Question

type the correct answer in each box. use numerals instead of words. if necessary, use / for the fraction bar(s).
a line passes through point (-2, 5) and has a slope of \\(\frac{2}{3}\\). points a(x, 3) and b(-2, y) lie on the line.
the value of x is \\(square\\), and the value of y is \\(square\\).

Explanation:

Step1: Recall slope formula

The slope \( m \) between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \( m = \frac{y_2 - y_1}{x_2 - x_1} \). We know the line passes through \((-2,5)\) with slope \( \frac{2}{3} \).

Step2: Find \( x \) for point \( A(x,3) \)

Using the slope formula with \((x_1,y_1)=(-2,5)\) and \((x_2,y_2)=(x,3)\), \( \frac{2}{3} = \frac{3 - 5}{x - (-2)} \). Simplify: \( \frac{2}{3} = \frac{-2}{x + 2} \). Cross - multiply: \( 2(x + 2)=-2\times3 \), \( 2x+4 = - 6 \), \( 2x=-10 \), \( x=-5 \).

Step3: Find \( y \) for point \( B(-2,y) \)

Using the slope formula with \((x_1,y_1)=(-2,5)\) and \((x_2,y_2)=(-2,y)\), \( \frac{2}{3}=\frac{y - 5}{-2-(-2)} \). But \( -2-(-2)=0 \), wait, no, wait, the point \( B \) is \((-2,y)\), and the other point is \((-2,5)\)? Wait, no, the line passes through \((-2,5)\), and point \( B \) is \((-2,y)\). Wait, no, maybe I made a mistake. Wait, the line has slope \( \frac{2}{3} \), passes through \((-2,5)\), and point \( A(x,3) \), point \( B(-2,y) \). Wait, for point \( B(-2,y) \), using the slope formula with \((-2,5)\) and \((-2,y)\) is wrong because the x - coordinates are the same, which would be a vertical line, but the slope is \( \frac{2}{3} \) (non - vertical). Wait, no, the point \( B \) should be \((-2,y)\)? Wait, no, maybe the problem is: the line passes through \((-2,5)\), slope \( \frac{2}{3} \), point \( A(x,3) \) and point \( B(-2,y) \). Wait, for point \( B(-2,y) \), let's use the slope formula with \((-2,5)\) and \((-2,y)\) is incorrect. Wait, no, maybe the point \( B \) is \((-2,y)\) and the other point is \((x,3)\)? Wait, no, let's re - do.

Wait, the line has slope \( m=\frac{2}{3} \), passes through \((x_0,y_0)=(-2,5)\). The equation of the line is \( y - y_0=m(x - x_0) \), so \( y - 5=\frac{2}{3}(x + 2) \).

For point \( A(x,3) \): substitute \( y = 3 \) into the line equation: \( 3 - 5=\frac{2}{3}(x + 2) \), \( - 2=\frac{2}{3}(x + 2) \), multiply both sides by \( \frac{3}{2} \): \( - 3=x + 2 \), \( x=-5 \).

For point \( B(-2,y) \): substitute \( x=-2 \) into the line equation: \( y - 5=\frac{2}{3}(-2 + 2) \), \( y - 5 = 0 \), \( y = 5 \)? Wait, no, that can't be. Wait, maybe the point \( B \) is not \((-2,y)\) with \( x=-2 \), maybe a typo? Wait, no, the original problem says "Points \( A(x, 3) \) and \( B(-2, y) \) lie on the line". Wait, the line passes through \((-2,5)\) and has slope \( \frac{2}{3} \). So the line equation is \( y=mx + b \), \( 5=\frac{2}{3}(-2)+b \), \( 5=-\frac{4}{3}+b \), \( b = 5+\frac{4}{3}=\frac{15 + 4}{3}=\frac{19}{3} \). So the equation is \( y=\frac{2}{3}x+\frac{19}{3} \).

For point \( A(x,3) \): \( 3=\frac{2}{3}x+\frac{19}{3} \), multiply both sides by 3: \( 9 = 2x+19 \), \( 2x=9 - 19=-10 \), \( x=-5 \).

For point \( B(-2,y) \): \( y=\frac{2}{3}(-2)+\frac{19}{3}=\frac{-4 + 19}{3}=\frac{15}{3}=5 \)? Wait, no, that's the same as the original point. Wait, maybe the point \( B \) is \((x,y)\) with \( x eq - 2 \). Wait, maybe the problem has a typo, but according to the calculation for \( x \):

Using slope formula between \((-2,5)\) and \((x,3)\):

\( m=\frac{3 - 5}{x-(-2)}=\frac{-2}{x + 2}=\frac{2}{3} \)

Cross - multiply: \( 2(x + 2)=-2\times3\)

\( 2x+4=-6\)

\( 2x=-10\)

\( x = - 5\)

For \( y \): using slope formula between \((-2,5)\) and \((-2,y)\) is wrong. Wait, maybe the point \( B \) is \((x,y)\) with \( x eq - 2 \), but the problem says \( B(-2,y) \). Wait, no, maybe I misread the problem. Wait, the line passes through \((-2,5)\), slope \( \frac{2}{3} \), points \( A(x,3) \) and \( B(-2,y) \) are on the line. Wait, for point \( B(-2,y) \), since the x - coordinate is - 2 (same as the given point \((-2,5)\)), the line would be vertical if \( x=-2 \) for two points, but the slope is \( \frac{2}{3} \) (non - vertical). So there must be a mistake in my understanding. Wait, no, the point \( B \) should be \((-2,y)\)? Wait, no, maybe the problem is: the line passes through \((-2,5)\), slope \( \frac{2}{3} \), and point \( A(x,3) \), and point \( B(-2,y) \) is on the line. Wait, let's recast the line equation: \( y - 5=\frac{2}{3}(x + 2) \).

For point \( A(x,3) \): \( 3-5=\frac{2}{3}(x + 2)\Rightarrow - 2=\frac{2}{3}(x + 2)\Rightarrow x + 2=-3\Rightarrow x=-5 \).

For point \( B(-2,y) \): \( y - 5=\frac{2}{3}(-2 + 2)\Rightarrow y - 5 = 0\Rightarrow y = 5 \). But that's the same as the original point. Wait, maybe the point \( B \) is \((-2,y)\) is a mistake, and it should be \((x,y)\) with \( x eq - 2 \). But according to the problem, we proceed with the calculation for \( x \) as - 5 and \( y \) as 5 (even though it seems odd, maybe a typo in the problem's point \( B \) coordinates).

Answer:

The value of \( x \) is \(-5\), and the value of \( y \) is \(5\)