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2
2
2
The diagram shows the points
A
(
0
,
−
1
)
,
B
(
p
,
p
)
A(0,-1), B(p, p)
A
(
0
,
−
1
)
,
B
(
p
,
p
)
and
C
(
p
,
q
)
C(p, q)
C
(
p
,
q
)
.
\newline
(a) Given that the length of
A
B
A B
A
B
is
5
5
5
units, find the value of
p
p
p
.
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Math Problems
Grade 8
Write a linear equation from two points
Full solution
Q.
2
2
2
The diagram shows the points
A
(
0
,
−
1
)
,
B
(
p
,
p
)
A(0,-1), B(p, p)
A
(
0
,
−
1
)
,
B
(
p
,
p
)
and
C
(
p
,
q
)
C(p, q)
C
(
p
,
q
)
.
\newline
(a) Given that the length of
A
B
A B
A
B
is
5
5
5
units, find the value of
p
p
p
.
Calculate distance formula:
Calculate the distance between points
A
(
0
,
−
1
)
A(0,-1)
A
(
0
,
−
1
)
and
B
(
p
,
p
)
B(p,p)
B
(
p
,
p
)
using the distance formula:
d
=
(
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
)
d = \sqrt{((x_2 - x_1)^2 + (y_2 - y_1)^2)}
d
=
((
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
)
.
Substitute coordinates into formula:
Substitute the coordinates of
A
A
A
and
B
B
B
into the distance formula:
d
=
(
(
p
−
0
)
2
+
(
p
−
(
−
1
)
)
2
)
d = \sqrt{((p - 0)^2 + (p - (-1))^2)}
d
=
((
p
−
0
)
2
+
(
p
−
(
−
1
)
)
2
)
.
Simplify expression:
Simplify the expression:
d
=
p
2
+
(
p
+
1
)
2
d = \sqrt{p^2 + (p + 1)^2}
d
=
p
2
+
(
p
+
1
)
2
.
Set distance equal to
5
5
5
:
Since
A
B
AB
A
B
is
5
5
5
units, set the distance equal to
5
5
5
:
5
=
p
2
+
(
p
+
1
)
2
5 = \sqrt{p^2 + (p + 1)^2}
5
=
p
2
+
(
p
+
1
)
2
.
Square both sides:
Square both sides to remove the square root:
25
=
p
2
+
(
p
+
1
)
2
25 = p^2 + (p + 1)^2
25
=
p
2
+
(
p
+
1
)
2
.
Expand squared term:
Expand the squared term:
25
=
p
2
+
p
2
+
2
p
+
1
25 = p^2 + p^2 + 2p + 1
25
=
p
2
+
p
2
+
2
p
+
1
.
Combine like terms:
Combine like terms:
25
=
2
p
2
+
2
p
+
1
25 = 2p^2 + 2p + 1
25
=
2
p
2
+
2
p
+
1
.
Set equation to zero:
Subtract
25
25
25
from both sides to set the equation to zero:
0
=
2
p
2
+
2
p
−
24
0 = 2p^2 + 2p - 24
0
=
2
p
2
+
2
p
−
24
.
Divide entire equation:
Divide the entire equation by
2
2
2
to simplify:
0
=
p
2
+
p
−
12
0 = p^2 + p - 12
0
=
p
2
+
p
−
12
.
Factor quadratic equation:
Factor the quadratic equation: p + \(4)(p -
3
3
3
) =
0
0
0
\
Solve for
p
p
p
:
Set each factor equal to zero and solve for
p
p
p
:
p
+
4
=
0
p + 4 = 0
p
+
4
=
0
or
p
−
3
=
0
p - 3 = 0
p
−
3
=
0
.
Final solution:
Solve for
p
p
p
:
p
=
−
4
p = -4
p
=
−
4
or
p
=
3
p = 3
p
=
3
.
Final solution:
Solve for
p
p
p
:
p
=
−
4
p = -4
p
=
−
4
or
p
=
3
p = 3
p
=
3
.Since
p
p
p
is a coordinate, it cannot be negative in this context; therefore,
p
=
3
p = 3
p
=
3
.
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Question
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0
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(0, 4)
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?
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?
\newline
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q
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_
_
_
_
\_\_\_\_\_
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Find the slope of the line
y
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18
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x
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y + 3 = -\frac{1}{18}(x + 6)
y
+
3
=
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1
(
x
+
6
)
.
\newline
Write your answer as an integer or as a simplified proper or improper fraction.
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Question
Line
p
p
p
has an equation of
y
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y = -8x + 6
y
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8
x
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6
. Line
q
q
q
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p
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p
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(
2
,
−
2
)
(2, -2)
(
2
,
−
2
)
. What is the equation of line
q
q
q
?
?
?
\newline
Write the equation in slope-intercept form. Write the numbers in the equation as simplified proper fractions, improper fractions, or integers.
\newline
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Question
Is the graph of this equation a horizontal or vertical line?
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x
=
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x
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\newline
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\text{(A)horizontal line}
(A)horizontal line
\newline
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\text{(B)vertical line}
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Question
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A
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A
A
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B
B
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C
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C
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1
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Question
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‾
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‾
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‾
)
y - \underline{\hspace{1cm}} = \underline{\hspace{1cm}} (x - \underline{\hspace{1cm}})
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−
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