Question

Consider the line y=5x-4

Find the equation of the line that is perpendicular to this line and passes through the points (-4,5)

Find the equation of the line that is parallel to this line and passes through the point (-4,5)

Note: equation need for both the perpendicular and parallel question

Answer 1

Explanation:

1.

The equation of the line is y=mx+b

The equation of the line that is perpendicular to this line is y=(-1/m)x+c

y=5x-4

Hence

`\displaystyle\\y=-(1)/(5) x+c \\\\y=-0.2x+c\ \ \ \ \ (1)\\\\Let's \ substitute\ coordinate \ (-4,5)\ x=-4\ and\ y=5 \ into\ formula\ (1):\\\\5=-0.2(-4)+c\\\\5=0,8+c\\\\5-0.8=0.8+c-0.8\\\\4.2=c\\\\Thus,\ c=4.2\\\\Hence, \ y=-0,2x+4.2`

2.

The equation of the line is y=mx+b

The equation of the line that is parallel to this line is y=mx+b

y=5x-4

Hence

y=5x+b (2)

Let's substitute coordinate (-4,5) x=-4 and y=5 into formula (2):

5=5(-4)+b

5=-20+b

5+20=-2-+b+20

25=b

Thus, b=25

Hence, y=5x+25

Answer 2

point slope: y - y1 =m(x-x1)

Find the equation of the line that is perpendicular to this line and passes through the points (-4,5)

slope of perpendicular lines are negative reciprocals of each other

y=5x-4 has slope of 5 so perpendicular line will have slope of -1/5

equation will be y -5 = -1/5(x+4).

Find the equation of the line that is parallel to this line and passes through the point (-4,5)

parallel lines have the same slope

equation will be y - 5 = 5(x+4)