# Where on the line x - y = 8 does the perpendicular from (2, 1) intersect.

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### 2 Answers

From the given line equation: x-y=8, express it in slope-intercept form (y=mx+b).

x-y =8

+y +y

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x = 8+y

-8 -8

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x -8 =y Or y = 1x-8 where slope `m_1 = 1.`

Note that perpendicular lines follows` m_2 = -1/m_1`

Then `m_2=- 1/(-1)=-1`

Determine the other line equation using `m_2=-1` and that will pass through point (2,1).

Plug-in the values in y=mx+b

1 = (-1)(2)+b

1= -2 +b

+2 +2

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3 =b

Line equation: y=-1x+3 or x+y =3 based from `m_2=-1` and b =3

the two lines are x-y=8 and x+y =3.

Applying elimination method to solve for x.

x-y=8 Add the to equations.

+ x+y =3

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2x = 11 Cancels out y's since -y+y = 0.

`(2x)/2=11/2` Divide both sides by 2 to isolate x.

`x = 11/2`

To solve for y, subtract the equations as:

x -y = 8 Or x -y =8

- ( x +y =3) -------> + (-x -y = -3 ). Subtraction rules of signs.

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-2y = 5

`(-2y)/-2=5/2 ` Divide both sides by -2 to isolate y.

`y = - 5/2`

Intersection point: ` (11/2, - 5/2)`

This is the same as (5.5, -2.5)

The point of intersection of the perpendicular drawn from (2, 1) on the line x - y = 8 has to be determined.

Writing the equation of the given line in the form y = mx + c gives its slope. x - y = 8 => y = x - 8. The slope of the line is 1. A line perpendicular to this line has slope -1.

Let the point of intersection be (A, B). The slope of the line joining (A, B) and (2, 1) is (1 - B)/(2 - A) = -1. As (A, B) lies on x - y = 8, A - B = 8.

Use the equations (1 - B)/(2 - A) = -1 and A - B = 8 to solve for A and B.

(1 - B)/(2 - A) = -1

Substitute A = B + 8

=> (1 - B)/(2 - B - 8) = -1

=> (1 - B) = -1*(-B - 6)

=> (1 - B) = B + 6

=> 2B = -5

=> B = -2.5

A = 5.5

**The required point of intersection is (5.5, -2.5)**