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2. a table of values is given for a line in the xy -coordinate plane. w…

Question

  1. a table of values is given for a line in the xy -coordinate plane.

write the equation of the line

  1. perpendicular lines intersect at 90° angles. line j is perpendicular to the line shown in the graph and passes through point a. write the equation for line j

Explanation:

Step1: Find slope of first - line

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. Using the points $(0,-5)$ and $(2,-3)$ from the table, $m_1=\frac{-3-(-5)}{2 - 0}=\frac{-3 + 5}{2}=\frac{2}{2}=1$.

Step2: Find equation of first - line

The slope - intercept form is $y=mx + b$. Using the point $(0,-5)$ (where $x = 0$ and $y=-5$), we get $b=-5$. So the equation of the first line is $y=x - 5$.

Step3: Find slope of perpendicular line

If two lines are perpendicular, the product of their slopes is $-1$. Let the slope of line $j$ be $m_2$. Since $m_1 = 1$, then $m_2=-1$.

Step4: Assume point $A$ is $(x_0,y_0)$

Let's assume the coordinates of point $A$ are $(x_0,y_0)$. Using the point - slope form $y - y_0=m_2(x - x_0)$, the equation of line $j$ is $y - y_0=-(x - x_0)$ or $y=-x+(x_0 + y_0)$.

Since the coordinates of point $A$ are not given in the problem statement, if we assume $A=(a,b)$, the equation of line $j$ is $y=-x+(a + b)$. If we assume the line in the graph has equation $y = 0$ (a horizontal line), and a perpendicular line to a horizontal line is a vertical line. If point $A$ has coordinates $(x_1,y_1)$, the equation of line $j$ is $x=x_1$.

If we consider only the first part of finding the equation of the line from the table:

Answer:

$y=x - 5$