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On a coordinate plane, a line goes through (negative 1, 1) and (0, negative 3). a point is at (negative 4, negative 3) and (0, negative 3). what is the equation, in point-slope form, of the line that is perpendicular to the given line and passes through the point (−4, −3)? y 3 = −4(x 4) y 3 = –one-fourth(x 4) y 3 = one-fourth(x 4) y 3 = 4(x 4)

Sagot :

The equation of the required line would be x - 4y = 8.

What is the slope-point form of the line?

Point-slope is the general form y-y₁=m(x-x₁) for linear equations. It emphasizes the slope of the line and a point on the line (that is not the y-intercept).

A line goes through the point (-1, 1) and (0,-3)

so the slope of the given line would be:

                     [tex]m=\frac{y_2-y_1}{x_2-x_1}\\\\m=\frac{-3-1}{0-(-1)}\\\\m=-4[/tex]

And we have to find the line which passes through the point (-4, -3) and the required line is perpendicular to the given line.

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

So the slope of the required line would be = 1/4

By using a slope-point form of the line, we get

    [tex]y - y_1=m(x-x_1)\\\\y-(-3)=\frac{1}{4}(x-(-4))\\\\4(y+3)=x+4\\\\4y+12=x+4\\\\x-4y=8[/tex]

Hence, the equation of the required line would be x - 4y = 8.

To learn more about the slope-point form of the line, visit:

https://brainly.com/question/24907633

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