Step 1
Find the slope of the line PQ
we know that
the formula to calculate the slope between two points is equal to
\(m=\frac{y2-y1}{x2-x1}\)
we have
\(P(-3,-3)\\Q(3,-1)\)
substitute the values
\(m=\frac{-1+3}{3+3}\)
\(m=\frac{2}{6}\)
\(m=\frac{1}{3}\)
Step 2
Find the equation of the line that passes through point R ans is parallel to line PQ
we know that
If two lines are parallel, then their slopes are the same
The equation of the line into point-slope form is equal to
\(y-y1=m(x-x1)\)
we have
\(m=\frac{1}{3}\)
\((x1,y1)=R(1,1)\)
substitute the values
\(y-1=\frac{1}{3}(x-1)\)
Step 3
we know that
The point on the y-axis that is on the line \(y-1=\frac{1}{3}(x-1)\) is equal to the y-intercept of the line
The y-intercept is the value of y when the value of x is equal to zero
so
For \(x=0\) find the value of y in the linear equation
\(y-1=\frac{1}{3}(0-1)\)
\(y-1=-\frac{1}{3}\)
\(y=1-\frac{1}{3}\)
\(y=\frac{2}{3}\)
The point that is on the y-axis is \((0,\frac{2}{3})\)
therefore
the answer is the option A
\((0,\frac{2}{3})\)
see the attached figure to better understand the problem
