Statement 1: Since we're referring to a line parallel to line XY, the slopes will be identical. We can use the points provided to calculate the slope:
   

    \(\displaystyle \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-3}{3+1}=\frac{2}{4}\)

We can simplify the slope to just \(\displaystyle \frac{1}{2}\).

Statement 2: Finding the slope of a line parallel to line XY is really straightforward when given the equation of a line.

       \(\displaystyle y=mx+b\)

Where \(\displaystyle m\) is the slope and \(\displaystyle b\) the y-intercept.

In this case, our \(\displaystyle m\) value is \(\displaystyle \frac{1}{2}\).\

Each statement alone is sufficient to answer the question.

Recall that parallel lines have the same slope and that slope can be calculated from any two points.

Statement I gives us a point on \(\displaystyle g(t)\)

Statement II gives us the x-intercept, a.k.a. the point \(\displaystyle (290,0)\).

Therfore, using both statements, we can find the slope of \(\displaystyle g(t)\) and any line parallel to it.

\(\displaystyle m=\frac{y_1-y_2}{x_1-x_2}\)

\(\displaystyle m=\frac{15-290}{14}=-\frac{275}{14}\)

In this case, I and II give us a ton of information about the size of the circle, but we have no clue as to its location.

In order to know the slope of the tangent line, we need to know location of the circle, so we cannot solve this problem.

Points are collinear if they lie on the same line.  Here A, B, and C are collinear if the line AB is the same as the line AC.  In other words, the slopes of line AB and line AC must be the same.  Statement 2 gives us the two slopes, so we know that Statement 2 is sufficient.  Statement 1 also gives us all of the information we need, however, because we can easily find the slopes from the vertices.  Therefore both statements alone are sufficient.

By using I) we know that the given point is on the line of the equation.

So I) is sufficient.

II) gives us the y coordinate of the minimum. In a quadratic equation, this is what  "c" represents.

Therefore, c=-80 and II) is also sufficient.

Statement I gives us our function.

Statement II gives us a clue to find the value of \(\displaystyle a\). \(\displaystyle a\) is five more than 3 times the y-intercept of \(\displaystyle g(j)\). So, we can find the following:

\(\displaystyle a=3(-14)+5=-37\)

To see if the point \(\displaystyle (-37,5)\) is on the line \(\displaystyle g(j)\), plug it into the function:

\(\displaystyle 5=15\cdot(-37)-14\)

This is not a true statement, so the point is not on the line.

Consider linear functions h(t) and g(t). 

I) \(\displaystyle h(t)\perp g(t)\) at the point \(\displaystyle (6,4)\)

II) \(\displaystyle g(t)=2t-8\)

Is the point \(\displaystyle (10,4)\) on the line h(t)?

We can use II) and I) to find the slope of h(t)

Recall that perpendicular lines have opposite reciprocal slope. Thus, the slope of h(t) must be \(\displaystyle \frac{-1}{2}\)

Next, we know that h(t) must pass through (6,4), so lets us that to find the y-intercept:

\(\displaystyle 4=\frac{-1}{2}*6+b\)

\(\displaystyle b=7\)

\(\displaystyle h(t)=\frac{-1}{2}t+7\)

Next, check if (10,4) is on h(t) by plugging it in.

\(\displaystyle 4\neq \frac{-1}{2}*10+7=2\)

So, the point is not on the line, and we needed both statements to know.

Statement 1 allows you to define the equation of line m, but does not provide enough information to solve for \(\displaystyle \tiny b\).  There are still 3 variables \(\displaystyle \left ( x,y,b\right )\) and only two different equations to solve.

if \(\displaystyle 2y-4x=b\), statement 2 supplies enough information to solve for b by substitution if \(\displaystyle \left ( 4,b\right )\) is on the line.   

\(\displaystyle 2y-4x=b\)

\(\displaystyle 2b-4(4)=b\)

\(\displaystyle 2b-16=b\)

\(\displaystyle 2b-b=16\)

\(\displaystyle b=16\)

To find the equation of a linear function, we need some combination of slope and a point.

Statement I gives us a clue to find the slope of the desired function. It must be the opposite reciprocal of the slope of \(\displaystyle h(j)\). This makes the slope of \(\displaystyle d(g)\) equal to \(\displaystyle -\frac{1}{5}\)

Statement II gives us a point on our desired function, \(\displaystyle (6,0)\).

Using slope-intercept form, we get the following:

\(\displaystyle 0=-\frac{1}{5}\cdot 6+b\)

\(\displaystyle b=\frac{6}{5}\)

So our equation is as follows

\(\displaystyle y=-\frac{1}{5}x+\frac{6}5\)

Given that the square of a negative is still positive, it is possible for a to have an x-intercept that is negative, while still having a positive slope. The example above shows how the square of the x-intercept for line a could be greater, while having still giving line a a slope that is less than that of b.