EEPROM Hex File Generation with SigCon Architect
1.
The point
(
t
, 2
t
)
lies on the line
4
x
− 3
y
+ 6 = 0
. Find the value of
t
.
Substitute the coordinates of (
t
, 2
t
) in for
x
and
y
:
4
x
− 3
y
+ 6 = 0
4(
t
) − 3(2
t
) + 6 = 0
4
t
− 6
t
+ 6 = 0
−2
t
+ 6 = 0
6 = 2
t
=
t
3 =
t
2.
Investigate whether the lines
l
: 2
x
− 3
y
+ 7 = 0
and
m
: 6
x
+ 4
y
− 5 = 0
are perpendicular.
l
:
2
x
− 3
y
+ 7 = 0
a
= 2,
b
= −3
Slope =
=
Slope
L
=
m
:
6
x
+ 4
y
− 5 = 0
a
= 6,
b
= 4
Slope =
Slope
L
=
=
If the lines are perpendicular,
m
l
×
m
m
= −1
m
l
×
m
m
=
= −1
Therefore,
l
m
3.
Find the equation of the line
k
, which passes through the point (3, −2)
and is perpendicular to the line passing through (3, 3) and (5, 7).
Find the slope of the line passing through (3,3) and (5,7)
Slope =
(3, 3)
(5, 7)
(
x
1
,
y
1
)
(
x
2
,
y
2
)
Slope =
=
= 2
3.
Find the equation of the line
k
, which passes through the point (3, −2)
and is perpendicular to the line passing through (3, 3) and (5, 7).
Since the lines are perpendicular:
m
1
×
m
2
= −1
2 ×
m
2
= −1
m
2
=
(
x
1
,
y
1
) = (3, −2)
m
=
(
y
−
y
1
) =
m
(
x
−
x
1
)
(
y
− (−2)) =
(
y
+ 2) =
2(
y
+ 2) = −1(
x
− 3)
2
y
+ 4 = −
x
+ 3
x
+ 2
y
+ 1 = 0
4.
A
(−2, 3) and
B
(2, 6) are two points.
(i)
Plot the points
A
and
B
on a coordinate diagram.
4.
A
(−2, 3) and
B
(2, 6) are two points.
(ii)
Find the equation of
AB
(−2, 3)
(2, 6)
(
x
1
,
y
1
)
(
x
2
,
y
2
)
Slope =
Slope =
=
=
(
x
1
,
y
1
) = (−2, 3) Slope =
0 = 3
x
− 4
y
+ 18
4.
A
(−2, 3) and
B
(2, 6) are two points.
(iii)
k
is the line 3
x
+ 2
y
− 9 = 0. Show that
k
passes through the midpoint of [
AB
].
A
(−2, 3)
B
(2, 6)
(
x
1
,
y
1
)
(
x
2
,
y
2
)
Midpoint
=
=
=
=
4.
A
(−2, 3) and
B
(2, 6) are two points.
(iii)
k
is the line 3
x
+ 2
y
− 9 = 0. Show that
k
passes through the midpoint of [
AB
].
Substitute the coordinates of the midpoint into the equation of the line,
k
:
k
:
3
x
+ 2
y
− 9 = 0
0 + 9 − 9 = 0
0 = 0
Midpoint is on
k
4.
A
(−2, 3) and
B
(2, 6) are two points.
(iv)
Investigate if
k
is perpendicular to
AB.
k
:
3
x
+ 2
y
− 9 = 0
Slope =
Slope =
=
a
= 3,
b
= 2
4.
A
(−2, 3) and
B
(2, 6) are two points.
(iv)
Investigate if
k
is perpendicular to
AB.
If the lines are perpendicular lines then
m
1
×
m
2
= −1
Therefore, the lines are not perpendicular.
5.
The point (3, 2) is on the line
k
:
ax
+
y
− 8 = 0
.
Find the value of
a
.
Substitute the coordinates of (3, 2) in for
x
and
y
:
k
:
ax
+
y
− 8 = 0
a
(3) + 2 − 8 = 0
3
a
− 6 = 0
3
a
= 6
a
= 2
(i)
5.
