201k views
4 votes
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

2 Answers

5 votes


y = 3x - 5 \\ perpendicular \: line \: equation \\ y = ( - 1)/(3) x + b \\ it \: passes \: thrugh(6 \: \: 5) \\ so \\ 5 = ( - 1)/(3) * 6 + b \\ b = 7 \\ so \: equation \: of \: line \: \: y = ( - 1)/( 3) x+ 7 \\ 3y + x = 7

User Rahoolm
by
7.1k points
5 votes

Answer:

y=−x/3+7 .

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.

User Eshirazi
by
7.3k points