Use the point-slope form of the linear equation: y - 2 = 2(x+1) What is the equation in standard form of a perpendicular line that passes through 5,1? What is the x-intercept of the perpendicular line?   Will you please show the steps for both problems.

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The linear equation y-2=2(x+1) is in point slope form where the slope is 2 and (-1,2) is a point on the line.  The first question asks to find the equation of the perpendicular line that passes through the point (5,1).  By definition, the perpendicular line will have a slope that...

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The linear equation y-2=2(x+1) is in point slope form where the slope is 2 and (-1,2) is a point on the line.  The first question asks to find the equation of the perpendicular line that passes through the point (5,1).  By definition, the perpendicular line will have a slope that is the negative reciprocal.  In this case, the negative reciprocal of 2 is -1/2.  Since we have both the slope and a point, let's write the equation in point slope form:

`y-1=-1/2(x-5)`

Now let's convert it to standard form:

2(y-1)=-(x-5)

2y-2=-x+5

2y+x=7

 

In order to find the x-intercept of this line, set the value of y to 0 and solve for x:

2(0)+x=7

x=7

The x-intercept is 7.

The graphs of both lines are shown below.  The original equation is in black and the perpendicular equation is in green.  Notice that the green line crosses the x-axis at 7 (the x-intercept).

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