What is the equation of the line that is perpendicular to the like y=-4x + 7 and passes through the point (8,2)?

Respuesta :

Answer:

[tex]y=\frac{1}{4}x[/tex]

Step-by-step explanation:

Given equation of line:

[tex]y=-4x+7[/tex]

To find the equation of line perpendicular to the line of the given equation and passes through point (8,2).

Applying slope relationship between perpendicular lines.

[tex]m_1=-\frac{1}{m_2}[/tex]

where [tex]m_1[/tex] and [tex]m_2[/tex] are slopes of perpendicular lines.

For the given equation in the form [tex]y=mx+b[/tex] the slope [tex]m_2[/tex]can be found by comparing [tex]y=-4x+3[/tex] with standard form.

∴ [tex]m_2=-4[/tex]

Thus slope of line perpendicular to this line [tex]m_1[/tex] would be given as:

[tex]m_1=-\frac{1}{-4}[/tex]

∴ [tex]m_1=\frac{1}{4}[/tex]

The line passes through point (8,2)

Using point slope form:

[tex]y_-y_1=m(x_-x_1)[/tex]

Where [tex](x_1,y_1)\rightarrow (8,2)[/tex] and [tex]m=m_1=\frac{1}{4}[/tex]

So,

[tex]y-2=\frac{1}{4}(x-8)[/tex]

Using distribution.

[tex]y-2=(\frac{1}{4}x)-(\frac{1}{4}\times 8)[/tex]

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

Adding 2 to both sides.

[tex]y-2+2=\frac{1}{4}x-2+2[/tex]

[tex]y=\frac{1}{4}x[/tex]

Thus the equation of line in standard form is given by:

[tex]y=\frac{1}{4}x[/tex]

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