The point (3, 2) is on the line
k
:
ax
+
y
− 8 = 0
.
Find the points where the line
k
intersects the
x
and
y
axes.
At the
y
-axis,
x
= 0:
2
x
+
y
− 8 = 0
2(0) +
y
− 8 = 0
0 +
y
− 8 = 0
y
= 8
y
-intercept = (0, 8)
(ii)
At the
x
-axis,
y
= 0:
2
x
+
y
− 8 = 0
2
x
+ 0 − 8 = 0
2
x
= 8
x
= 4
x
-intercept = (4, 0)
5.
The point (3, 2) is on the line
k
:
ax
+
y
− 8 = 0
.
Hence sketch the line
k.
(iii)
6.
A
(0, 0),
B
(−4, 1) and
C
(2, 8) are the vertices of a triangle.
Prove that the triangle is right-angled at the point
A
.
(i)
If the triangle is right-angled then [
AB
] is perpendicular to [
AC
],
so
m
1
×
m
2
= − 1.
6.
A
(0, 0),
B
(−4, 1) and
C
(2, 8) are the vertices of a triangle.
Prove that the triangle is right-angled at the point
A
.
(i)
Slope of [
AB
]:
(0, 0) (−4, 1)
(
x
1
,
y
1
) (
x
2
,
y
2
)
Slope of [
AC
]:
(0, 0) (2, 8)
(
x
1
,
y
1
) (
x
2
,
y
2
)
Slope =
m
AB
=
=
Slope =
6.
A
(0, 0),
B
(−4, 1) and
C
(2, 8) are the vertices of a triangle.
Prove that the triangle is right-angled at the point
A
.
(i)
Investigate if:
m
AB
×
m
AC
= −1
−1 = −1
Therefore
AB
AC
, showing the triangle is a right-angled triangle at the
point
A
.
6.
A
(0, 0),
B
(−4, 1) and
C
(2, 8) are the vertices of a triangle.
Find the area of the triangle
ABC
.
(ii)
Area
=
=
=
=
=
= 17 units
2
(0, 0) (−4, 1) (2, 8)
(
x
1
,
y
1
) (
x
2
,
y
2
)
7.
A
(−3, 4),
B
(5, 6)
and
C
(1, −2)
are three points.
P
is the midpoint of
[
AB
]
and
Q
is the
midpoint of
[
AC
]
.
Find the coordinates of
P
and the coordinates of
Q
.
(i)
A
(−3, 4)
B
(5, 6)
(
x
1
,
y
1
)
(
x
2
,
y
2
)
Midpoint =
7.
A
(−3, 4),
B
(5, 6)
and
C
(1, −2)
are three points.
P
is the midpoint of
[
AB
]
and
Q
is the
midpoint of
[
AC
]
.
Find the coordinates of
P
and the coordinates of
Q
.
(i)
A
(−3, 4)
C
(1, −2)
(
x
1
,
y
1
)
(
x
2
,
y
2
)
Midpoint =
7.
A
(−3, 4),
B
(5, 6)
and
C
(1, −2)
are three points.
P
is the midpoint of
[
AB
]
and
Q
is the
midpoint of
[
AC
]
.
Plot
A
,
B
,
C
,
P
and
Q
on a coordinate diagram and show the line segments
[
BC
] and [
PQ
] on your diagram.
(ii)
7.
A
(−3, 4),
B
(5, 6)
and
C
(1, −2)
are three points.
P
is the midpoint of
[
AB
]
and
Q
is the
midpoint of
[
AC
]
.
Investigate whether
PQ
is parallel to
BC
.
(iii)
If the lines are parallel their slope will be equal.
P
(1, 5)
Q
(−1, 1)
(
x
1,
y
1
) (
x
2
,
y
2
)
B
(5, 6)
C
(1, −2)
(
x
1
,
y
1
)
(
x
2
,
y
2
)
7.
