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).