Write the equation of the line that passes through the point (-1,7) and is perpendicular to the line

7 answers
Question:

Write the equation of the line that passes through the point (-1,7) and is perpendicular to the line x=4. explain how you determined this equation.​

Answers

The answer is: y = 7

Step-by-step explanation:

For x = 4, it is a vertical line where everything is x = 4. No matter what y is, x always equals 4.

For the line y = 7, it is a horizontal line, perpendicular to the line x = 4. The line y = 7 passes through the point (-1, 7) and all other similar points where y = 7.

-x+y=8

Step-by-step explanation:

Slope intercept form is y=my+b. (y2-y1)/(x2-x1)=m or slope, (10-7)/(2- -1)=3/3=1. Therefore m=1. Now we have y=1x+b. Plug in either point to solve for b. 7=1(-1)+b which simplifies to 7=-1+b. To solve for b, add -1 to both sides in order to isolate b. 7+1=8, so b=8. Now we have y=x+8. subtract x from both sides to get y-x=8 or -x+y=8.

The answer is y-7=-1(x-(-1))

Step-by-step explanation Because the formula is y-y1=m(x-x1) you ultimately get this equation.

[tex]7=-1\cdot (-1)+b[/tex]

[tex]y-4=-1\cdot (x-2)[/tex]

[tex]y-7=-1\cdot (x-(-1))[/tex]

[tex]y=-x+6[/tex]

Step-by-step explanation:

Find the slope of the line that passes through the points (-1,7) and (2.4):

[tex]\text{Slope}=\dfrac{7-4}{-1-2}=\dfrac{3}{-3}=-1[/tex]

So, the equation of the line is

[tex]y-y_1=-1\cdot (x-x_1),[/tex]

where [tex](x_1,y_1)[/tex] are the coordintaes of the point from the line

or

[tex]y=-1\cdot x+b,[/tex]

where b is y-intercept.

To find b, substitute x = -1 and y = 7:

[tex]7=-1\cdot (-1)+b\\ \\b=7-1=6\\ \\y=-x+6[/tex]

If [tex](x_1,y_1)=(-1,7),[/tex] then

[tex]y-7=-1\cdot (x-(-1))[/tex]

If [tex](x_1,y_1)=(2,4),[/tex] then

[tex]y-4=-1\cdot (x-2)[/tex]

So correct options are

[tex]7=-1\cdot (-1)+b[/tex]

[tex]y-4=-1\cdot (x-2)[/tex]

[tex]y-7=-1\cdot (x-(-1))[/tex]

[tex]y=-x+6[/tex]

bearing in mind that standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

[tex]\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{7})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{10}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{10-7}{2-(-1)}\implies \cfrac{3}{2+1}\implies \cfrac{3}{3}\implies 1[/tex]

[tex]\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-7=1[x-(-1)]\implies y-7=x+1 \\\\\\ y=x+8\implies \boxed{-x+y=8}\implies \stackrel{\textit{standard form}}{x-y=-8}[/tex]

just to point something out, is none of the options, however -x + y = 8, is one, though improper.

if we look at the equation y = -2x - 1, is already in slope-intercept form, therefore, [tex]\bf y=\stackrel{slope}{-2}x-1[/tex] has a slope of -2.


now, parallel lines have exactly equal slopes, therefore a parallel to that one above, will have also a slope of -2, so we're really looking for a line whose slope is -2 and runs through -1, 7.


[tex]\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{7})\qquad \qquad \qquad slope = m\implies -2\\\\\\\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-7=-2[x-(-1)]\\\\\\y-7=-2(x+1)\implies y-7=-2x-2\implies y=-2x+5[/tex]

its 3, 4, and 6

Step-by-step explanation:

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