Question

What is the equation of the line that passes through the point (5,4) and is perpendicular to the line whose equation is 2x + y = 3in standard form?

2x+y=3y=-2x+3
m=-2m=1/2
y-4=1/2(x-5)

Answer 1

The equation of the line that passes through the point (5,4) and is perpendicular to the line whose equation is 2x + y = 3 in standard form is x - 2y = - 3

Solution:

Given, line equation is 2x + y = 3

2x + y – 3 = 0 ----- eqn (1)

We have to find a line that is perpendicular to 2x + y – 3 = 0 and passing through (5, 4).

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`

`\text { Slope of a line } * \text { slope of perpendicular line }=-1`

`\begin{array}{l}{-2 * \text { slope of perpendicular line }=-1} \\\\ {\text { Slope of perpendicular line }=-1 * (1)/(-2)=(1)/(2)}\end{array}`

Now, slope of our required line = 1/2 and it passes through (5, 4)

The point slope form is given as:

`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. }`

`\text { Here in our problem, } \mathrm{m}=(1)/(2), \text { and }\left(\mathrm{x}_(1), \mathrm{y}_(1)\right)=(5,4)`

`y-4=(1)/(2)(x-5)`

Now let us convert to standard form:

The standard form of a line is just another way of writing the equation of a line.

The standard form of an equation is Ax + By = C. In this kind of equation, x and y are variables and A, B, and C are integers.

2(y – 4) = 1(x - 5)

2y – 8 = x – 5

x – 2y - 5 + 8 = 0

x - 2y + 3 = 0

x - 2y = - 3

Hence, the line equation in standard form is x - 2y = - 3