SLOPES AND LINEAR EQUATIONS PRACTICE FOR SAT

Problem 1 :

2y + 3x = 5

2y - 3x = 5

Which of the following describes the graph of the system of equations above in the xy-plane.

(a)  A single line    (b) Two parallel lines

(c)  Two perpendicular lines

(d)  Two distinct intersecting lines they are not perpendicular.

Solution :

The slopes are not equal, so they are not parallel lines.

The product of their slopes is not equal to -1. So, they are not perpendicular.

Then,

They are two distinct intersecting lines they are not perpendicular.

Problem 2 :

(i) Point B (not shown) is located by starting at A, moving 2 units down and moving 1 unit to the right. If a line is drawn through points A and B, what is the y-intercept of the line ?

(a)  1/2      (b)  1       (c)  2      (d)  4

(ii) Line l (not shown) contains point A and has a slope of 6. Which of the following points is on the line l.

(a)  (1, 6)      (b)  (2, 6)       (c)  (3, 6)      (d)  (6, 1)

Solution :

(i)  y-intercept of the line is 4.

(ii)  Equation of line A :

Slope = 4/2 ==> 2, y-intercept = 4

y = 2x + 4

Problem 3 :

The ordered pairs in the table above would form a line. Where would this line intersect the x-axis.

Solution :

Considering two points from the table.

(-1, 7) and (0, 5)

Slope (m)  = (y₂ - y₁)/(x₂ - x₁)

m = (5 - 7) / (0 + 1)

m = -2/1

m = -2

From the table, y-intercept is 5.

Equation of the line :

y = -2x + 5

To find x-intercept, we put y = 0

0 = -2x + 5

2x = 5

x = 5/2

x = 2.5

So, the x-intercept is 2.5.

Problem 4 :

In the xy plane, a line has an x-intercept of -2 and the y-intercept of -4. What is the slope of the line ?

(a)  -2      (b)  -1/2       (c)  1/2      (d)  2

Solution :

x-intercept = -2, y-intercept = -4

From the given information, it is clear the straight line must be a falling line. Starting from -4 go up 4 units and move left two units.

Rise = 4 and run = 2

Slope (m) = -4/2 ==>  -2

Problem 5 :

The graph of the function f contains the points (0, 3), (-2, 7) and (5, k). If the graph of f is a line. What must be the value of k ?

(a)  -13      (b)  -7       (c)  5      (d)  8

Solution :

Slope of the line passes through (0, 3) and (-2, 7)

= Slope of the line passes through (-2, 1) and (5, k).

Slope (m)  = (y₂ - y₁)/(x₂ - x₁)

(1) = (2)

-2 = (k - 7)/7

-14 = k - 7

-14 + 7 = k

k = -7

Problem 6 :

Line k has a negative slope and passes through the origin. If line m is perpendicular to like k, which of the following must be true ?

(a)  Line m passes through the origin.

(b) Line m does not pass through the origin.

(c)  Line m has a positive slope

(d)  Line m has negative slope

Solution :

Line k is having a negative slope and it must be a falling line. When we draw the line which is perpendicular to k, we will have a raising line. That will have positive slope.

Problem 7 :

The graph of the line in the xy plane passes through points (0, 0) and (1, 2). The graph of the second line passes through points (1, 2) and (k, 0). If the two lines are perpendicular, what is the value of k ?

Solution :

Slope of the line passes through two points (0, 0) and (1, 2)

m = (2 - 0) / (1 - 0)

m = 2 -----(1)

Slope of the second line passes through two points (1, 2) and (k , 0)

m = (0 - 2) / (k - 1)

m = -2/(k - 1) -----(2)

Since the first and second lines are perpendicular to each other, then m₁ x m₂ = -1

2[-2/(k - 1) ] = -1

-4/(k - 1) = -1

1/(k - 1) = 1

k - 1 = 1

k = 2

Problem 8 :

Line m is graphed in the xy-plane. If an equation for line m is 6y + 2x = 5, which of the following statements is true ?

(a) Line m has slope of -3

(b) Line m has slope of 1/3

(c)  The x-intercept of the line m is 5/2 and the y-intercept is 5/6.

(d) The x-intercept of line m is 5/6 and the y-intercept is 5/2.

Solution :

6y + 2x = 5

Writing the equation in slope intercept form, we get

6y = 5 - 2x

y = -2x/6 + 5/6

y = (-1/3)x + 5/6

Finding x and y-intercepts, we get

So, option d is correct.