\(\left(\text {Hint:} \frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b} \div \frac{c}{d}\right)\). $$ \left(-\frac{2}{9}, \frac{5}{18}\right) \text { and }\left(\frac{1}{18},-\frac{5}{9}\right) $$

Segment PQ has the given coordinates for one endpoint P and for its midpoint M. Find the coordinates of the other endpoint \(Q .\) (Hint: Represent \(Q\) by \((x, y)\) and write two equations using the midpoint formula, one involving \(x\) and the other involving \(y .\) Then solve for \(x\) and \(y .\) $$ P(5,8), M(8,2) $$

Find an equation of the line that satisfies the given conditions. (a) Write the equation in slope-intercept form. (b) Write the equation in standard form. Through \((-2,-2) ;\) parallel to \(-x+2 y=10\)

Find an equation of the line passing through the given points. (a) Write the equation in standard form. (b) Write the equation in slope-intercept form if possible. $$ (13,5) \text { and }(13,-1) $$
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