A
(−3, 4),
B
(5, 6)
and
C
(1, −2)
are three points.
P
is the midpoint of
[
AB
]
and
Q
is the
midpoint of
[
AC
]
.
Investigate whether
PQ
is parallel to
BC
.
(iii)
Parallel lines have equal slopes.
Since
m
PQ
= m
BC
,
PQ
||
BC
8.
A
(2, 3),
B
(1, −2
) and
C
(12, 1)
are the vertices of a triangle
ABC
.
Investigate if the triangle is right-angled.
(i)
Right-angled implies two sides are perpendicular to each other
A
(2, 3)
B
(1, −2)
(
x
1
,
y
1
)
(
x
2
,
y
2
)
A
(2, 3)
C
(12, 1)
(
x
1
,
y
1
)
(
x
2
,
y
2
)
8.
A
(2, 3),
B
(1, −2
) and
C
(12, 1)
are the vertices of a triangle
ABC
.
Investigate if the triangle is right-angled.
(i)
If the lines are perpendicular,
Therefore, the triangle is right-angled at the point
A
.
8.
A
(2, 3),
B
(1, −2
) and
C
(12, 1)
are the vertices of a triangle
ABC
.
The point (3,
k
) is on the line
AB
. Find the value of
k
.
(ii)
(
x
1
,
y
1
) = (2, 3)
m
= 5
(
y
–
y
1
) =
m
(
x
–
x
1
)
(
y
– 3) = 5(
x
– 2)
y
– 3 = 5
x
– 10
0 = 5
x
–
y
– 7
Find the equation of the line
AB
:
8.
A
(2, 3),
B
(1, −2
) and
C
(12, 1)
are the vertices of a triangle
ABC
.
The point (3,
k
) is on the line
AB
. Find the value of
k
.
(ii)
Substitute the coordinates of (3,
k
) in for
x
and
y
:
0 = 5
x
–
y
– 7
0 = 5(3) –
k
– 7
0 = 15 –
k
– 7
0 = 8 –
k
k
= 8
9.
The line
l
intersect the
x
-axis at (-4, 0) and the
y
-axis at (0, 3)
.
Find the slope of
l
.
(i)
(−4, 0)
(0, 3)
(
x
1
,
y
1
)
(
x
2
,
y
2
)
9.
The line
l
intersect the
x
-axis at (-4, 0) and the
y
-axis at (0, 3)
.
Find the equation of
l
.
(ii)
(
x
1
,
y
1
) = (−4, 0)
m
=
(
y
–
y
1
) =
m
(
x
–
x
1
)
9.
The line
m
passes through
(0, 0)
and is perpendicular to
l
.
Show the lines
l
and
m
on a coordinate diagram.
(iii)
Perpendicular lines
m
1
m
2
= −1
m
l
m
m
= −1
m
m
= −1
m
m
=
9.
Find the equation of
m
(iv)
(
x
1
,
y
1
) = (0, 0)
m
=
(
y
–
y
1
) =
m
(
x
–
x
1
)
(
y
– 0) =
3(
y
– 0) = −4(
x
– 0)
3
y
= −4
x
4
x
+ 3
y
= 0
The line
m
passes through
(0, 0)
and is perpendicular to
l
.
9.
The point (
t
, 4) is on the line
m
. Find the value of
t
.
(v)
Substitute the coordinates of (
t
, 4) into the line
m
:
4
x
+ 3
y
= 0
4(
t
) + 3(4) = 0
4
t
+ 12 = 0
4
t
= −12
t
= −3
The line
m
passes through
(0, 0)
and is perpendicular to
l
.
10.
The line
k
: 3
x
+ 2
y
− 12 = 0
intersects the
x
-axis at the point
P
and the
y
-axis at the
point
Q
.
Find the coordinates of the points
P
and
Q.
(i)
At
x
-axis,
y
= 0:
3
x
+ 2
y
– 12 = 0
3
x
+ 2(0) −12 = 0
3
x
= 12
x
= 4
x
-intercept,
P
= (4, 0)
At
y
-axis,
x
= 0:
3
x
+ 2
y
– 12 = 0
3(0) + 2
y
– 12 = 0
2
y
= 12
y
= 6
y
-intercept,
Q
= (0, 6)
10.
