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Which of the following is an equation of the line in the xy-plane that passes through the point (0, −3) and is perpendicular to the line y = −4x + 7?

equation can be rewritten as 4x+y=7

an equation perpendicular to this equation will be x-4Y=k..........(1)
as this line pass through point (0, -3), plug in the x & y in (1)
0-4(-3)=k and K=12

equation of the line will be x-4y=12 or y=x/4 -3

option D
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↧↧↧ Weekly Video Solution to the Problem Series ↧↧↧



We need to find equation of the line in the xy-plane that passes through the point (0, −3) and is perpendicular to the line y = −4x + 7

Let's find the slope of line y = −4x + 7

We know that generic equation of a line is y = mx + b (Where m is the slope of the line)
=> Slope of line y = −4x + 7 = -4

Now, the line whose equation we have to find is perpendicular to y = −4x + 7
=> Slope,m of this line can be found by m * -4 = -1
=> m = \(\frac{1}{4}\)

Equation of a line passing through the point (0, -3) and having a slope of \(\frac{1}{4}\) is given by

y - \(y_1\) = m * (x - \(x_1\))
=> y - (-3) = \(\frac{1}{4}\) * (x - 0)
=> y + 3 = \(\frac{1}{4}\) x
=> \(y =\frac{ 1}{4} x − 3\)

So, Answer will be D
Hope it helps!

Watch the following video to learn the Basics of Slope of a Line

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