Question

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Answer 1

Answer:

$d=3\sqrt{2}\ units$

Step-by-step explanation:

The complete question is

Line L contains points (3, 5) and (7, 9). Points P has coordinates (2, 10). Find the distance from P to L

step 1

Find the equation of the line L contains points (3,5) and (7,9)

Find the slope

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the given values

$m=\frac{9-5}{7-3}$

$m=\frac{4}{4}=1$

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=1\\point\ (3,5)$

substitute

$y-5=(1)(x-3)$

isolate the variable y

$y=x-3+5$

$y=x+2$ -----> equation A

step 2

Find the equation of the perpendicular line to the given line L  that passes through the point P

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal

so

The slope of the given line L  is $m=1$

The slope of the line perpendicular to the given line L  is

$m=-1$

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=-1\\point\ (2,10)$

substitute

$y-10=-(x-2)$

isolate the variable y

$y=-x+2+10$

$y=-x+12$ ----> equation B

step 3

Find the intersection point equation A and equation B

$y=x+2$ -----> equation A

$y=-x+12$ ----> equation B

solve the system by graphing

The intersection point is (5.7)

see the attached figure

step 4

we know that

The distance from point P to the the line L is equal to the distance between the point P and point (5,7)

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

(2,10) and (5,7)

substitute

$d=\sqrt{(7-10)^{2}+(5-2)^{2}}$

$d=\sqrt{(-3)^{2}+(3)^{2}}$

$d=\sqrt{18}\ units$

simplify

$d=3\sqrt{2}\ units$

see the attached figure to better understand the problem