Question

PLLLLEEEEEEAAASSSSSSSSEEEE

Answer 1

Given:

Line a passes through (2, 10) and (4, 13).

Line b passes through (4, 9) and (6, 12).

Line c passes through (2, 10) and (4, 9).

To find:

Which of the lines, if any are perpendicular.

Solution:

If a line passes through two points, then the slope of line is

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

Line a passes through (2, 10) and (4, 13).  So, slope of this line is

$m_a=\dfrac{13-10}{4-2}=\dfrac{3}{2}$

Line b passes through (4, 9) and (6, 12).  So, slope of this line is

$m_b=\dfrac{12-9}{6-4}=\dfrac{3}{2}$

Line c passes through (2, 10) and (4,9).  So, slope of this line is

$m_c=\dfrac{9-10}{4-2}=\dfrac{-1}{2}$

Product of slopes of to perpendicular lines is -1.

$m_a\cdot m_b=\dfrac{3}{2}\times \dfrac{3}{2}=\dfrac{9}{4}\neq -1$

$m_b\cdot m_c=\dfrac{3}{2}\times \dfrac{-1}{2}=\dfrac{-3}{4}\neq -1$

$m_a\cdot m_c=\dfrac{3}{2}\times \dfrac{-1}{2}=\dfrac{-3}{4}\neq -1$

Therefore, any of these lines are not perpendicular to each other.