The line
k
: 3
x
+ 2
y
− 12 = 0
intersects the
x
-axis at the point
P
and the
y
-axis at the
point
Q
.
Graph the line
k
.
(ii)
10.
The line
k
: 3
x
+ 2
y
− 12 = 0
intersects the
x
-axis at the point
P
and the
y
-axis at the
point
Q
.
Hence, or otherwise, find the area of the triangle formed by the
x
-axis,
y
-axis
and the line
k
.
(iii)
(0, 0)
(4, 0)
(0, 6)
(
x
1
,
y
1
)
(
x
2
,
y
2
)
Area =
=
=
=
= 12 units
2
11.
Given the line
m
: 3
x
− 4
y
+ 16 = 0
:
Find the slope of
m
3
x
– 4
y
+ 16 = 0
a
= 3,
b
= −4
(i)
11.
Given the line
m
: 3
x
− 4
y
+ 16 = 0
:
The line
n
is perpendicular to
m
and cuts the
x
-axis at the point (3, 0).
Find the equation of
n
.
(ii)
m
n
m
m
m
n
= −1
m
n
= −1
m
n
=
(
x
1
,
y
1
) = (3, 0)
m
n
=
(
y
–
y
1
) =
m
(
x
–
x
1
)
(
y
– 0) =
3(
y
– 0) = −4(
x
– 3)
3
y
= −4
x
+ 12
4
x
+ 3
y
– 12 = 0
11.
Given the line
m
: 3
x
− 4
y
+ 16 = 0
:
Find the coordinates of the point of intersection of the lines
m
and
n
.
(iii)
3
x
– 4
y
= −16
4
x
+ 3
y
= 12
9
x
– 12
y
= −48
16
x
+ 12
y
= 48
25
x
= 0
x
= 0
Let
x
= 0:
3
x
– 4
y
= −16
3(0) – 4
y
= −16
−4
y
= −16
y
= 4
Point of intersection = (0, 4)
(
3)
(
4)
12.
Given the line
l
: 2
x
– 5
y
+ 16 = 0
, find:
the coordinates of
A
, where
l
intersects the
x
-axis
(i)
At the
x
-axis,
y
= 0:
2
x
– 5
y
+ 16 = 0
2
x
– 5(0) + 16 = 0
2
x
= −16
x
= −8
A
: (−8, 0)
12.
Given the line
l
: 2
x
– 5
y
+ 16 = 0
, find:
the equation of the line
k
, of slope 2 and passing through
B
(0, −4)
(ii)
(
x
1
,
y
1
) = (0, −4)
m
k
= 2
(
y
–
y
1
) =
m
(
x
–
x
1
)
(
y
– (−4)) = 2(
x
– 0)
y
+ 4 = 2
x
0 = 2
x
–
y
– 4
12.
Given the line
l
: 2
x
– 5
y
+ 16 = 0
, find:
the point
C
, where
l
∩
k
= {C
}.
(iii)
2
x
– 5
y
= −16
2
x
–
y
= 4
–2
x
+ 5
y
= 16
2
x
–
y
= 4
4
y
= 20
y
= 5
Let
y
= 5:
2
x
–
y
= 4
2
x
– 5 = 4
2
x
= 9
x
=
C
=
= (4·5, 5)
(
−1)
"Generate custom Intel-format Hex file using SigCon Architect software to program EEPROM regardless of live device presence. Step-by-step guide for setting up SigCon Architect, loading device profiles, determining register settings, and writing EEPROM hex files."
Download Presentation
Please find below an Image/Link to download the presentation.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author.If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.
You are allowed to download the files provided on this website for personal or commercial use, subject to the condition that they are used lawfully. All files are the property of their respective owners.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author.
