Perpendicular lines questions with answers

Questions

1. Question 1Perpendicular gradient

   Find the gradient of a line perpendicular to the line $4x - y = 9$.

   Show the worked solution

   1. Step 1: Make y the subject to read the gradient: y = mx + c$-y = -4x + 9$$y = 4x - 9$$\text{Gradient} = 4$
   2. Step 2: Perpendicular gradients multiply to −1: flip and change the sign$m_1 \times m_2 = -1$$\text{Perpendicular gradient} = -\dfrac{1}{4}$
2. Question 2Perpendicular gradient

   Find the gradient of a line perpendicular to the line $4x + 3y = 12$.

   Show the worked solution

   1. Step 1: Make y the subject to read the gradient: y = mx + c$3y = -4x + 12$$y = \dfrac{-4}{3}x + 4$$\text{Gradient} = -\dfrac{4}{3}$
   2. Step 2: Perpendicular gradients multiply to −1: flip and change the sign$m_1 \times m_2 = -1$$\text{Perpendicular gradient} = \dfrac{3}{4}$
3. Question 3Line through a point

   Find the equation of the line which passes through the point $(3, 0)$ and is perpendicular to the line $x - 2y = 7$.

   Show the worked solution

   1. Step 1: Make y the subject to read the gradient: y = mx + c$-2y = -x + 7$$y = \dfrac{1}{2}x + \dfrac{-7}{2}$$\text{Gradient} = \dfrac{1}{2}$
   2. Step 2: Perpendicular gradients multiply to −1: flip and change the sign$m_1 \times m_2 = -1$$\text{Perpendicular gradient} = -2$
   3. Step 3: Through (3, 0): y − y₁ = m(x − x₁), then make y the subject$y - 0 = -2(x - 3)$$\text{y-intercept } c = 6$
4. Question 4Line through a point

   Find the equation of the line which passes through the point $(1, 2)$ and is perpendicular to the line $3x - 2y = 4$.

   Show the worked solution

   1. Step 1: Make y the subject to read the gradient: y = mx + c$-2y = -3x + 4$$y = \dfrac{3}{2}x - 2$$\text{Gradient} = \dfrac{3}{2}$
   2. Step 2: Perpendicular gradients multiply to −1: flip and change the sign$m_1 \times m_2 = -1$$\text{Perpendicular gradient} = -\dfrac{2}{3}$
   3. Step 3: Through (1, 2): y − y₁ = m(x − x₁), then make y the subject$y - 2 = -\dfrac{2}{3}(x - 1)$$\text{y-intercept } c = \dfrac{8}{3}$
5. Question 5Perpendicular bisector

   Find the equation of the perpendicular bisector of the line joining $A(2, -3)$ and $B(4, 5)$.

   Show the worked solution

   1. Step 1: It passes through the mid-point of AB: (average of the x's, average of the y's)$x = \dfrac{2 + 4}{2}$$= 3$$y = \dfrac{-3 + 5}{2}$$= 1$
   2. Step 2: It is perpendicular to AB: m₁ × m₂ = −1. Put the coordinates into the gradient formula$m_{AB} = \dfrac{y_2 - y_1}{x_2 - x_1}$$m_{AB} = \dfrac{5 - (-3)}{4 - 2}$$= 4$$\text{Perpendicular gradient} = -\dfrac{1}{4}$
   3. Step 3: y − y₁ = m(x − x₁) through the mid-point, then make y the subject$y = -\dfrac{1}{4}x + \dfrac{7}{4}$
6. Question 6Perpendicular bisector

   Find the equation of the perpendicular bisector of the line joining $A(2, 2)$ and $B(8, 6)$.

   Show the worked solution

   1. Step 1: It passes through the mid-point of AB: (average of the x's, average of the y's)$x = \dfrac{2 + 8}{2}$$= 5$$y = \dfrac{2 + 6}{2}$$= 4$
   2. Step 2: It is perpendicular to AB: m₁ × m₂ = −1. Put the coordinates into the gradient formula$m_{AB} = \dfrac{y_2 - y_1}{x_2 - x_1}$$m_{AB} = \dfrac{6 - 2}{8 - 2}$$= \dfrac{2}{3}$$\text{Perpendicular gradient} = -\dfrac{3}{2}$
   3. Step 3: y − y₁ = m(x − x₁) through the mid-point, then make y the subject$y = -\dfrac{3}{2}x + \dfrac{23}{2}$