Answers:
FALSE.
Explanations:
Given the linear relations 2x - 3y = 4 and y = -2/3 x + 5
Both equations are equations of a line. For the lines to be perpendicular, the product of their slope is -1
The standard equation of a line in slope-intercept form is expressed as
[tex]y=mx+b[/tex]
m is the slope of the line
For the line 2x - 3y = 4, rewrite in standard form
[tex]\begin{gathered} 2x-3y=4 \\ -3y=-2x+4 \\ y=\frac{-2}{-3}x-\frac{4}{3} \\ y=\frac{2}{3}x-\frac{4}{3} \end{gathered}[/tex]
Compare with the general equation
[tex]\begin{gathered} mx=\frac{2}{3}x \\ m=\frac{2}{3} \end{gathered}[/tex]
The slope of the line 2x - 3y = 4 is 2/3
For the line y = -2/3 x + 5
[tex]\begin{gathered} mx=-\frac{2}{3}x \\ m=-\frac{2}{3} \end{gathered}[/tex]
The slope of the line y = -2/3 x + 5 is -2/3
Take the product of their slope to determine whether they are perpendicular
[tex]\begin{gathered} \text{Product = }\frac{2}{3}\times-\frac{2}{3} \\ \text{Product = -}\frac{4}{9} \end{gathered}[/tex]
Since the product of their slope is not -1, hence the linear relations do not represent lines that are perpendicular. Hence the correct answer is FALSE
