Question

The line r passea through the
Points (-2, 2) and (5,8) and the lines
Passes through the points s (-8,4) and(-2,0).
Is the line perpendicular tos?​

Answer 1

Question:

• The line 'r' passes through the points (-2, 2) and (5,8) and the line 's' passes through the points (-8,4) and (-2,0) . Is line 'r' perpendicular to 's' ?

Concept to know:

• Two lines are said to be perpendicular if product of both the lines is equal to -1 , i.e $\sf m_{1}m_2=-1$

• $\sf m_{1}\; \textsf{is slope of line r }\; and \; m_2 \;is\; slope\; of \; line\; s$
• So, first of all we will find the slope of both lines, then we will multiply slopes of both the lines if we get -1 as a result then we can say both lines are perpendicular to each other.

$\red\bigstar\sf \ slope(m) = \dfrac{y_2-y_1}{x_2-x_1}$

$\longmapsto\sf \ slope\; of \; r (m_1) = \dfrac{y_2-y_1}{x_2-x_1}$

here,

• y₂ = 8
• y₁ = 2
• x₂ = 5
• x₁ = -2

Putting the values ,

$\longmapsto\sf\rm \ slope\; of \; r (m_1) \sf= \dfrac{8-2}{5-(-2)}\\\\\dashrightarrow \sf \ slope\; of \; r (m_1) \sf= \dfrac{6}{7}$

We got m₁ = 6/7, Now we will find m₂

$\longmapsto\sf \ slope\; of \; r (m_2) = \dfrac{y_2-y_1}{x_2-x_1}$

here,

• y₂ = 0
• y₁ = 4
• x₂ = -2
• x₁ = -8

Putting the values ,

$\longmapsto\sf\rm \ slope\; of \; s\; (m_2) \sf= \dfrac{0-4}{-2-(-8)}\\\\\dashrightarrow \sf \ slope\; of \; s\; (m_2) \sf= \dfrac{-4}{6} \\\\\dashrightarrow \sf \ slope\; of \; s\;(m_2) \sf= \dfrac{-2}{3}$

We got m₂ = -2/3

$\red\bigstar\; \textsf{Two lines are said to be perpendicular, if }\sf m_1m_2=-1$

$\longmapsto \sf m_1m_2 = \dfrac{6}{7}\times\dfrac{-2}{3} \\\\\dashrightarrow \sf m_1m_2 = -\dfrac{4}{7}$

Product of both slopes is not equal to -1.

Hence, Lines are not perpendicular.