It is given that the straight line l1: (k+1)x-y-8=0 is parallel to the straight line l2: (2k-1)x-y-3k=0. 1)find k 2)if the straight line l3 is perpendicular to l2 and has the same x-intercept as l2, find the equation of l3

Part 1:  k = 2

Part 2:  y = (-1/3)x - 6

Step-by-step explanation:

Hello!

Recall that parallel lines have the same slope.  Given  (k+1)x-y-8=0 and (2k-1)x-y-3k=0, we'll solve each equation for its slope in terms of k and then equate the two slopes to determine the value of k.

We can immediately read off the slope of (2k-1)x-y-3k=0:  It's the coefficient, 2k-1, of x.  Similarly, the slope of the other line is k+1.  Equating these quantities, we get:

2k - 1= k + 1.  Then k = 2, and the slope of the parallel lines is 2(2)-1, or 3.

Part 2:  A line perpendicular to these parallel lines has a slope of -1/3, which is the negative reciprocal of 3.

Let's look at L2, substitute 2 for k and find the equation of the line:

(2k-1)x-y-3k=0  =>  (2[2]-1)x-y-3[2]=0, or 3x - y - 6 = 0, or y = 3x - 6.  Here, the y-intercept is -6.

Finally, let's write the equation of L3:  It has the slope -1/3 and the y-intercept of -6:

y = (-1/3)x - 6

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)