Use the point-slope form of a line: y − y 1 = m ( x − x 1 ) .
Substitute the given point ( 1 , − 1 ) and slope − 5 into the point-slope form: y − ( − 1 ) = − 5 ( x − 1 ) .
Simplify the equation: y + 1 = − 5 x + 5 .
Solve for y to get the equation in slope-intercept form: y = − 5 x + 4 .
The equation of the line in slope-intercept form is y = − 5 x + 4 .
Explanation
Understanding the Problem We are given a point and a slope, and we want to find the equation of the line in slope-intercept form. The slope-intercept form of a line is given by y = m x + b , where m is the slope and b is the y-intercept.
Using Point-Slope Form We are given the point ( 1 , − 1 ) and the slope m = − 5 . We can use the point-slope form of a line, which is y − y 1 = m ( x − x 1 ) , where ( x 1 , y 1 ) is the given point.
Substituting Values Substitute the given point ( 1 , − 1 ) and slope m = − 5 into the point-slope form:
y − ( − 1 ) = − 5 ( x − 1 )
Simplifying the Equation Simplify the equation:
y + 1 = − 5 x + 5
Solving for y Solve for y to get the equation in slope-intercept form:
y = − 5 x + 5 − 1
y = − 5 x + 4
Final Equation The equation of the line in slope-intercept form is y = − 5 x + 4 .
Examples
Understanding linear equations is crucial in many real-world applications. For instance, if you are tracking the depreciation of a car, the value of the car decreases linearly over time. If the car's initial value is $20,000 and it depreciates at a rate of 2 , 000 p erye a r , t h ec a r ′ s v a l u e y a f t er x ye a rsc anb e m o d e l e d b y t h ee q u a t i o n y = -2000x + 20000 . S imi l a r l y , in p h ys i cs , t h e d i s t an ce t r a v e l e d a t a co n s t an t s p ee d c anb e m o d e l e d u s in g a l in e a re q u a t i o n . I f a t r ain t r a v e l s a t a co n s t an t s p ee d o f 80 mi l es p er h o u r , t h e d i s t an ce y i tt r a v e l s in x h o u rs i s g i v e nb y y = 80x$. These examples demonstrate how linear equations help us understand and predict real-world phenomena.