Question

vera wants to graph a line that passes through (0, 2) and has a slope of \\(\frac{2}{3}\\). which points could vera use to graph the line? select three options. \\(\square\\) (-3, 0) \\(\square\\) (-2, -3) \\(\square\\) (2, 5) \\(\square\\) (3, 4) \\(\square\\) (6, 6)

Explanation:

First, we know the equation of a line in slope - intercept form is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y - intercept. Given that the line passes through \((0,2)\), the y - intercept \(b = 2\), and the slope \(m=\frac{2}{3}\). So the equation of the line is \(y=\frac{2}{3}x + 2\).

Step 1: Check the point \((- 3,0)\)

Substitute \(x=-3\) and \(y = 0\) into the equation \(y=\frac{2}{3}x+2\).
Left - hand side (LHS): \(y = 0\)
Right - hand side (RHS): \(\frac{2}{3}\times(-3)+2=-2 + 2=0\)
Since \(LHS = RHS\), the point \((-3,0)\) lies on the line.

Step 2: Check the point \((-2,-3)\)

Substitute \(x = - 2\) and \(y=-3\) into the equation \(y=\frac{2}{3}x + 2\).
LHS: \(y=-3\)
RHS: \(\frac{2}{3}\times(-2)+2=-\frac{4}{3}+2=\frac{-4 + 6}{3}=\frac{2}{3}
eq-3\)
So the point \((-2,-3)\) does not lie on the line.

Step 3: Check the point \((2,5)\)

Substitute \(x = 2\) and \(y = 5\) into the equation \(y=\frac{2}{3}x+2\).
LHS: \(y = 5\)
RHS: \(\frac{2}{3}\times2+2=\frac{4}{3}+2=\frac{4 + 6}{3}=\frac{10}{3}\approx3.33
eq5\)
Wait, we made a mistake here. Let's recalculate:
\(y=\frac{2}{3}x + 2\), when \(x = 2\), \(y=\frac{2}{3}\times2+2=\frac{4 + 6}{3}=\frac{10}{3}\approx3.33\), which is wrong. Wait, maybe we can use the slope formula. The slope between \((0,2)\) and \((x,y)\) is \(\frac{y - 2}{x-0}=\frac{y - 2}{x}\). We know the slope is \(\frac{2}{3}\), so \(\frac{y - 2}{x}=\frac{2}{3}\), or \(3(y - 2)=2x\), or \(3y-6 = 2x\), or \(3y=2x + 6\), or \(y=\frac{2}{3}x + 2\) (same as before).

Let's use the slope formula to check the point \((3,4)\):
The slope between \((0,2)\) and \((3,4)\) is \(\frac{4 - 2}{3-0}=\frac{2}{3}\), which is equal to the given slope. So \((3,4)\) is on the line.

Check the point \((6,6)\):
The slope between \((0,2)\) and \((6,6)\) is \(\frac{6 - 2}{6-0}=\frac{4}{6}=\frac{2}{3}\), which is equal to the given slope. So \((6,6)\) is on the line.

Wait, let's re - check \((2,5)\) with the slope formula:
Slope between \((0,2)\) and \((2,5)\) is \(\frac{5 - 2}{2-0}=\frac{3}{2}
eq\frac{2}{3}\), so \((2,5)\) is not on the line.

Re - check \((-3,0)\) with slope formula: slope between \((0,2)\) and \((-3,0)\) is \(\frac{0 - 2}{-3-0}=\frac{-2}{-3}=\frac{2}{3}\), correct.

Check \((3,4)\): slope between \((0,2)\) and \((3,4)\) is \(\frac{4 - 2}{3-0}=\frac{2}{3}\), correct.

Check \((6,6)\): slope between \((0,2)\) and \((6,6)\) is \(\frac{6 - 2}{6-0}=\frac{4}{6}=\frac{2}{3}\), correct.

So the correct points are \((-3,0)\), \((3,4)\), \((6,6)\)

Answer:

A. \((-3, 0)\)
D. \((3, 4)\)
E. \((6, 6)\)