Uhhh- The straight line L, has equation y = 3X - 4.
The straight line L2, is perpendicular to L, and passes through the point (9, 5). Find an equation of line L2. Show your working out.

The equation of required line is .

Y = mX+c

Given, The equation of straight line L.

Y=3X-4

compareing  with the standard equation of line

Y=mX+c

slope of the equation of line L is = 3.

The product of slopes of two perpendicular lines is -1.

The slope of perpendicular line is  and it passes through the point (9,5). So the equation of the line is

5 = -3 + c

5+3 = c

c = 8

Therefore, the equation of required line is .

Y = mX+c

Y =

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Step-by-step explanation:

y = mx + c, where m is the slope of the line and c is the y-intercept.

We have y = 3x - 4 as line L.

Slope of line L = 3

=> Slope of line L2 = -1/3

We have y = -1/3 x + c as our line L2 equation.

When x = 9, y = 5.

=> (5) = -1/3 * (9) + c

=> 5 = c - 3, c = 8

Hence the answer is y = -1/3 x + 8.

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8