Perpendicular Lines 1
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Question 1 of 8
1. Question
Are the two lines at right angles with each other?
Hint
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Perpendicular lines have gradients which are negative reciprocals of one another.Gradient Intercept Form: `y=mx+b`
- `m` is the gradient of the line
- `b` is the y-intercept (where the line cuts the y-axis)
The gradient is given by the coefficient of `x` or the value of `m`.`y` `=` `-1/2x+4` `m` `=` `-1/2` `y` `=` `2x-3` `m` `=` `2` The gradients are negative reciprocals of each other, so the lines are perpendicular or are at right angles with each other.The lines are perpendicular. -
Question 2 of 8
2. Question
>Find the equation of a line that passes through `(3,2)` and is perpendicular to `y=x+5`
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Point Gradient Form: `y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))`
- `color(tomato)(m)` is the gradient of the line
- `(\color(royalblue)(x_1,y_1) )` is a point that lies on the line
Remember
The gradient of perpendicular lines are negative reciprocals of each other.First, identify the gradient of the given equation.In gradient intercept form `(y= color(tomato)(m)x+b)`, `color(tomato)(m)` is the gradient.`y` `=` `color(tomato)(1)x+5` `m_1` `=` `1` Get the negative reciprocal of the `m_1` by flipping it upside down and changing the sign.`m_1` `=` `1` `=` `1/1` Flip the number upside down `m_2` `=` `-1` Change the sign Use the point gradient formula to find the equation.Point: `(x_1,y_1)=(3,2)`Gradient: `m_2=-1``y- color(royalblue)(y_1)` `=` `color(tomato)(m)(x- color(royalblue)(x_1))` Point Gradient Formula `y- color(royalblue)(2)` `=` `color(tomato)(-1)(x- color(royalblue)(3))` Substitute values `y-2` `=` `-x+3` `y-2 color(crimson)(+2)` `=` `-x+3 color(crimson)(+2)` Add `2` to both sides `y` `=` `-x+5` Simplify `y=-x+5` -
Question 3 of 8
3. Question
>Find the equation of a line that passes through `(3,7)` and is perpendicular to `y=3x-2`
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Point Gradient Form: `y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))`
- `color(tomato)(m)` is the gradient of the line
- `(\color(royalblue)(x_1,y_1) )` is a point that lies on the line
Remember
The gradient of perpendicular lines are negative reciprocals of each other.First, identify the gradient of the given equation.In gradient intercept form `(y= color(tomato)(m)x+b)`, `color(tomato)(m)` is the gradient.`y` `=` `color(tomato)(3)x+2` `m_1` `=` `3` Get the negative reciprocal of the `m_1` by flipping it upside down and changing the sign.`m_1` `=` `3` `=` `1/3` Flip the number upside down `m_2` `=` `-1/3` Change the sign Use the point gradient formula to find the equation.Point: `(x_1,y_1)=(3,7)`Gradient: `m_2=-1/3``y- color(royalblue)(y_1)` `=` `color(tomato)(m)(x- color(royalblue)(x_1))` Point gradient Formula `y- color(royalblue)(7)` `=` `color(tomato)(-1/3)(x- color(royalblue)(3))` Substitute values `y-7` `=` `-1/3x+1` `y-7 color(crimson)(+7)` `=` `-1/3x+1 color(crimson)(+7)` Add `7` to both sides `y` `=` `-1/3x+8` Simplify `y=-1/3x+8` -
Question 4 of 8
4. Question
Find the equation of a line that passes through `(6,4)` and is perpendicular to `3x-4y+8=0`Hint
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Point Gradient Form: `y-``y_1``=``m``(x-``x_1``)`
- `m` is the gradient of the line
- `(x_1,y_1)` is a point that lies on the line
Remember
