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)

Respuesta :

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,  

[tex]\text { Slope of a line }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-2}{1}=-2[/tex]

[tex]\text { Slope of a line } \times \text { slope of perpendicular line }=-1[/tex]

[tex]\begin{array}{l}{-2 \times \text { slope of perpendicular line }=-1} \\\\ {\text { Slope of perpendicular line }=-1 \times \frac{1}{-2}=\frac{1}{2}}\end{array}[/tex]

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

The point slope form is given as:

[tex]y-y_{1}=m\left(x-x_{1}\right)[/tex]

[tex]\text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on the line. }[/tex]

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

[tex]y-4=\frac{1}{2}(x-5)[/tex]

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

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