Problem 1 :
A line in the xy-plane passes through the origin and has a slope of 1/7 . Which of the following points lies on the line?
a) (0, 7) b) (1, 7) c) (7, 7) d) (14, 2)
Solution :
One of the points on the straight line is origin (0, 0).
Slope = 1/7
Slope of the line joining the points
Slope (m) = (y_{2} - y_{1}) / (x_{2} - x_{1})
(0, 0) and (0, 7) = (7 - 0) / (0 - 0) = 7/0 = infinity It is not 1/7 |
(0, 0) and (1, 7) = (7 - 0) / (1 - 0) = 7/1 = 7 It is not 1/7 |
(0, 0) and (7, 7) = (7 - 0) / (7 - 0) = 7/7 = 1 It is not 1/7 |
(0, 0) and (14, 2) = (2 - 0) / (14 - 0) = 2/14 = 1/7 |
So, the point (14, 2) lies on the line.
Problem 2 :
In the xy-plane above, line A is parallel to line k. What is the value of p ?
a) 4 b) 5 c) 8 d) 10
Solution :
Slope of the line joining the points (-5, 0) and (0, 2)
m = (2 - 0) / (0 - (-5))
= 2/5 -----(1)
Slope of the line joining the points (0, -4) and (p, 0)
m = (0 - (-4)) / (p - 0)
= (0 + 4) / p
= 4/p -----(2)
Since the lines are parallel, their slopes will be equal.
(1) = (2)
2/5 = 4/p
2p = 5(4)
2p = 20
p = 10
So, the answer is option d.
Problem 3 :
The graph of a line in the xy-plane has slope 2 and contains the point (1, 8). The graph of a second line passes through the points (1, 2) and (2, 1). If the two lines intersect at the point (a, b), what is the value of a + b ?
a) 4 b) 3 c) −1 d) −4
Solution :
Equation of the first line :
Slope = 2, Point (1, 8)
y - y_{1} = m(x - x_{1})
y - 8 = 2(x - 1)
y = 2x - 2 + 8
y = 2x + 6 -----(1)
Equation of second line :
(1, 2) and (2, 1)
Slope = (y_{2} - y_{1}) / (x_{2} - x_{1})
= (1 - 2) / (2 - 1)
= -1
(y - 2) = -1 (x - 1)
y - 2 = -1x + 1
y = -1x + 1 + 2
y = -x + 3 -----(2)
(1) = (2)
2x + 6 = -x + 3
2x + x = 3 - 6
3x = -3
x = -1
Applying the value of x in (1), we get
y = -(-1) + 3
y = 4
Problem 4 :
The graph of the linear function f is shown in the
xy-plane above. The slope of the graph of the linear
function g is 4 times the slope of the graph of f. If
the graph of g passes through the point (0, −4) , what
is the value of g(9) ?
a) 5 b) 9 c) 14 d) 18
Solution :
Slope of line f :
(0, 3) and (2, 4) are the points lie on the line
slope = (4 - 3) / (2 - 0)
= 1/2
Slope of linear function g = 4 (1/2) = 2
Line g passes through the point (0, -4)
y - (-4) = 2(x - 0)
y + 4 = 2x
y = 2x - 4
g(x) = 2x - 4
g(9) = 2(9) - 4
= 14
So, option c is correct.
Problem 5 :
Line A in the xy-plane contains points from each of Quadrants II, III, and IV, but no points from Quadrant I. Which of the following must be true?
a) The slope of line A is undefined.
b) The slope of line A is zero.
c) The slope of line A is positive.
d) The slope of line A is negative
Solution :
It must be a falling line, then its slope will be negative.
Problem 6 :
In the xy-plane, the line determined by the points (2, k) and (k, 32) passes through the origin. Which of the following could be the value of k ?
a) 0 b) 4 c) 8 d) 16
Solution :
(2, k) (k, 32) and (0, 0) lies on the same line.
If the points are collinear, then
(2, k) (k, 32) Slope (m) = (y_{2} - y_{1}) / (x_{2} - x_{1}) = (32 - k) / (k - 2) ----(1) |
(k, 32) (0, 0) Slope (m) = (y_{2} - y_{1}) / (x_{2} - x_{1}) = (0 - 32) / (0 - k) = 32/k ----(2) |
(1) = (2)
(32 - k) / (k - 2) = 32/k
k(32 - k) = 32(k - 2)
32k - k^{2} = 32k - 64
k^{2 }+ 32k - 32k - 64 = 0
k^{2 }- 64 = 0
(k - 8) (k + 8) = 0
k = 8 and k = -8
So, the value of k is 8.
Problem 8 :
What are the slope and y-intercept of the line
5x + 4y + 3 = 0?
Solution :
5x + 4y + 3 = 0
4y = -5x - 3
Dividing by 4 on both sides,
y = -5x/4 - (3/4)
Slope m = -5/4 and y-intercept = -3/4
Problem 9 :
In the xy-plane above, ABCD is a square and point E is the center of the square. The coordinates of points C and E are (7, 2) and (1, 0), respectively. Which of the following is an equation of the line that passes through points B and D ?
a) y = −3x − 1 b) y = −3(x − 1)
c) y = (−1/3)x + 4 d) y = − (1/3)x − 1
Solution :
The shape shown above is a square, then the diagonals will be perpendicular to each other.
Finding slope of line passes through the points C and E :
m = (0 - 2) / (1 - 7)
= -2/(-6)
= 1/3
Slope of the line passes through B which is perpendicular to the other diagonal :
= -3
Equation of the line :
y - 0 = -3(x - 1)
y = -3(x - 1)
So, option b is correct.
Problem 10 :
−2x + 3y = 6
In the xy-plane, the graph of which of the following equations is perpendicular to the graph of the equation above?
a) 3x + 2y = 6 b) 3x + 4y = 6 c) 2x + 4y = 6
d) 2x + 6y = 3
Solution :
−2x + 3y = 6
If two lines are perpendicular, then the product of their slopes will be equal to -1.
Slope of the given line :
−2x + 3y = 6
3y = 2x + 6
y = (2/3)x + 2
Slope of the given line = 2/3
Option a :
3x + 2y = 6
2y = -3x + 6
y = (-3/2)x + 3
Slope of this line = -3/2
So, these two lines are perpendicular to each other.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM