Respuesta :
Step 1
Find the slope of the line KL
we know that
The formula to calculate the slope between two points is equal to
[tex]m=\frac{y2-y1}{x2-x1}[/tex]
we have
[tex]K(-6,8)\ L(6,0)[/tex]
substitute the values
[tex]m=\frac{0-8}{6+6}[/tex]
[tex]m=\frac{-8}{12}[/tex]
[tex]m=-\frac{2}{3}[/tex]
Remember that
If two lines are parallel . then their slope are equal
we are going to calculate the slope of the line between point M and the different points. If the slope is equal to the slope of the line KL, then the point could be on line that is parallel to the line KL and passes through the point M.
Step 2
Find the slope of the points
[tex]M(-4,-2)\ A(-10,0)[/tex]
substitute the values in the formula
[tex]m=\frac{0+2}{-10+4}[/tex]
[tex]m=\frac{2}{-6}[/tex]
[tex]m=-\frac{1}{3}[/tex]
The slope of the line MA is not equal to the slope of the line KL
[tex]-\frac{1}{3}\neq -\frac{2}{3}[/tex]
therefore
The point A could not be on line that is parallel to the line KL and passes through the point M
Step 3
Find the slope of the points
[tex]M(-4,-2)\ B(-6,2)[/tex]
substitute the values in the formula
[tex]m=\frac{2+2}{-6+4}[/tex]
[tex]m=\frac{4}{-2}[/tex]
[tex]m=-2[/tex]
The slope of the line MB is not equal to the slope of the line KL
[tex]-2\neq -\frac{2}{3}[/tex]
therefore
The point B could not be on line that is parallel to the line KL and passes through the point M
Step 4
Find the slope of the points
[tex]M(-4,-2)\ C(0,-6)[/tex]
substitute the values in the formula
[tex]m=\frac{-6+2}{0+4}[/tex]
[tex]m=\frac{-4}{4}[/tex]
[tex]m=-1[/tex]
The slope of the line MC is not equal to the slope of the line KL
[tex]-1\neq -\frac{2}{3}[/tex]
therefore
The point C could not be on line that is parallel to the line KL and passes through the point M
Step 5
Find the slope of the points
[tex]M(-4,-2)\ D(8,-10)[/tex]
substitute the values in the formula
[tex]m=\frac{-10+2}{8+4}[/tex]
[tex]m=\frac{-8}{12}[/tex]
[tex]m=-\frac{2}{3}[/tex]
The slope of the line MD is equal to the slope of the line KL
[tex]-\frac{2}{3}=-\frac{2}{3}[/tex]
therefore
The point D could be on line that is parallel to the line KL and passes through the point M
the answer is
[tex](8,-10)[/tex]
The line which is parallel to the line KL passes through the point [tex](8,-10)[/tex] i.e, [tex]\fbox{\begin\\\ \bf option 4\\\end{minisapce}}[/tex].
Further explanation:
From the given figure in the question it is observed that the line KL has a [tex]x[/tex]-intercept and a [tex]y[/tex]- intercept.
From the given figure it is observed that the line intersect the [tex]x[/tex]-axis at the point [tex](6,0)[/tex] and intersects the [tex]y[/tex]-axis at the point [tex](0,4)[/tex].
This implies that the [tex]x[/tex]-intercept is [tex](6,0)[/tex] and the [tex]y[/tex]-intercept is [tex](0,4)[/tex].
Consider the point [tex](6,0)[/tex] as [tex](x_{1},y_{1})[/tex] and [tex](0,4)[/tex] as [tex](x_{2},y_{2})[/tex].
The slope of a line which passes through the points [tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex] is calculated as follows:
[tex]\fbox{\begin\\\ \math m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\end{minispace}}[/tex]
Substitute the value of [tex]x_{1},x_{2},y_{1}[/tex] and [tex]y_{2}[/tex] in the above equation.
[tex]\begin{aligned}m&=\dfrac{4-0}{0-6}\\&=\dfrac{-2}{3}\end{aligned}[/tex]
Therefore, the slope of the line KL is[tex]\dfrac{-2}{3}[/tex].
Since, the slope of line KL is [tex]\dfrac{-2}{3}[/tex] so, the slope of the line which is parallel to KL is also [tex]\dfrac{-2}{3}[/tex] because two parallel lines always have equal slope.
It is given that the line which is parallel to the line KL passes through the point M.
From the given figure it is observed that the coordinate for point M is [tex](-4,-2)[/tex].
