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Veseljchak [2.6K]
4 years ago
5

Find the equation in slope intercept form and standard form of the line that passes through (4,-3) and is perpendicular to 3x-y=

5.
Mathematics
1 answer:
Tcecarenko [31]4 years ago
5 0

The given line is y = 3x - 5 after adding Y and subtracting 5 from both sides.

The slope of this given line is 3.

Therefore, the slope of the perpendicular line is -1/3, as it must be the negative reciprocal.

The general form of a line equation in slope intercept form is y = Mx+B where M is the slope and B is the intercept.

Solving for B is: B = y- Mx

So the intercept of the perpendicular line with slope M=-1/3 and passing through (x=4, y=-3) is

y M * x

B = -3 - (-1/3)*4 =

-3 + 1/3*4 = <-- subtracting the negative is the same as adding the positive; definition of subtraction

-3 + 4/3 = <-- multiplies the fractions first per order of mixed operations

-9/3 + 4/3 <-- common denominator is 3

= -5/3

So the equation of the perpendicular line is y = -1/3X + -5/3 = -1/3X-5/3

Notice when X=4, y = -1/3(4) - 5/3 = -4/3 - 5/3 = -9/3 = -3 as expected

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