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
{3x+2y=4 /*5
{4x+5y=17 /*(-2)
{15x+10y=20
<u>{-8x-10y=-34 </u>(+)
<em>7x=-14</em><em />
<em>x=-2</em>
15x+10y=20
15*(-2)+10y=20
-30+10y=20
10y=50
<em>y=5</em>
Answer: The required values are
x = 12 units, ST = 60 units and SU = 120 units.
Step-by-step explanation: Given that T is the midpoint of SU, where
ST = 5x and TU = 3x + 24.
We are to find the values of x, ST and SU.
Since T is the midpoint of SU, so we get
So, the value of x is 12.
Therefore,
and
Thus, the required values are
x = 12 units, ST = 60 units and SU = 120 units.
H = 2f / (m+1)
[multiply by (m+1)]
h(m+1) = 2f
[divide by 2]
f = (h (m + 1)) / 2
3b / (b+2) = 12 / (b+2)
[multiply by (b+2)]
3b = 12
[divide by 3]
b = 4
3 / (6x + 1) / 2 = 8 / (x + 4) / 3
[multiply both denominators to mike one denominator]
3 / 8(6x+1) = 8 / 3(x+4)
[expand brackets]
3 / (48x + 8) = 8 / (3x + 12)
[multiply by (48x + 8)]
3 = 8(48x + 8) / (3x+12)
[multiply by (3x+12)]
3(3x +12) = 4(48x + 8)
[simplify]
9x + 36 = 192x + 32
173x = 4
x = 4 / 173