E N D
Presentation Transcript
CHAPTER 14 Coordinate Geometry I: The Line Solutions: Revision - Section A
14 Revision and Exam Style Questions: Section A 1. The point (t, 2t) lies on the line 4x 3y + 6 = 0. Find the value of t. Substitute the coordinates of (t, 2t) in for x and y: 4x 3y + 6 = 0 4(t) 3(2t) + 6 = 0 4t 6t + 6 = 0 2t + 6 = 0 6 = 2t 6 2 = t 3 = t
14 Revision and Exam Style Questions: Section A 2. Investigate whether the lines l: 2x 3y + 7 = 0 and m: 6x + 4y 5 = 0 are perpendicular. l: 2x 3y + 7 = 0 m: 6x + 4y 5 = 0 a = 2, b= 3 a = 6, b = 4 -a -a Slope = Slope = b b 6 4 3 2 -2 -3 2 3 = = SlopeL = SlopeL = If the lines are perpendicular, ml mm= 1 ml mm =2 3 -3 = 1 2 Therefore, l m
14 Revision and Exam Style Questions: Section A 3. Find the equation of the line k, which passes through the point (3, 2) and is perpendicular to the line passing through (3, 3) and (5, 7). Find the slope of the line passing through (3,3) and (5,7) y2- y1 x2- x1 Slope = (3, 3) (5, 7) (x1, y1) (x2, y2) 7-3 5-3 4 2 Slope = = = 2
14 Revision and Exam Style Questions: Section A 3. Find the equation of the line k, which passes through the point (3, 2) and is perpendicular to the line passing through (3, 3) and (5, 7). Since the lines are perpendicular: m1 m2= 1 2 m2= 1 m2 = -1 2 (x1, y1) = (3, 2) -1 m = 2 (y y1) = m(x x1) -1 2(x-3) -1 2(x-3) (y ( 2)) = (y + 2) = 2(y+ 2) = 1(x 3) 2y+ 4 = x + 3 x + 2y + 1 = 0
14 Revision and Exam Style Questions: Section A 4. A ( 2, 3) and B (2, 6) are two points. Plot the points A and B on a coordinate diagram. (i)
14 Revision and Exam Style Questions: Section A 4. A ( 2, 3) and B (2, 6) are two points. Find the equation of AB (ii) (x1, y1) = ( 2, 3) Slope = 3 ( (y-3)=3 4(x-(-2)) (y-3)=3 4(x+2) 4 y-3 ( 4y-12=3x+6 ( 2, 3) (2, 6) 4 (x1, y1) (x2, y2) )=m x-x1 ( ) y- y1 Slope = y2- y1 x2- x1 6-3 2-(-2) Slope = )=3 x+2 ( ) 3 = 2+2 3 4 0 = 3x 4y + 18 =
14 Revision and Exam Style Questions: Section A 4. A ( 2, 3) and B (2, 6) are two points. k is the line 3x + 2y 9 = 0. Show that k passes through the midpoint of [AB]. (iii) A( 2, 3) B (2, 6) (x1, y1) (x2, y2) x1+x2 2 , y1+ y2 Midpoint = 2 -2+2 2 , 3+6 = 2 0 2, 9 = 2 0, 9 = 2
14 Revision and Exam Style Questions: Section A 4. A ( 2, 3) and B (2, 6) are two points. k is the line 3x + 2y 9 = 0. Show that k passes through the midpoint of [AB]. (iii) Substitute the coordinates of the midpoint into the equation of the line, k: k: 3x + 2y 9 = 0 -9=0 3(0)+29 2 0 + 9 9 = 0 Midpoint is on k 0 = 0
14 Revision and Exam Style Questions: Section A 4. A ( 2, 3) and B (2, 6) are two points. Investigate if k is perpendicular to AB. (iv) k: 3x + 2y 9 = 0 Slope = -a b a = 3, b = 2 Slope = -a b = -3 2
14 Revision and Exam Style Questions: Section A 4. A ( 2, 3) and B (2, 6) are two points. Investigate if k is perpendicular to AB. (iv) If the lines are perpendicular lines then m1 m2= 1 3 4 -3 2=-1 -9 8 -1 Therefore, the lines are not perpendicular.
