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A straight line, L, is perpendicular to the straight line y=3x-5 and passes through the point (6,5)
Find the equation of L
Please help ASAP

Respuesta :

Answer:

y=−x/3+7 .

Step-by-step explanation:

The equation of the line in the slope-intercept form is y=3x−5.

The slope of the perpendicular line is negative inverse: m=−13.

So, the equation of the perpendicular line is y=−x3+a.

To find a, we use the fact that the line should pass through the given point: 5=(−13)⋅(6)+a.

Thus, a=7.

Therefore, the equation of the line is y=−x/3+7.

[tex]y = 3x - 5 \\ perpendicular \: line \: equation \\ y = \frac{ - 1}{3} x + b \\ it \: passes \: thrugh(6 \: \: 5) \\ so \\ 5 = \frac{ - 1}{3} \times 6 + b \\ b = 7 \\ so \: equation \: of \: line \: \: y = \frac{ - 1}{ 3} x+ 7 \\ 3y + x = 7[/tex]

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