Consider the line y=7x-6. Find the equation of the line that is perpendicular to this line and passes through the point (6, 2).
Find the equation of the line that is parallel to this line and passes through the point (6, 2).

Case 1: 7y + x - 20 = 0 (Perpendicular)

Case 2: y - 7x + 40 = 0 (Parallel)

Step-by-step explanation:

Case 1: Perpendicular

Equation of the given line y = 7x - 6

Comparing with the general equation y = mx + c; m = 7

For two lines to be perpendicular, the product of their gradients is (-1) or we can also say; gradient of one line must be the negative inverse of the other gradient.

So the gradient of a line perpendicular to y = 7x - 6 should be (-1/7)

Now solving for the equation of the line using the formula

y - y = m(x - x ) where (x , y ) are the coordinates of the point.

substituting..

y - 2 = (-1/7)(x - 6)

7(y - 2) = -1(x - 6)

7y - 14 = -x + 6

7y + x -14 - 6 = 0

7y + x - 20 = 0 is the equation of the line perpendicular to y = 7x - 6 and passes through the point (6,2)

Case 2: Parallel

Equation of the given line y = 7x - 6

Comparing with the general equation y = mx + c; m = 7

For two lines to be parallel, they must have equal gradients.

So the gradient of a line parallel to y = 7x - 6 should be 7

Now solving for the equation of the line using the formula

y - y = m(x - x ) where (x , y ) are the coordinates of the point.

substituting..

y - 2 = 7(x - 6)

y - 2 = 7x - 42

y - 7x -2 + 42 = 0

y - 7x + 40 = 0 is the equation of the line parallel to y = 7x - 6 and passes through the point (6,2)

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

i think is g(1)=-1.

hope that i can