Name:
Date:
y2 - y1
Slope = tan θ= x - x
2
1
1. Find slope of a line which makes 45°
with the positive direction of X-axis
Given: θ = 45°
To-find: m = ?
Solution:m = tan θ (Slope Formula)
m = tan 45°
m=1
The slope of the line is 1
∴
2. Find slope of a line which makes
30° with the X-axis
Given:
To-find:
Solution:
∴
3. Find the slope of the line passing 4. Find the slope of the line
through the points A (-3, 5), B (4, -1)
passing through A (2, 3), B (4, 7)
Given: A (-3, 5) and B (4, -1)
Given:
x1 = -3
To-find:
y1 = 5
Solution:
x2 = 4
y2 = -1
To-find: m(A, B) = ?
Solution:
y2 - y 1
m(A, B) = x - x (Slope Formula) 5. Find slope of a line which makes
2
1
60° with the X-axis.
-1
5
m(A, B) =
6. What is the slope of the X-axis?
4 - (-3)
7. What is the slope of the Y-axis?
m(A, B) = -6
8. Find slope of the line passing
7
through P (-3, 1), Q (5, -2)
The slope of the line is -6/7
∴
∴
9. If slope of the line joining P(k, 0) and 10.Find k, if R(1, -1), S (-2, k) and
Q(-3, -2) is 2/7 then find k.
slope of line RS is -2.
Given: P(k, 0) and Q(-3, -2)
Given:
x1 = k
To-find:
y1 = 0
Solution:
x2 = -3
y2 = -2
m(P, Q) = 2/7
To-find:
k=?
y -y
11. Find k, if B(k, -5), C (1, 2) and
Solution: m(A, B) = x2 - x1 (Slope Formula)
slope of the line is 7.
2
1
2 = -2 - 0
12. Find slope of the line passing
7 -3 - k
through C (5, -2) , D (7, 3)
-6 - 2k = -14
k=4
13. Find slope of the line passing
through L (-2, -3) , M (-6, -8)
The value of k is 4
∴
∴
Name:
Date:
Collinearity
1. Show that points P(-2, 3), Q(1, 2),
R(4, 1) are collinear.
Given: P(-2, 3), Q(1, 2) and R(4, 1)
To-prove: collinearity
=> m(P, Q) = m(Q, R)
y2 - y 1
Solution: m = x2 - x1 (Slope Formula)
m(P, Q) = 2 - 3
1 - (-2)
m(P, Q) = -1/3
m(Q, R) = 1 - 2
4-1
m(Q, R) = -1/3
m(P, Q) = m(Q, R)
∴
Hence Proved
2. Check if A(-1, -1), B(0, 1), C(1, 3)
are collinear
3. Check if D(-2, -3), E(1, 0), F(2, 1)
are collinear
4. Check if L(2, 5), M(3, 3), N(5, 1)
are collinear
5. Check if P(2, -5), Q(1, -3), R(-2, 3)
are collinear
6. Check if R(1, -4), S(-2, 2), T(-3, 4)
are collinear
7. Check if A(1, -3), B(2, -5), C(-4, 7)
are collinear
8. Check if L(-2, 3), M(1, -3), N(5, 4)
are collinear
9. Check if R(0, 3), D(2, 1), S(3, -1)
are collinear
Parallelogram
1. A(6, 1), B(8, 2), C(9, 4), D(7, 3) are
the vertices of ▭ABCD, show that
▭ABCD is a parallelogram.
2. Show that A (-4, -7),B (-1, 2), C
(8, 5) and D (5, -4) are the vertices
of a parallelogram.
Given: A(6, 1), B(8, 2), C(9, 4), D(7, 3) 3. Show that points P(1,-2), Q(5,2),
R(3,-1), S(-1,-5) are the vertices
To-prove: ▭ABCD is a parallelogram
of a parallelogram.
y2 - y 1
(Slope
Formula)
Solution: m = x2 - x1
4. Show that the PQRS formed by
P(2,1), Q(-1,3), R(-5,-3), S(-2,-5)
is a parallelogram.
5. Show that A(-4, -2), B(-3, -7),
C(3, -2), and D(2, 3) are the
vertices of a parallelogram.
6. Show that points P(2, -2), Q(7, 3),
R(11, -1) and S (6, -6) are vertices
of a parallelogram.
Hence Proved