14 Revision and Exam Style Questions: Section A 5. The point (3, 2) is on the line k: ax + y 8 = 0. Find the value of a. (i) Substitute the coordinates of (3, 2) in for x and y: k: ax + y 8 = 0 a(3) + 2 8 = 0 3a 6 = 0 3a = 6 a = 2
14 Revision and Exam Style Questions: Section A 5. The point (3, 2) is on the line k: ax + y 8 = 0. Find the points where the line k intersects the x and y axes. (ii) At the y-axis, x = 0: At the x-axis, y = 0: 2x + y 8 = 0 2x + y 8 = 0 2(0) + y 8 = 0 2x+ 0 8 = 0 0 + y 8 = 0 2x = 8 y = 8 x = 4 y-intercept = (0, 8) x-intercept = (4, 0)
14 Revision and Exam Style Questions: Section A 5. The point (3, 2) is on the line k: ax + y 8 = 0. Hence sketch the line k. (iii)
14 Revision and Exam Style Questions: Section A 6. A (0, 0), B ( 4, 1) and C (2, 8) are the vertices of a triangle. Prove that the triangle is right-angled at the point A. (i) If the triangle is right-angled then [AB] is perpendicular to [AC], so m1 m2= 1.
14 Revision and Exam Style Questions: Section A 6. A (0, 0), B ( 4, 1) and C (2, 8) are the vertices of a triangle. Prove that the triangle is right-angled at the point A. (i) Slope of [AB]: Slope of [AC]: (0, 0) ( 4, 1) (x1, y1) (x2, y2) (0, 0) (2, 8) (x1, y1) (x2, y2) Slope = y2- y1 Slope = y2- y1 x2- x1 x2- x1 mAC=8-0 mAB= 1-0 2-0 -4-0 =8 = 1 2 -4 =4
14 Revision and Exam Style Questions: Section A 6. A (0, 0), B ( 4, 1) and C (2, 8) are the vertices of a triangle. Prove that the triangle is right-angled at the point A. (i) Investigate if: mAB mAC= 1 -1 4 4=-1 1 = 1 Therefore AB AC, showing the triangle is a right-angled triangle at the point A.
14 Revision and Exam Style Questions: Section A 6. A (0, 0), B ( 4, 1) and C (2, 8) are the vertices of a triangle. Find the area of the triangle ABC. (ii) 1 2| x1y2-x2y1| 1 2|-4(8)-2(1)| Area = (0, 0) ( 4, 1) (2, 8) (x1, y1) (x2, y2) = =1 2|-32-2| =1 2|-34| =1 2(34) = 17 units2
14 Revision and Exam Style Questions: Section A 7. A ( 3, 4), B (5, 6) and C (1, 2) are three points. Pis the midpoint of [AB] and Qis the midpoint of [AC]. Find the coordinates of P and the coordinates of Q. (i) x1+x2 2 , y1+ y2 A( 3, 4) (x1, y1) B (5, 6) (x2, y2) Midpoint = 2 =-3+5 , 4+6 2 2 ( 2 2, 10 = 2 ) P = 1, 5
14 Revision and Exam Style Questions: Section A 7. A ( 3, 4), B (5, 6) and C (1, 2) are three points. Pis the midpoint of [AB] and Qis the midpoint of [AC]. Find the coordinates of P and the coordinates of Q. (i) Midpoint = x1+x2 , y1+ y2 A( 3, 4) (x1, y1) (x2, y2) C(1, 2) 2 2 ( =-3+1 , 4+-2 2 2 =-3+1 , 4-2 2 2 =-2 2, 2 2 ) Q = -1, 1
14 Revision and Exam Style Questions: Section A 7. A ( 3, 4), B (5, 6) and C (1, 2) are three points. Pis the midpoint of [AB] and Qis the midpoint of [AC]. Plot A, B, C, P and Q on a coordinate diagram and show the line segments [BC] and [PQ] on your diagram. (ii)
