please help Use the distance formula d=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√ to find the length of the line with endpoints at (1,3) and (3,1)? Question 2 options: 8√ 8 32−−√ 32 Hide hint for Question 2 Question 3 (1 point) Question 3 Saved What is the slope of the line that is perpendicular to a line whose equeation is 3y = -4x + 2 ? Question 3 options: 34 -34 43 -43

To find the length of the line with endpoints at (1,3) and (3,1), we will use the distance formula:

d = √((x2 − x1)^2 + (y2 − y1)^2)

Let's take the coordinates (1, 3) as (x1, y1) and the coordinates (3, 1) as (x2, y2).

Now let's substitute these values into the distance formula:

d = √((3 − 1)^2 + (1 − 3)^2) d = √((2)^2 + (-2)^2) d = √(4 + 4) d = √8 d = 2√2

Therefore, the length of the line segment is 2√2.

For the second question, the slope of a line perpendicular to another line can be found by taking the negative reciprocal of the original line's slope.

The equation of the original line is given as 3y = -4x + 2.

First, we need to express this in the slope-intercept form y = mx + b, where m is the slope.

Let's solve for y:

3y = -4x + 2 y = (-4/3)x + 2/3

So the slope (m) of this line is -4/3. The negative reciprocal of this slope is 3/4, because you flip the fraction to get -3/4 and then change the sign to get 3/4. Therefore, the slope of the line that is perpendicular to the line 3y = -4x + 2 is 3/4.