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Determine which equation is parallel to line jk and which is perpendicular to line jk.

Answer 1

Answer:

5x + 3y = 13 parallel.

6x - 10y = 7 perpendicular.

Step-by-step explanation:

Two lines with slopes $m_1$ and $m_2$ are parallel when $m_1=m_2$ and are perpendicular when $m_1*m_2=-1$

Now determine the slope of all the lines

The line jk passes through the points

(-5,5) and (1,-5) so its slope is

$\frac{-5-5}{1-(-5)}=\frac{-10}{6}=-\frac{5}{3}$

To determine the slope of the lines in the blue rectangles, isolate y from each one and the coefficient of x is the slope

5x-3y = 8 ------> y = (5/3)x + 8/3 ------> slope 5/3

Neither parallel nor perpendicular.

6x+10y = 11 ------> y = (-6/10)x + 11/10 = (-3/5)x + 11/10 ------> slope -3/5

Neither parallel nor perpendicular.

5x + 3y = 13 ------> y = (-5/3)x + 13/3 ------> slope -5/3

This line is parallel

6x - 10y = 7 ------> y = (6/10)x - 7/10 = (3/5)x -7/10 ------> slope 3/5

Since (-5/3)(3/5) = -1 this line is perpendicular.

Answer 2

The equation that is parallel to line jk is; 5x + 3y = 13 and the equation that is perpendicular to line jk is 6x - 10y = 7.

What are the slopes of different lines?

The slope of parallel lines is same. Slopes of perpendicular lines are negative reciprocal of each other.

Now to determine the slope of all the lines

The line jk passes through the points (-5,5) and (1,-5) so its slope is

m = (-5 -5 )/ (1 +5 )

m = -10/ 6

m = -5/3

For the line;

5x-3y = 8

y = (5/3)x + 8/3

slope = 5/3

Neither parallel nor perpendicular.

For the line;

6x+10y = 11

y = (-6/10)x + 11/10

y = (-3/5)x + 11/10

slope = -3/5

Neither parallel nor perpendicular.

For the line;

5x + 3y = 13

y = (-5/3)x + 13/3

slope = -5/3

This line is parallel

For the line;

6x - 10y = 7

y = (6/10)x - 7/10

y = (3/5)x -7/10

slope = 3/5

Since, (-5/3)(3/5) = -1 this line is perpendicular.

Hence, The equation that is parallel to line jk is; 5x + 3y = 13 and the equation that is perpendicular to line jk is 6x - 10y = 7.

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