14 Revision and Exam Style Questions: Section A 7. A ( 3, 4), B (5, 6) and C (1, 2) are three points. Pis the midpoint of [AB] and Qis the midpoint of [AC]. Investigate whether PQ is parallel to BC. (iii) If the lines are parallel their slope will be equal. P (1, 5) Q( 1, 1) B (5, 6) C(1, 2) (x1, y1) (x2, y2) (x1, y1) (x2, y2) Slope=y2- y1 Slope=y2- y1 x2-x1 =1-5 -1-1 =-4 -2 x2-x1 =-2-6 1-5 =-8 -4 Slope of PQ=2 Slope of BC =2
14 Revision and Exam Style Questions: Section A 7. A ( 3, 4), B (5, 6) and C (1, 2) are three points. Pis the midpoint of [AB] and Qis the midpoint of [AC]. Investigate whether PQ is parallel to BC. (iii) Parallel lines have equal slopes. Since mPQ = mBC , PQ || BC
14 Revision and Exam Style Questions: Section A 8. A (2, 3), B (1, 2) and C (12, 1) are the vertices of a triangle ABC. Investigate if the triangle is right-angled. (i) Right-angled implies two sides are perpendicular to each other A (2, 3) B(1, 2) (x1, y1) (x2, y2) A (2, 3) C (12, 1) (x1, y1) (x2, y2) Slope=y2- y1 Slope=y2- y1 x2-x1 =-2-3 1-2 =-5 -1 x2-x1 =1-3 12-2 =-2 10 Slope of AB=5 Slope of AC =-1 5
14 Revision and Exam Style Questions: Section A 8. A (2, 3), B (1, 2) and C (12, 1) are the vertices of a triangle ABC. Investigate if the triangle is right-angled. (i) m1 m2=-1 If the lines are perpendicular, 5 -1 5=-1 -1=-1 \AB^ AC Therefore, the triangle is right-angled at the point A.
14 Revision and Exam Style Questions: Section A 8. A (2, 3), B (1, 2) and C (12, 1) are the vertices of a triangle ABC. The point (3, k) is on the line AB. Find the value of k. (ii) Find the equation of the line AB: (x1, y1) = (2, 3) m = 5 (y y1) = m(x x1) (y 3) = 5(x 2) y 3 = 5x 10 0 = 5x y 7
14 Revision and Exam Style Questions: Section A 8. A (2, 3), B (1, 2) and C (12, 1) are the vertices of a triangle ABC. The point (3, k) is on the line AB. Find the value of k. (ii) Substitute the coordinates of (3, k) in for x and y: 0 = 5x y 7 0 = 5(3) k 7 0 = 15 k 7 0 = 8 k k = 8
14 Revision and Exam Style Questions: Section A 9. The line l intersect the x-axis at (-4, 0) and the y-axis at (0, 3). Find the slope of l. (i) Slope=y2- y1 ( 4, 0) (x1, y1) (0, 3) (x2, y2) x2-x1 3-0 0- -4 ( ) 3 0+4 =3 4 = =
14 Revision and Exam Style Questions: Section A 9. The line l intersect the x-axis at (-4, 0) and the y-axis at (0, 3). Find the equation of l. (ii) 3 4 (x1, y1) = ( 4, 0) m= (y y1) = m(x x1) ( ) )=3 ( x -4 ( ) y 0 4 )=3 ( ( ) x+4 y 0 4 4 y 0 ( )=3 x+4 ( ) 4y =3x+12 0=3x 4y+12
14 Revision and Exam Style Questions: Section A 9. The line m passes through (0, 0) and is perpendicular to l. Show the lines l and m on a coordinate diagram. (iii) Perpendicular lines m1 m2= 1 ml mm= 1 3 4 mm= 1 -4 mm = 3
14 Revision and Exam Style Questions: Section A 9. The line m passes through (0, 0) and is perpendicular to l. Find the equation of m (iv) -4 (x1, y1) = (0, 0) m = (y y1) = m(x x1) 3 -4 ( ) x 0 (y 0) = 3 3(y 0) = 4(x 0) 3y= 4x 4x + 3y = 0
