PLEASE HELP!!! Write the equation of a line that goes through the point (6, 3) and is perpendicular to the line y = 4x + 1.

y−3=4(x−6)

y−3=−14(x−6)

y−6=4(x−3)

y−6=−14(x−3)

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absor201 absor201 · Answer 1 · 2019-12-16T16:58:19+01:00

Equation of line that goes through the point (6, 3) and is perpendicular to the line y = 4x + 1 is: [tex]y-3 = -\frac{1}{4}(x-6)[/tex]

Step-by-step explanation:

Given equation of line is:

[tex]y = 4x+1[/tex]

As the equation is in slope-intercept form, the coefficient of x will be the slope of line

Let m1 be the slope of the given line

m1 = 4

As we know, product of slopes of two perpendicular lines is -1.

Let m2 be the slope of required line

[tex]m_1.m_2 = -1\\4 . m_2 = -1\\m_2 = -\frac{1}{4}[/tex]

Point-slope form of equation is:

[tex]y-y_1=m_2(x-x_1)[/tex]

Here

(x1,y1) = (6,3)

Putting the point and slope in point-slope form

[tex]y-3 = -\frac{1}{4}(x-6)[/tex]

Hence,

Equation of line that goes through the point (6, 3) and is perpendicular to the line y = 4x + 1 is: [tex]y-3 = -\frac{1}{4}(x-6)[/tex]

Keywords: Point-slope form, Equation of line

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