Problem 50
Question
Find the slope and y-intercept of the line, and draw its graph. $$ -3 x-5 y+30=0 $$
Step-by-Step Solution
Verified Answer
Slope: \(-\frac{3}{5}\); Y-intercept: 6; Graph by plotting (0,6) and using slope.
1Step 1: Rewrite the Equation in Slope-Intercept Form
The slope-intercept form of a line is given by the equation \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. Start by manipulating the given equation to match this form. The given equation is \(-3x - 5y + 30 = 0\). To isolate \( y \), first move \(-3x\) and \(30\) to the other side by adding them: \(-5y = 3x - 30\).
2Step 2: Solve for y
Continue solving for \( y \) by dividing every term by \(-5\) to get \( y = -\frac{3}{5}x + 6\). This is the slope-intercept form of the line where the coefficient of \( x \) represents the slope \( m \) and the constant term is the y-intercept \( b \).
3Step 3: Identify Slope and Y-Intercept
From the slope-intercept form \( y = -\frac{3}{5}x + 6 \), identify that the slope \( m \) is \( -\frac{3}{5} \) and the y-intercept \( b \) is \( 6 \).
4Step 4: Graph the Line
Begin by plotting the y-intercept \( (0, 6) \) on the graph. Use the slope \( -\frac{3}{5} \), which means "down 3 units, right 5 units," to find another point. From \( (0, 6) \), move down 3 units and 5 units to the right to reach \( (5, 3) \). Draw a straight line through these points to graph the line.
Key Concepts
Y-InterceptSlopeGraphing Linear Equations
Y-Intercept
A line's y-intercept is where it crosses the y-axis. This point is pivotal in graphing linear equations, as it provides a starting point to plot the line. In slope-intercept form, expressed as \( y = mx + b \), the \( b \) value is the y-intercept.
To understand further, consider that when \( x = 0 \), the equation simplifies to \( y = b \). This means the line intercepts the y-axis at \( (0, b) \).
In our example equation \( y = -\frac{3}{5}x + 6 \), \( b \) is \( 6 \). Consequently, the y-intercept is at the point \( (0, 6) \).
This point is always plotted first when graphing, helping guide the graph's orientation and position.
To understand further, consider that when \( x = 0 \), the equation simplifies to \( y = b \). This means the line intercepts the y-axis at \( (0, b) \).
In our example equation \( y = -\frac{3}{5}x + 6 \), \( b \) is \( 6 \). Consequently, the y-intercept is at the point \( (0, 6) \).
This point is always plotted first when graphing, helping guide the graph's orientation and position.
Slope
The slope of a line indicates its steepness and direction. Represented by \( m \) in the slope-intercept form \( y = mx + b \), slope reflects how much the y-value changes for a given change in the x-value.
For instance, a slope of \( -\frac{3}{5} \) signifies that for every 5 units you move to the right along the x-axis, you move 3 units down on the y-axis. A positive slope means the line ascends, while a negative slope implies it descends.
For instance, a slope of \( -\frac{3}{5} \) signifies that for every 5 units you move to the right along the x-axis, you move 3 units down on the y-axis. A positive slope means the line ascends, while a negative slope implies it descends.
- Zero slope: Line is horizontal.
- Undefined slope: Line is vertical.
Graphing Linear Equations
Graphing linear equations involves plotting a line on a coordinate plane that reflects the equation's slope and y-intercept.
This process starts with transforming the equation into the slope-intercept form \( y = mx + b \). For our line, the equation is \( y = -\frac{3}{5}x + 6 \).
Steps to graphing:
This process starts with transforming the equation into the slope-intercept form \( y = mx + b \). For our line, the equation is \( y = -\frac{3}{5}x + 6 \).
Steps to graphing:
- First, plot the y-intercept. Here, the intercept is \( (0, 6) \).
- Use the slope to find another point. For \( -\frac{3}{5} \), start at \( (0, 6) \), move 3 units down and 5 units right to arrive at \( (5, 3) \).
- Draw a straight line through these points.
Other exercises in this chapter
Problem 49
\(49-54\) . Find the center and radius of the circle, and sketch its graph. $$ x^{2}+y^{2}=9 $$
View solution Problem 49
Find a point on the \(y\) -axis that is equidistant from the points \((5,-5)\) and \((1,1) .\)
View solution Problem 50
\(49-54\) . Find the center and radius of the circle, and sketch its graph. $$ x^{2}+y^{2}=5 $$
View solution Problem 50
Find the lengths of the medians of the triangle with vertices \(A(1,0), B(3,6),\) and \(C(8,2) .\) (A median is a line segment from a vertex to the midpoint of
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