See, please, suggested decision:
When the line intersects the xy-plane, then parameter z=0, when does the xz - y=0 and when does yz, x=0. All the details are in attachment.
If it's possible, check arithmetic part.
To solve this, you’ll first need to solve for their slopes.
The slope for line Q is y2-y1/x2-x1 = -8-(-2)/-8-(-10) = -3
We know that the lines are perpendicular so the negative reciprocal of -3 is 1/3
The equation you get it y = 1/3x + b.
Now you will need to solve for b by substituting in the first ordered pair of line R.
2 = 1/3(1) + b.
Once you solve for b, you should get 5/3 and y = 1/3x + 5/3
Now, to find a, you will need to substitute in 10 from the second ordered pair into x in your new equation.
y = 1/3(10) + 5/3.
Your solution should be 5.
So your answer is: a = 5
6/8. multiply 3/4 by 2 and there is your answer.
6x + 52 > 130
Subtract 52 from both sides:
6x > 78
Divide both sides by 6:
X > 78/6
X > 13