The gradients of perpendicular lines are negative reciprocals of each other.First, convert the equation into gradient-intercept form and identify the gradient.In gradient-intercept form `(y=``m``x+b)`, `m` is the gradient.`3x-4y+8` `=` `0` `3x-4y+8` `+4y` `=` `0` `+4y` Add `4y` on both sides `3x+8` `=` `4y` `3/4x+8/4` `=` `4/4y` Divide all terms by `4` `3/4 x + 2` `=` `y` `y` `=` `3/4``x+2` Identify the gradient `m_1` `=` `3/4` Get the negative reciprocal of the `m_1` by flipping it upside down and changing the sign.`m_1` `=` `3/4` `=` `-4/3` Flip the number upside down `m_2` `=` `-4/3` Change the sign Use the Point Gradient Formula to find the equation.Point: `(x_1,y_1)=(6,4)`Gradient: `m_2=2``y-``y_1` `=` `m``(x-``x_1``)` Point Gradient Formula `y-``4` `=` `-4/3``(x-``6``)` Substitute values `y-4` `=` `-4/3x+8` `y-4` `+4` `=` `-4/3x + 8` `+4` Add `4` to both sides `y` `=` `-4/3x +12` Simplify `y=-4/3x+12` -
Question 5 of 8
5. Question
Check if the line passing through `R(-2,2)` and `S(1,5)` is perpendicular to the line passing through `T(1,2)` and `U(4,-1)`.Hint
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Gradient Formula
$$m=\frac{\color{#9a00c7}{y_2}-\color{#9a00c7}{y_1}}{\color{#00880a}{x_2}-\color{#00880a}{x_1}}$$Remember
To prove the perpendicularity of two lines, the product of their gradients should be equal to `-1`First, solve for the gradient of each line using the gradient formulaLine 1`(``x_1``,``y_1``)` `=` `S(``-2``,``2``)` `(``x_2``,``y_2``)` `=` `R(``1``,``5``)` `m_1` `=` $$\frac{\color{#9a00c7}{y_2}-\color{#9a00c7}{y_1}}{\color{#00880a}{x_2}-\color{#00880a}{x_1}}$$ Gradient Formula `=` $$\frac{\color{#9a00c7}{5}-\color{#9a00c7}{2}}{\color{#00880a}{1}-(\color{#00880a}{-2})}$$ Substitute values `=` `3/3` Simplify `=` `1` Line 2`(``x_1``,``y_1``)` `=` `T(``1``,``2``)` `(``x_2``,``y_2``)` `=` `U(``4``,``-1``)` `m_2` `=` $$\frac{\color{#9a00c7}{y_2}-\color{#9a00c7}{y_1}}{\color{#00880a}{x_2}-\color{#00880a}{x_1}}$$ Gradient Formula `=` $$\frac{\color{#9a00c7}{-1}-\color{#9a00c7}{2}}{\color{#00880a}{4}-\color{#00880a}{1}}$$ Substitute values `=` `-3/3` Simplify `=` `-1` To prove that these two lines are perpendicular, check if the product of the two gradients is equal to `-1`.`m_1 times m_2` `=` `1 times -1` `=` `-1` Therefore, Line `1` and Line `2` are perpendicular.Perpendicular -
Question 6 of 8
6. Question
>Find the equation of a line that passes through `(2,2)` and is perpendicular to `y=-2x+6`
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Point Gradient Form: `y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))`
- `color(tomato)(m)` is the gradient of the line
- `(\color(royalblue)(x_1,y_1) )` is a point that lies on the line
Remember
The gradient of perpendicular lines are negative reciprocals of each other.First, identify the gradient of the given equation.In gradient intercept form `(y= color(tomato)(m)x+b)`, `color(tomato)(m)` is the gradient.`y` `=` `color(tomato)(-2)x+6` `m_1` `=` `-2` Get the negative reciprocal of the `m_1` by flipping it upside down and changing the sign.`m_1` `=` `-2` `=` `-1/2` Flip the number upside down `m_2` `=` `1/2` Change the sign Use the point gradient formula to find the equation.Point: `(x_1,y_1)=(2,2)`Gradient: `m_2=1/2``y- color(royalblue)(y_1)` `=` `color(tomato)(m)(x- color(royalblue)(x_1))` Point Gradient Formula `y- color(royalblue)(2)` `=` `color(tomato)(1/2)(x- color(royalblue)(2))` Substitute values `y-2` `=` `1/2x-1` `y-2 color(crimson)(+2)` `=` `1/2x-1 color(crimson)(+2)` Add `2` to both sides `y` `=` `1/2x+1` Simplify `y=1/2x+1` -
