Perpendicular Lines 1
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Question 1 of 8
1. Question
Are the two lines at right angles with each other?- 1.
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2.
No
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Chapters- Chapters
Perpendicular lines have gradients which are negative reciprocals of one another.Gradient Intercept Form: y=mx+by=mx+b
- mm is the gradient of the line
- bb is the y-intercept (where the line cuts the y-axis)
The gradient is given by the coefficient of xx or the value of mm.yy == -12x+4−12x+4 mm == -12−12 yy == 2x-32x−3 mm == 22 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)(3,2) and is perpendicular to y=x+5y=x+5
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1.
y=x+15y=x+15 -
2.
y=-3x+5y=−3x+5 -
3.
y=2x+3y=2x+3 -
4.
y=-x+5y=−x+5
Correct
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Point Gradient Form: y-y1=m(x-x1)y−y1=m(x−x1)
- mm is the gradient of the line
- (x1,y1)(x1,y1) 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=mx+b)(y=mx+b), mm is the gradient.yy == 1x+51x+5 m1m1 == 11 Get the negative reciprocal of the m1m1 by flipping it upside down and changing the sign.m1m1 == 11 == 1111 Flip the number upside down m2m2 == -1−1 Change the sign Use the point gradient formula to find the equation.Point: (x1,y1)=(3,2)(x1,y1)=(3,2)Gradient: m2=-1m2=−1y-y1y−y1 == m(x-x1)m(x−x1) Point Gradient Formula y-2y−2 == -1(x-3)−1(x−3) Substitute values y-2y−2 == -x+3−x+3 y-2+2y−2+2 == -x+3+2−x+3+2 Add 22 to both sides yy == -x+5−x+5 Simplify y=-x+5y=−x+5 -
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Question 3 of 8
3. Question
>Find the equation of a line that passes through (3,7)(3,7) and is perpendicular to y=3x-2y=3x−2
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1.
y=-13x+8y=−13x+8 -
2.
y=3x+4y=3x+4 -
3.
y=-x+83y=−x+83 -
4.
y=-x+3y=−x+3
Correct
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Incorrect
Point Gradient Form: y-y1=m(x-x1)y−y1=m(x−x1)
- mm is the gradient of the line
- (x1,y1)(x1,y1) 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=mx+b)(y=mx+b), mm is the gradient.yy == 3x+23x+2 m1m1 == 33 Get the negative reciprocal of the m1m1 by flipping it upside down and changing the sign.m1m1 == 33 == 1313 Flip the number upside down m2m2 == -13−13 Change the sign Use the point gradient formula to find the equation.Point: (x1,y1)=(3,7)(x1,y1)=(3,7)Gradient: m2=-13m2=−13y-y1y−y1 == m(x-x1)m(x−x1) Point gradient Formula y-7y−7 == -13(x-3)−13(x−3) Substitute values y-7y−7 == -13x+1−13x+1 y-7+7y−7+7 == -13x+1+7−13x+1+7 Add 77 to both sides yy = -13x+8 Simplify y=-13x+8 -
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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-
1.
y=-43x-12 -
2.
y=34x+12 -
3.
y=43x+12 -
4.
y=-43x+12
Hint
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Chapters- Chapters
Point Gradient Form: y-y1=m(x-x1)
- m is the gradient of the line
- (x1,y1) 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=mx+b), m is the gradient.3x-4y+8 = 0 3x-4y+8 +4y = 0 +4y Add 4y on both sides 3x+8 = 4y 34x+84 = 44y Divide all terms by 4 34x+2 = y y = 34x+2 Identify the gradient m1 = 34 Get the negative reciprocal of the m1 by flipping it upside down and changing the sign.m1 = 34 = -43 Flip the number upside down m2 = -43 Change the sign Use the Point Gradient Formula to find the equation.Point: (x1,y1)=(6,4)Gradient: m2=2y-y1 = m(x-x1) Point Gradient Formula y-4 = -43(x-6) Substitute values y-4 = -43x+8 y-4 +4 = -43x+8 +4 Add 4 to both sides y = -43x+12 Simplify y=-43x+12 -
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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).-
1.
