we know that
if two lines are perpendicular
then the product of their slopes is equal to minus one
so
[tex]m1*m2=-1[/tex]
in this problem we have
[tex]m1=-\frac{4}{5}[/tex]
the value of m2 must be equal to
[tex]m2=-\frac{1}{m1} =-\frac{1}{(-4/5)}=\frac{5}{4}[/tex]
we know that
The formula to calculate the slope between two points is equal to
[tex]m=\frac{(y2-y1)}{(x2-x1)}[/tex]
Find the slopes of each of ordered pairs and compare with m2
case a) [tex](-2,0)\ and\ (2,5)[/tex]
Substitute in the formula
[tex]m=\frac{(5-0)}{(2+2)}[/tex]
[tex]m=\frac{(5)}{(4)}[/tex]
The slope is equal to m2
so
The ordered pair case a) could be points on a line that is perpendicular to the given line
case b) [tex](-4,5)\ and\ (4,-5)[/tex]
Substitute in the formula
[tex]m=\frac{(-5-5)}{(4+4)}[/tex]
[tex]m=\frac{(-10)}{(8)}[/tex]
[tex]m=-\frac{5}{4}[/tex]
The slope is not equal to m2
so
The ordered pair case b) could not be points on a line that is perpendicular to the given line
case c) [tex](-3,4)\ and\ (2,0)[/tex]
Substitute in the formula
[tex]m=\frac{(0-4)}{(2+3)}[/tex]
[tex]m=\frac{(-4)}{(5)}[/tex]
The slope is not equal to m2
so
The ordered pair case c) could not be points on a line that is perpendicular to the given line
case d) [tex](1,-1)\ and\ (6,-5)[/tex]
Substitute in the formula
[tex]m=\frac{(-5+1)}{(6-1)}[/tex]
[tex]m=\frac{(-4)}{(5)}[/tex]
The slope is not equal to m2
so
The ordered pair case d) could not be points on a line that is perpendicular to the given line
case e) [tex](2,-1)\ and\ (10,9)[/tex]
Substitute in the formula
[tex]m=\frac{(9+1)}{(10-2)}[/tex]
[tex]m=\frac{(10)}{(8)}[/tex]
[tex]m=\frac{5}{4}[/tex]
The slope is equal to m2
so
The ordered pair case e) could be points on a line that is perpendicular to the given line
therefore
the answer is
[tex](-2,0)\ and\ (2,5)\\(2,-1)\ and\ (10,9)[/tex]
The points on the line are:
[tex]\text{A}. (-2,0) and (2,5)\\\text{E}. (2,1) and (10,9)[/tex]
Step-by-step explanation:
Given information:
The slope of the line = -4/5
Now as we know that the two lines are perpendicular
So,
[tex]m_1\times m_2=-1\\[/tex]
So, the slope of the second line is :
[tex]m_2=5/4[/tex]
Now , check for the options given:
And slope [tex]m= \frac{y_2-y_1}{x_2-x_1}[/tex]
(A) The slope of the line passing through the points (-2,0) and (2,5) is :
[tex]m=(5-0)/(2+2)[/tex]
[tex]m=5/4[/tex]
Hence the slope is equal to [tex]m_2[/tex].
Similarly,
(B)The slope of the line passing through the points (-4,5) and (4,-5) is :
[tex]m=(-5-5)/(4+4)[/tex]
[tex]m=-5/4[/tex]
Hence the slope not is equal to [tex]m_2[/tex].
(C) The slope of the line passing through the points (-3,4) and (2,0) is :
[tex]m=(0-4)/(2+3)[/tex]
[tex]m=-4/5[/tex]
Hence the slope is not equal to [tex]m_2[/tex].
(D) The slope of the line passing through the points (1,-1) and (6,5) is :
[tex]m=(-5+1)/(6-5)[/tex]
[tex]m=-4/5[/tex]
Hence the slope is not equal to [tex]m_2[/tex].
(E) The slope of the line passing through the points (2,-1) and (10,9) is :
[tex]m=(9+1)/(10-2)[/tex]
[tex]m=5/4[/tex]
Hence the slope is equal to [tex]m_2[/tex].
Hence, the points on the lines are:
[tex]\text{A}. (-2,0) and (2,5)\\\text{E}. (2,1) and (10,9)[/tex]
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