Question

Write an equation in slope-intercept form for the line that passes through (5,0) and is perpendicular to the line described by y=-5/2x+6

Answer 1

For this case we have to;

We have that an equation in slope-intercept form is given by:

$y = mx + b$

Where:

m is the slope

b is the cut point with the y axis

Also, by definition, two lines are perpendicular when the product of their slopes is -1. That is:$m_ {1} * m_ {2} = - 1$

We have the line as data: $y_ {1} = \frac {-5} {2} x + 6$

Then $m_ {1} = \frac {-5} {2}$

We found$m_ {2}$:

$m_ {1} * m_ {2} = - 1$

$\frac {-5} {2} * m_ {2} = - 1$

$m_ {2} = \frac {-1} {(\frac {-5} {2})}$

$m_ {2} = \frac {(2) (- 1)} {(- 5) (1)}$

$m_ {2} = \frac {2} {5}$

Thus, $y_ {2} = \frac {2} {5} x_ {2} + b_ {2}$

We must find $b_ {2}$:

We know that $y_ {2}$ passes through the point$(x_ {2}, y_ {2}) = (5,0)$

We substitute the point in the equation of $y_ {2}$:

$0 = \frac {2} {5} (5) + b_ {2}\\0 = 2 + b_ {2}\\b_ {2} = - 2$

Thus, $y_ {2} = \frac {2} {5} x_ {2} -2$

Then the equation in slope-intercept for the line that passes through (5,0) and is perpendicular to the line described by $y_ {1} = \frac {-5} {2} x_{1} + 6$ is: $y_ {2} = \frac {2} {5} x_ {2} -2$

Answer:

$y_ {2} = \frac {2} {5} x_ {2} -2$