JohnWatkins345
contestada

Which of these is a point-slope equation of the line that is perpendicular to y-3=4(x-10) and passes through(-3,7)

A. y+7=4(x-3)
B. y-7= -1/4(x+3)
C. y+7= -1/4(x-3)
D. y-7= -4x(x+3)

Respuesta :

evgeniylevi
Try this option:
1. rule: if y+a₁=k₁(x+b₁) ⊥ y+a₂=k₂(x+b₂), then k₁*k₂= -1.
for unknown line k= -1/4.
2. substituting the coordinates (-3;7) into the equations, it is possible to find that 'y-7' and 'x+3'.
3. at the end, if to make up the equation: y-7= -1/4(x+3).

Answer: B
yoodyannapolis

The point-slope equation of the line that is perpendicular to y-3=4(x-10) and passes through(-3,7) is:

B. [tex]y-7=\frac{-1}{4} (x+3)\\[/tex]

The slope intercept form is y=mx+b where m is slope.

We have given perpendicular line as:

y-3=4(x-10)

y-3=4x-40

y=4x-37

The slope intercept form of the given line is

y=4x-37

Slope of this line is m=4

Now, as we know the two perpendicular lines have negative reciprocal slope.

Thus the other line have slope [tex]m=-\frac{1}{4}[/tex]

Now. the equation of line passing through point (-3,7) and having slope [tex]\frac{-1}{4}[/tex] is:

[tex]y-7=\frac{-1}{4} (x+3)\\[/tex]

Therefore the correct option is B.

Learn more: brainly.com/question/18595188

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