Question

Write an equation of the line that passes through (-5,2) and is perpendicular to y+3=2x

Answer 1

The line that passes through (-5,2) and is perpendicular to y+3=2x is
`y=(-x)/(2)+(-1)/(2)`

Solution:

Given, line equation is y + 3 = 2x ⇒ 2x – y – 3 = 0

We have to find a line that is perpendicular to y + 3 = 2x and passing through (-5, 2).

Now, let us find the slope of the given line,

`\text { Slope of a line }=\frac{-x \text { coefficient }}{y \text { coefficient }}=(-2)/(-1)=2`

We know that, slope of a line
`*` slope of perpendicular line = -1

Then, 2
`*` slope of perpendicular line = -1

`\rightarrow \text { slope of perpendicular line }=-1 * (1)/(2)=-(1)/(2)`

Now, slope of our required line =
`(-1)/(2)` and it passes through (-5, 2)

We know that, point slope form is

`y-y_(1)=m\left(x-x_(1)\right) \text { where } m \text { is slope and }\left(x_(1), y_(1)\right) \text { is point on the line. }`

Here in our problem,
`m=-(1)/(2), \text { and }\left(x_(1), y_(1)\right)=(-5,2)`

Then, line equation
`\rightarrow y-2=-(1)/(2)(x-(-5))`

2(y – 2) = -1(x + 5)

2y – 4 = -x – 5

x + 2y + 5 – 4 = 0

`y=(-x)/(2)+(-1)/(2)`

Thus the point slope form is
`y=(-x)/(2)+(-1)/(2)`