Question 7 of 8
7. Question
>Find the equation of a line that passes through `(-3,1)` and is perpendicular to `y=4x+1`
Correct
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Point Gradient Form: `y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))`
- `color(tomato)(m)` is the gradient of the line
- `(\color(royalblue)(x_1,y_1) )` is a point that lies on the line
Remember
The gradient of perpendicular lines are negative reciprocals of each other.First, identify the gradient of the given equation.In gradient intercept form `(y= color(tomato)(m)x+b)`, `color(tomato)(m)` is the gradient.`y` `=` `color(tomato)(4)x+1` `m_1` `=` `4` Get the negative reciprocal of the `m_1` by flipping it upside down and changing the sign.`m_1` `=` `4` `=` `1/4` Flip the number upside down `m_2` `=` `-1/4` Change the sign Use the point gradient formula to find the equation.Point: `(x_1,y_1)=(-3,1)`Gradient: `m_2=-1/4``y- color(royalblue)(y_1)` `=` `color(tomato)(m)(x- color(royalblue)(x_1))` Point Gradient Formula `y- color(royalblue)(1)` `=` `color(tomato)(-1/4)(x- color(royalblue)(-3))` Substitute values `y-1` `=` `-1/4x-3/4` `y-1 color(crimson)(+1)` `=` `-1/4x-3/4 color(crimson)(+1)` Add `1` to both sides `y` `=` `-1/4x+1/4` Simplify `y=-1/4x+1/4` -
Question 8 of 8
8. Question
>Find the equation of a line that passes through `(-6,4)` and is perpendicular to `3y=2x-3`
Correct
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Point Gradient Form: `y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))`
- `color(tomato)(m)` is the gradient of the line
- `(\color(royalblue)(x_1,y_1) )` is a point that lies on the line
Remember
The gradient of perpendicular lines are negative reciprocals of each other.First, convert the equation into gradient-intercept form and identify the gradient.In gradient-intercept form `(y=``m``x+b)`, `m` is the gradient.`3y` `=` `2x-3` `y` `=` `2/3x-3/3` Divide all terms by `3` `y` `=` `2/3x-1` Then, identify the gradient of the given equation.In gradient intercept form `(y= color(tomato)(m)x+b)`, `color(tomato)(m)` is the gradient.`y` `=` `color(tomato)(2/3)x-1` `m_1` `=` `2/3` Get the negative reciprocal of the `m_1` by flipping it upside down and changing the sign.`m_1` `=` `2/3` `=` `3/2` Flip the number upside down `m_2` `=` `-3/2` Change the sign Use the point gradient formula to find the equation.Point: `(x_1,y_1)=(-6,4)`Gradient: `m_2=-3/2``y- color(royalblue)(y_1)` `=` `color(tomato)(m)(x- color(royalblue)(x_1))` Point Gradient Formula `y- color(royalblue)(4)` `=` `color(tomato)(-3/2)(x- color(royalblue)(-6))` Substitute values `y-4` `=` `-3/2x-9` `y-4 color(crimson)(+4)` `=` `-3/2x-9 color(crimson)(+4)` Add `4` to both sides `y` `=` `-3/2x-5` Simplify `y=-3/2x-5`
Quizzes
- Distance Between Two Points 1
- Distance Between Two Points 2
- Distance Between Two Points 3
- Midpoint of a Line 1
- Midpoint of a Line 2
- Midpoint of a Line 3
- Gradient of a Line 1
- Gradient of a Line 2
- Gradient Intercept Form: Graph an Equation 1
- Gradient Intercept Form: Graph an Equation 2
- Gradient Intercept Form: Write an Equation 1
- Determine if a Point Lies on a Line
- Graph Linear Inequalities 1
- Graph Linear Inequalities 2
- Convert Between General Form and Gradient Intercept Form 1
- Convert Between General Form and Gradient Intercept Form 2
- Point Gradient and Two Point Formula 1
- Point Gradient and Two Point Formula 2
- Parallel Lines 1
- Parallel Lines 2
- Perpendicular Lines 1
- Perpendicular Lines 2