Use the point-slope form y − y 1 = m ( x − x 1 ) with m = − 9 2 and ( 4 , − 5 ) to find the point-slope equation.
Substitute the given values into the point-slope form: y − ( − 5 ) = − 9 2 ( x − 4 ) , which simplifies to y + 5 = − 9 2 ( x − 4 ) .
Convert the point-slope form to slope-intercept form y = m x + b by distributing and isolating y .
Simplify the equation to find the slope-intercept form: y = − 9 2 x − 9 37 .
The equation of the line in point-slope form is y + 5 = − 9 2 ( x − 4 ) , and the equation of the line in slope-intercept form is y = − 9 2 x − 9 37 .
Explanation
Understanding the Problem We are given the slope of a line, m = − 9 2 , and a point it passes through, ( 4 , − 5 ) . Our goal is to find the equation of this line in both point-slope form and slope-intercept form.
Using Point-Slope Form The point-slope form of a line is given by the equation y − y 1 = m ( x − x 1 ) , where m is the slope and ( x 1 , y 1 ) is a point on the line. We can plug in the given values m = − 9 2 and ( x 1 , y 1 ) = ( 4 , − 5 ) into this equation.
Point-Slope Equation Substituting the given values, we get y − ( − 5 ) = − 9 2 ( x − 4 ) . Simplifying this, we have y + 5 = − 9 2 ( x − 4 ) . This is the equation of the line in point-slope form.
Finding Slope-Intercept Form Now, let's find the equation of the line in slope-intercept form, which is given by y = m x + b , where m is the slope and b is the y-intercept. We can start from the point-slope form we found earlier: y + 5 = − 9 2 ( x − 4 ) .
Distributing the Slope First, distribute the slope on the right side of the equation: y + 5 = − 9 2 x + 9 8 .
Isolating y Next, isolate y by subtracting 5 from both sides: y = − 9 2 x + 9 8 − 5 . To combine the constants, we need a common denominator. Since 5 = 9 45 , we have y = − 9 2 x + 9 8 − 9 45 .
Slope-Intercept Equation Finally, simplify the equation: y = − 9 2 x − 9 37 . This is the equation of the line in slope-intercept form.
Final Answer Therefore, the equation of the line in point-slope form is y + 5 = − 9 2 ( x − 4 ) , and the equation of the line in slope-intercept form is y = − 9 2 x − 9 37 .
Examples
Understanding linear equations is crucial in many real-world applications. For instance, if you are tracking the depreciation of an asset, a linear equation can model the decrease in value over time. Imagine a car that loses a fixed amount of its value each year. If the car was initially worth $20,000 and depreciates by 2 , 000 e a c h ye a r , t h ec a r ′ s v a l u ec anb e m o d e l e d b y t h ee q u a t i o n y = -2000x + 20000 , w h ere y i s t h e v a l u eo f t h ec a r a f t er x$ years. This allows you to predict the car's value at any point in time. Similarly, linear equations are used in calculating simple interest, determining the cost of services with a fixed rate, and modeling various other real-life scenarios where a quantity changes at a constant rate.
The equation of the line in point-slope form is y + 5 = − 9 2 ( x − 4 ) . In slope-intercept form, the equation is y = − 9 2 x − 9 37 . This demonstrates the relationship between the slope and the y-intercept based on the given point.
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