Perpendicular -
2.
Not Perpendicular
Hint
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Great Work!
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Chapters- Chapters
Gradient Formula
m=y2−y1x2−x1Remember
To prove the perpendicularity of two lines, the product of their gradients should be equal to -1First, solve for the gradient of each line using the gradient formulaLine 1(x1,y1) = S(-2,2) (x2,y2) = R(1,5) m1 = y2−y1x2−x1 Gradient Formula = 5−21−(−2) Substitute values = 33 Simplify = 1 Line 2(x1,y1) = T(1,2) (x2,y2) = U(4,-1) m2 = y2−y1x2−x1 Gradient Formula = −1−24−1 Substitute values = -33 Simplify = -1 To prove that these two lines are perpendicular, check if the product of the two gradients is equal to -1.m1×m2 = 1×-1 = -1 Therefore, Line 1 and Line 2 are perpendicular.Perpendicular -
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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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1.
y=2x+4 -
2.
y=x-1 -
3.
y=x+2 -
4.
y=12x+1
Correct
Keep Going!
Incorrect
Point Gradient Form: y-y1=m(x-x1)
- m is the gradient of the line
- (x1,y1) 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=mx+b), m is the gradient.y = -2x+6 m1 = -2 Get the negative reciprocal of the m1 by flipping it upside down and changing the sign.m1 = -2 = -12 Flip the number upside down m2 = 12 Change the sign Use the point gradient formula to find the equation.Point: (x1,y1)=(2,2)Gradient: m2=12y-y1 = m(x-x1) Point Gradient Formula y-2 = 12(x-2) Substitute values y-2 = 12x-1 y-2+2 = 12x-1+2 Add 2 to both sides y = 12x+1 Simplify y=12x+1 -
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Question 7 of 8
7. Question
>Find the equation of a line that passes through (-3,1) and is perpendicular to y=4x+1
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1.
y=-14x+14 -
2.
y=-4x+4 -
3.
y=-x+1 -
4.
y=-x+14
Correct
Keep Going!
Incorrect
Point Gradient Form: y-y1=m(x-x1)
- m is the gradient of the line
- (x1,y1) 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=mx+b), m is the gradient.y = 4x+1 m1 = 4 Get the negative reciprocal of the m1 by flipping it upside down and changing the sign.m1 = 4 = 14 Flip the number upside down m2 = -14 Change the sign Use the point gradient formula to find the equation.Point: (x1,y1)=(-3,1)Gradient: m2=-14y-y1 = m(x-x1) Point Gradient Formula y-1 = -14(x--3) Substitute values y-1 = -14x-34 y-1+1 = -14x-34+1 Add 1 to both sides y = -14x+14 Simplify y=-14x+14 -
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Question 8 of 8
8. Question
>Find the equation of a line that passes through (-6,4) and is perpendicular to 3y=2x-3
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1.
y=x-5 -
2.
y=-3x-10 -
3.
y=-3x-5 -
4.
y=-32x-5
Correct
Keep Going!
Incorrect
Point Gradient Form: y-y1=m(x-x1)
- m is the gradient of the line
- (x1,y1) 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=mx+b), m is the gradient.3y = 2x-3 y = 23x-33 Divide all terms by 3 y = 23x-1 Then, identify the gradient of the given equation.In gradient intercept form (y=mx+b), m is the gradient.y = 23x-1 m1 = 23 Get the negative reciprocal of the m1 by flipping it upside down and changing the sign.m1 = 23 = 32 Flip the number upside down m2 = -32 Change the sign Use the point gradient formula to find the equation.Point: (x1,y1)=(-6,4)Gradient: m2=-32y-y1 = m(x-x1) Point Gradient Formula y-4 = -32(x--6) Substitute values y-4 = -32x-9 y-4+4 = -32x-9+4 Add 4 to both sides y = -32x-5 Simplify y=-32x-5 -
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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