1) find the equation of the line parallel to

x-5y=6 and through (4,-2).

2) find the equation of the line perpendicular to y= -2/5x + 3 and through (2,-1)

answer you so

1) The equation of the line parallel to  x-5y=6 and through (4,-2) is 5y = x -14

2) The equation of the line perpendicular to y= -2/5x + 3 and through (2,-1) is  2y = 5x -12

Solution:

1) find the equation of the line parallel to  x-5y=6 and through (4,-2).

Given, line equation is x – 5y = 6  

We have to find the line equation that is parallel to given line and passing through the point (4, -2)

Now, let us find slope of the given line.

Now, we know that, slope of parallel lines are equal.

So, slope of required line is 1/5 and it passes through (4, -2)

Now, using point slope form

Hence, the line equation is 5y = x -14

2) find the equation of the line perpendicular to y= -2/5x + 3 and through (2,-1)

We have to find the line equation that is perpendicular to given line and passing through the point (2, -1)

Now, let us find slope of the given line.

Now, we know that, product of slopes of perpendicular lines equals to -1.

So, slope of required line slope of given line = -1

slope of required line =

And it passes through (2, -1)

Now, using point slope form

Hence, the line equation is 2y = 5x -12

triangle abc is reflected on the line x=-6.5 then reflected again on the x axis

step-by-step explanation: