Identify perpendicular lines Geometry

The points \left(0,-1\right) and \left(0.5{,}0\right) are on the line d_1. Therefore, the slope of the line is:

m_1= \dfrac{0-\left(-1\right)}{0.5-0}=2

The points \left(0{,}1.5\right) and \left(2{,}0\right) are on the line d_2. Therefore, the slope of the line is:

m_2= \dfrac{0-\left(1.5\right)}{2-0}=-0.75

We have:

m_1 \times m_2 = 2 \times -0.75 = 1.5

Since m_1 \times m_2 \ne -1, we can deduce that the lines are not perpendicular.

The points \left(0,-3\right) and \left(3{,}0\right) are on the line d_1. Therefore, the slope of the line is:

m_1= \dfrac{0-\left(-3\right)}{3-0}=\dfrac{3}{3}=1

The points \left(0{,}3\right) and \left(3{,}0\right) are on the line d_2. Therefore, the slope of the line is:

m_2= \dfrac{0-3}{3-0}=\dfrac{-3}{3}=-1

We have:

m_1 \times m_2 = 1 \times -1 = -1

Since m_1 \times m_2 =- 1, we deduce that the lines are perpendicular.

The points \left(2{,}1\right) and \left(0,-3\right) are on the line d_1. Therefore, the slope of the line is:

m_1= \dfrac{-3-1}{0-2}=\dfrac{-4}{-2}=2

The points \left(0{,}4\right) and \left(4{,}0\right) are on the line d_2. Therefore, the slope of the line is:

m_2= \dfrac{0-4}{4-0}=-1

We have:

m_1 \times m_2 = 2 \times-1 = -2

Since m_1 \times m_2 \ne 1, we deduce that the lines are not perpendicular.

The points \left(0{,}2\right) and \left(-1,-1\right) are on the line d_1. Therefore, the slope of the line is:

m_1= \dfrac{-1-2}{-1-0}=\dfrac{-3}{-1}=3

The points \left(0{,}2\right) and \left(6{,}0\right) are on the line d_2. Therefore, the slope of the line is:

m_2= \dfrac{0-2}{6-0}=\dfrac{-2}{6}=-\dfrac{1}{3}

We have:

m_1 \times m_2 = 3 \times -\dfrac{1}{3} = -1

Since m_1 \times m_2=-1 , we deduce that the lines are perpendicular.

The points \left(1{,}0\right) and \left(0{,}1\right) are on the line d_1. So the slope of the line is:

m_1= \dfrac{1-0}{0-1}=-1

The points \left(0{,}2\right) and \left(-2{,}0\right) are on the line d_2. So the slope of the line is:

m_2= \dfrac{0-2}{-2-0}=1

Hence we have:

m_1 \times m_2 = 1 \times -1 = -1

Since m_1 \times m_2=-1 we can deduce that the lines are perpendicular.

The points \left(0,-2\right) and \left(2{,}0\right) are on the line d_1. Therefore, the slope of the line is:

m_1= \dfrac{0-\left(-2\right)}{2-0}=\dfrac{2}{2}=1

The points \left(-2,-1\right) and \left(-3,-3\right) are on the line d_2. Therefore, the slope of the line is:

m_2= \dfrac{-3-\left(-1\right)}{-3-\left(-2\right)}=\dfrac{-2}{-1}=2

We have:

m_1 \times m_2 = 2 \times 1 = 2

Since m_1 \times m_2\neq-1 , we can deduce that the lines are not perpendicular.

The points \left(0{,}3\right) and \left(-3{,}0\right) are on the line d_1. Therefore, the slope of the line is:

m_1= \dfrac{0-3}{-3-0}=\dfrac{-3}{-3}=1

The points \left(0,-2\right) and \left(-2{,}0\right) are on the line d_2. Therefore, the slope of the line is:

m_2= \dfrac{0-\left(-2\right)}{-2-0}=\dfrac{2}{-2}=-1

We have:

m_1 \times m_2 = 1 \times -1 = -1

Since m_1 \times m_2=-1 , we can deduce that the lines are perpendicular.