The point slope form of a line is as follows:
[tex]\fbox{\begin\\\ \math (y-y_{1})=m(x-x_{1})\\\end{minispace}}[/tex]
To obtain the equation of the line which is parallel to the line KL substitute [tex]-4[/tex] for [tex]x_{1}[/tex], [tex]-2[/tex] for [tex]y_{1}[/tex] and [tex]\dfrac{-2}{3}[/tex] for [tex]m[/tex] in the above equation,.
[tex]\begin{aligned}(y+2)&=\dfrac{-2}{3}(x+4)\\y+2&=\dfrac{-2x}{3}-\dfrac{8}{3}\\y&=\dfrac{-2x}{3}-\dfrac{14}{3}\end{aligned}[/tex]
Therefore, the equation of the line which is parallel to the line KL is as follows:
[tex]\fbox{\begin\\\ \math y=\dfrac{-2x}{3}-\dfrac{14}{3}\\\end{minispace}}[/tex]
Label the above equation as equation (1).
[tex]y=\dfrac{-2x}{3}-\dfrac{14}{3}[/tex] (1)
Option1:
As per the option 1 the line [tex]y=\dfrac{-2x}{3}-\dfrac{14}{3}[/tex] passes through the point [tex](-10,0)[/tex].
Substitute [tex]-10[/tex] for [tex]x[/tex] in equation (1).
[tex]\begin{aligned}y&=\dfrac{-2\times (-10)}{3}-\dfrac{14}{3}\\&=\dfrac{20}{3}-\dfrac{14}{3}\\&=\dfrac{6}{3}\\&=2\end{aligned}[/tex]
As per the above calculation it is concluded that the line [tex]y=\dfrac{-2x}{3}-\dfrac{14}{3}[/tex] passes through the point [tex](-10,2)[/tex].
Therefore, the option 1 is incorrect.
Option 2:
As per the option 2 the line [tex]y=\dfrac{-2x}{3}-\dfrac{14}{3}[/tex] passes through the point [tex](-6,2)[/tex].
Substitute [tex]-6[/tex] for [tex]x[/tex] in equation (1).
[tex]\begin{aligned}y&=\dfrac{-2\times (-6)}{3}-\dfrac{14}{3}\\&=\dfrac{12}{3}-\dfrac{14}{3}\\&=\dfrac{-2}{3}\\&=\dfrac{-2}{3}\end{aligned}[/tex]
As per the above calculation it is concluded that the line [tex]y=\dfrac{-2x}{3}-\dfrac{14}{3}[/tex] passes through the point [tex](-6,\frac{-2}{3})[/tex].
Therefore, option 2 is incorrect.
Option 3:
As per the option 3 the line [tex]y=\dfrac{-2x}{3}-\dfrac{14}{3}[/tex] passes through the point [tex](0,-6)[/tex].
Substitute [tex]0[/tex] for [tex]x[/tex] in equation (1).
[tex]\begin{aligned}y&=\dfrac{-2\times 0}{3}-\dfrac{14}{3}\\&=-\dfrac{14}{3}\end{aligned}[/tex]
As per the above calculation it is concluded that the line [tex]y=\dfrac{-2x}{3}-\dfrac{14}{3}[/tex] passes through the point [tex](0,\frac{-14}{3})[/tex].
Therefore, option 3 is incorrect.
Option 4:
As per the option 4 the line [tex]y=\dfrac{-2x}{3}-\dfrac{14}{3}[/tex] passes through the point [tex](8,-10)[/tex].
Substitute [tex]8[/tex] for [tex]x[/tex] in equation (1).
[tex]\begin{aligned}y&=\dfrac{-2\times 8}{3}-\dfrac{14}{3}\\&=-\dfrac{16}{3}-\dfrac{14}{3}\\&=-\dfrac{30}{3}\\&=-10\end{aligned}[/tex]
As per the above calculation it is concluded that the line [tex]y=\dfrac{-2x}{3}-\dfrac{14}{3}[/tex] passes through the point [tex](8,-10)[/tex].
Therefore, option 4 is correct.
Figure 1 (attached in the end) shows that the line KL and the line [tex]y=-\dfrac{2x}{3}-\dfrac{14}{3}[/tex] are parallel.
Thus, the line which is parallel to the line KL passes through the point [tex](8,-10)[/tex] i.e, [tex]\fbox{\begin\\\ \bf option 4\\\end{minispace}}[/tex].
Learn more:
1. A problem on greatest integer function https://brainly.com/question/8243712
2. A problem to find radius and center of circle https://brainly.com/question/9510228
Answer details:
Grade: High school
Subject: Mathematics
Chapter: Lines
Keywords: Line, slope, intercept, x-intercept, y-intercept, slope intercept form, point slope form, curve, graph, intersects, equation, linear equation, (8,-10), y=-2x/3-14/3, KL, passes through M.
