Perpendicular lines: equations 1
H
*
1
Here are two perpendicular lines.
Point P has coordinates (1, 3)
y
2
Here are two perpendicular lines.
Point P has coordinates (1, 4)
y
y = 2x + 1
Line A
P
y=
1
x
4
Line A
P
x
O
1a
2a
Calculate the gradient of Line A.
2
×
x
O
Calculate the gradient of Line A.
1
= –1
1
Gradient of
Line A
y = 2x + 1
1b Use your answer to 1a to find the
equation of Line A.
Gradient of
Line A
1
×
+
x
×
+
Equation of Line A:
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= –1
4
Gradient of
x
×
y=
Gradient of
y=
Gradient of
Line A
1
x
4
2b Use your answer to 1a to find the
equation of Line A.
Gradient of
Line A
y
x
3
1
×
+
x
×
+
Equation of Line A:
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y
4
y=
1
Perpendicular lines: equations 2
H
*
1
Here are two perpendicular lines.
Point P has coordinates (1, 4)
2
Here are two perpendicular lines.
Point P has coordinates (1, 2)
y
y
P
Line A
P
Line A
x
O
1a Calculate the gradient of OP.
x
y
–
–
0
0
2a Calculate the gradient of OP.
y
–
–
=
÷
1b Calculate the gradient
2b Calculate the gradient
of Line A.
of Line A.
×
= –1
×
1c Find the equation
y
1 ×
4
of Line A.
x
y=
y
×
x
×
Equation of Line A:
= –1
2c Find the equation
of Line A.
x
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x
=
÷
x
x
O
×
Equation of Line A:
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y=
2
Perpendicular lines: equations 3
H
*
1
Here are two perpendicular lines.
Point P has coordinates (2, 1)
2
Here are two perpendicular lines.
Point P has coordinates (6, 2)
y
y
Line A
P
P
x
O
Line A
1a Calculate the gradient of OP.
x
y
–
–
0
0
2a Calculate the gradient of OP.
y
–
–
=
÷
1b Calculate the gradient
2b Calculate the gradient
of Line A.
of Line A.
×
= –1
×
1c Find the equation
y
2 ×
1
of Line A.
x
y=
y
×
x
×
Equation of Line A:
= –1
2c Find the equation
of Line A.
x
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x
=
÷
x
x
O
×
Equation of Line A:
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y=
3
Perpendicular lines: equations
Test
H
*
1
Here are two perpendicular lines.
Point P has coordinates (4, 2)
y
Line A
P
x
O
1a Calculate the gradient of OP.
Answer
1b Calculate the gradient Line A.
Answer
1c Find the equation of Line A.
Answer
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Perpendicular lines: equations
Answers
H
*
Perpendicular lines: equations 1
2a x 1–0=1
y 2–0=2
2÷1=2
2
1
1 a × − = −1
1
2
2b
The gradient of line A is −
1 b 1× −
x×−
1
2
2 c 1× −
x×−
1
1
1
1
1
→ – x → +3 → – x + 3
2
2
2
2
2
1
1
y= – x+2
2
2
1a x 2–0=2
y 1–0=1
1
4
× − = −1
4
1
1÷2=
1b
2b 1➝ ×–4 ➝ –4 ➝ +8 ➝ 4
x ➝ × – 4 ➝ – 4x ➝ + 8 ➝ – 4x + 8
1
2
× − = −1
2
1
x ➝ × – 2 ➝ – 2x ➝ + 5 ➝ – 2x + 5
y = – 2x + 5
Perpendicular lines: equations 2
2a x 6–0=6
1a x 1–0=1
y 2–0=2
y 4–0=4
4÷1=4
2÷6=
4
1
× − = −1
1
4
x×−
1
2
1c 2➝ ×–2 ➝ –4 ➝ +5 ➝ 1
y = – 4x + 8
1 c 1× −
1
1
1
1
1
→ – x → +2 → – x + 2
2
2
2
2
2
Perpendicular lines: equations 3
The gradient of line A is – 4
1b
1
1
1
→ – → +2 → 2
2
2
2
1
1
1
→ – → +3 → 4
2
2
2
1
1
y= – x+3
2
2
2a
2
1
× − = −1
1
2
2b
1
1
1
→ – → +4 → 4
4
4
4
1
3
1
3
× − = −1
3
1
2 c 6 ➝ × – 3 ➝ – 18 ➝ + 20 ➝ 2
1
1
1
1
1
→ – x → +4 → – x + 4
4
4
4
4
4
x ➝ × – 3 ➝ – 3x ➝ + 20 ➝ – 3x + 20
y = – 3x + 20
1
1
y= – x+4
4
4
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5
Perpendicular lines: equations
Answers
H
*
Perpendicular lines: equations Test
1a x 4–0=4
y 2–0=2
2÷4=
1b
1
2
1
2
× − = −1
2
1
1 c 4 ➝ × – 2 ➝ – 8 ➝ + 10 ➝ 2
x ➝ × – 2 ➝ – 2x ➝ + 10 ➝ – 2x + 10
y = – 2x + 10
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6