Question

Write an equation of the line that is perpendicular to 5x+20y=10 and passes through the point (8,3)

Answer 1

Write an equation of the line that is perpendicular to 5x+20y=10 and passes through the point (8,3)

Answer 2

Answer: The required equation of the line is
`y=4x-29.`

Step-by-step explanation: We are given to write the equation of the line that is perpendicular to the following line and passes through the point (8, 3).

`5x+20y=10~~~~~~~~~~~~~~~~~~~~~~~~~(i)`

We know that

the slope-intercept form of a straight line is given by

`y=mx+c,`

where m is the slope and c is the y-intercept of the line.

From equation (i), we have

`5x+20y=10\\\\\Rightarrow 20y=-5x+10\\\\\\\Rightarrow y=-(5)/(20)x+(10)/(20)\\\\\\\Rightarrow y=-(1)/(4)x+(1)/(2).`

So,

`\textup{slope, }m=-(1)/(4)~~\textup{and}~~\textup{y-intercept, }c=(1)/(2)`

Since the product of the slopes of two perpendicular lines is - 1.

Let, m' be the slope of the line perpendicular to line (i).

Then, we must have

`m* m'=-1\\\\\\\Rightarrow -(1)/(4)* m'=-1\\\\\Rightarrow m'=4.`

Since the line passes through the point (8, 3), so its equation will be

`y-3=m'(x-8)\\\\\Rightarrow y-3=4(x-8)\\\\\Rightarrow y=4x-32+3\\\\\Rightarrow y=4x-29.`

Thus, the required equation of the line is
`y=4x-29.`