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graph a line with a slope of -\\frac{3}{2} that contains the point (4, …

Question

graph a line with a slope of -\frac{3}{2} that contains the point (4, - 3).

Explanation:

Step1: Use point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $m$ is the slope and $(x_1,y_1)$ is a point on the line. Given $m =-\frac{3}{2}$ and $(x_1,y_1)=(4,-3)$, we have $y+3 =-\frac{3}{2}(x - 4)$.

Step2: Expand the equation

Expand the right - hand side: $y+3=-\frac{3}{2}x+6$.

Step3: Solve for $y$

Subtract 3 from both sides to get the slope - intercept form $y =-\frac{3}{2}x + 3$.

Step4: Find another point

Let $x = 2$. Then $y=-\frac{3}{2}\times2 + 3=-3 + 3=0$. So another point on the line is $(2,0)$.

Step5: Graph the line

Plot the points $(4,-3)$ and $(2,0)$ on the coordinate plane and draw a straight line passing through them.

Answer:

Graph the line by plotting the points $(4,-3)$ and $(2,0)$ and drawing a straight line through them.