14 Revision and Exam Style Questions: Section A 9. The line m passes through (0, 0) and is perpendicular to l. The point (t, 4) is on the line m. Find the value of t. (v) Substitute the coordinates of (t, 4) into the line m: 4x + 3y = 0 4(t) + 3(4) = 0 4t + 12 = 0 4t= 12 t= 3
14 Revision and Exam Style Questions: Section A 10. The line k: 3x + 2y 12 = 0 intersects the x-axis at the point Pand the y-axis at the point Q. Find the coordinates of the points P and Q. (i) At y-axis, x = 0: At x-axis, y = 0: 3x + 2y 12 = 0 3x + 2y 12 = 0 3(0) + 2y 12 = 0 3x+ 2(0) 12 = 0 2y = 12 3x = 12 y = 6 x = 4 y-intercept, Q = (0, 6) x-intercept, P = (4, 0)
14 Revision and Exam Style Questions: Section A 10. The line k: 3x + 2y 12 = 0 intersects the x-axis at the point Pand the y-axis at the point Q. Graph the line k. (ii)
14 Revision and Exam Style Questions: Section A 10. The line k: 3x + 2y 12 = 0 intersects the x-axis at the point Pand the y-axis at the point Q. Hence, or otherwise, find the area of the triangle formed by the x-axis, y-axis and the line k. (iii) (0, 0) (4, 0) (x1, y1) (x2, y2) (0, 6) Area = 1 2x1y2-x2y1 = 1 24(6)-0(0) = 1 224-0 = 1 224 = 12 units2
14 Revision and Exam Style Questions: Section A 11. Given the line m: 3x 4y + 16 = 0: Find the slope of m (i) 3x 4y + 16 = 0 a = 3, b = 4 Slope=-a b =-3 -4 Slope=3 4
14 Revision and Exam Style Questions: Section A 11. Given the line m: 3x 4y + 16 = 0: The line n is perpendicular to m and cuts the x-axis at the point (3, 0). Find the equation of n. (ii) -4 m n mm mn= 1 3 4 (x1, y1) = (3, 0) mn = 3 mn= 1 (y y1) = m(x x1) -4 -4 ( ) x 3 (y 0) = mn = 3 3 3(y 0) = 4(x 3) 3y= 4x + 12 4x + 3y 12 = 0
14 Revision and Exam Style Questions: Section A 11. Given the line m: 3x 4y + 16 = 0: Find the coordinates of the point of intersection of the lines m and n. (iii) ( 3) ( 4) 3x 4y= 16 Let x = 0: 4x + 3y = 12 3x 4y= 16 9x 12y= 48 3(0) 4y= 16 16x + 12y = 48 4y= 16 25x = 0 y = 4 x = 0 Point of intersection = (0, 4)
14 Revision and Exam Style Questions: Section A 12. Given the line l: 2x 5y + 16 = 0, find: the coordinates of A, where l intersects the x-axis (i) At the x-axis,y = 0: 2x 5y + 16 = 0 2x 5(0) + 16 = 0 2x= 16 x= 8 A: ( 8, 0)
14 Revision and Exam Style Questions: Section A 12. Given the line l: 2x 5y + 16 = 0, find: the equation of the line k, of slope 2 and passing through B (0, 4) (ii) (x1, y1) = (0, 4) mk = 2 (y y1) = m(x x1) (y ( 4)) = 2(x 0) y + 4 = 2x 0 = 2x y 4
14 Revision and Exam Style Questions: Section A 12. Given the line l: 2x 5y + 16 = 0, find: the point C, where l k = {C}. (iii) ( 1) 2x 5y= 16 Let y = 5: 2x y = 4 2x y = 4 2x + 5y = 16 2x 5 = 4 2x y = 4 2x = 9 9 2 4y = 20 x = y = 5 9 = (4 5, 5) 2, 5 C =
