Mathematics, 22.10.2019 18:50 cleef6451

Determine if jk and lm are parallel, perpendicular, or neither j(1,9),k(7,4),l(8,13),m(-2,1)

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Answer from: Iamchill5998

The slope of JK is the negative reciprocal of the slope of LM, they are therefore perpendicular.

Note the following key-points regarding parallel lines and perpendicular lines:

The slope of two lines that are parallel to each other is the same.If the slope of one line is 4, the slope of the other would be 4 if they are parallel lines.The slope of perpendicular lines are always a negative reciprocal of each other.If a line has slope value of 3/2, and the slope of another line is -2/3, both lines are perpendicular.

Let's find the slope of JK and LM to know if they are perpendicular, parallel or not:

Given:

$j(1,9)\\\\k(7,4)\\\\l(8,13)\\\\m(-2,1)$

Find the slope of JK:

Slope formula = $\frac{y_2-y_1}{x_2-x_1}$

Let,

$j(1,9) = (x_1, y_1)\\\\k(7,4) = (x_2, y_2)$

Plug in the values

$\frac{4 - 9}{7-1} = \frac{-5}{6}$

The slope of JK is $-\frac{5}{6}$

Find the slope LM:

Slope formula = $\frac{y_2-y_1}{x_2-x_1}$

Let,

$L(8, 13) = (x_1, y_1)\\\\M(-2,1) = (x_2, y_2)$

Plug in the values

$\frac{1-13}{-2 -8} = \frac{-12}{-10} = \frac{6}{5}$

The slope of LM is $\frac{6}{5}$

The slope of JK, $-\frac{5}{6}$ is the negative reciprocal of the slope of LM, $\frac{6}{5}$ , therefore, JK and LM are perpendicular.

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Answer from: evazquez

"perpendicular"

Step-by-step explanation:

We need to find slope (m) of both line JK and LM

Slope has formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

Lets find slope of JK:

$m=\frac{4-9}{7-1}\\m=-\frac{5}{6}$

Lets find slope of LM:

$m=\frac{y_2-y_1}{x_2-x_1}\\m=\frac{1-13}{-2-8}\\m=\frac{6}{5}$

we know if we multiply two lines' slope and get -1, they are perpendicular. Let's check:

$-\frac{5}{6}*\frac{6}{5}=-1$

Hence, they are perpendicular

Answer from: Quest

75%

step-by-step explanation:

Answer from: Quest

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step-by-step explanation:

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Determine if jk and lm are parallel, perpendicular, or neither j(1,9),k(7,4),l(8,13),m(-2,1)...

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