Intermediate Algebra Problems With Answers  Sample 6  Equations of Lines
A set of intermediate algebra problems, related to equations of lines, with answers, are presented. The solutions are at the bottom of the page.

Find the equation of the line that passes through the point (0 , 6) and has a slope equal to 5.

Find the equation of the line that passes through the point (2 , 3) and has a slope equal to 2.

Is there a line that passes through all the following 3 points: A(2 , 3), B(4 , 7) and C(5 , 6).

Find the equation of the line through the point (3 , 0) and is perpendicular to the line through the points (2 , 1) and (4 , 5).

What are the values of m and w so that the line with equation y  mx = w passes through the points (0 , 4) and (2 , 0)?

Find the equation of the line through the point (4 , 5) and is perpendicular to the x axis.

Find the equation of the line through the point (6 , 1) and is perpendicular to the y axis.

The graph shown below is that of a linear function f. Use the graph to approximate f(120).
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Is the line with equation 0.5 y + 0.5 x = 5 perpendicular to the line shown in the graph below? Explain your answer.
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If the two lines with equations 2x  3y = 5 and 4x + 5y = 7 are graphed in the same system of axes, will they have a point of intersection? If yes what are its coordinates?

Find the equations of two perpendicular lines with a point of intersection at (2 , 3) and one of the lines passes through the origin of the system of axes.

Find the equations of a line that passes through the origin and is tangent to the circle with center at (2 , 4) and radius equal to 1.

Which of the points A(3 , 0) , B(2 , 5), C(0 , 2) lie on the graph of the line whose equation is given by
 3 y + 2x + 6 = 0
Answers to the Above Questions

given slope and y intercept, slope intercept form of the equation is: y = 5x  6

given slope and a point, point slope form of the equation is: y  3 = 2(x + 2) or in slope intercept form: y =  2x  1

For a line to pass through all the 3 points, the points must be collinear.
slope through AB = (7  3) / (4  2) = 2 and slope through BC = (6  7) / (5  4) = 1
The slopes through AB and BC are not equal therefore A, B and C are not collinear and no single line passes through all 3 points.

We first find the slope m through the points (2 , 1) and (4 , 5).
m = (5  1) / (4  2) = 2
The slope s of the line perpendicular to a line of slope 2 is found by solving: m s =  1 . Hence s =  1/2
The equation of the line is given by
y  0 = (1/2)(x  3) or y = (1 / 2) x + 3 / 2

The line with equation y  mx = w passes through the point (0 , 4), substitute x by 0 and y by 4 in the equation to obtain:
4  m(0) = W
Solve for W: W = 4
The line with equation y  mx = w passes through the point (2 , 0), substitute x by 2 and y by 0 in the equation to obtain:
0  2 m = w
Solve for m: m = W / (2) = 4 / (2) =  2

The equation of a line perpendicular to the x axis is parallel to the y axis and is of the form: y = constant. The equation of line parallel to the y axis and through the point (4 , 5) has the equation: y = 5.

The equation of a line perpendicular to the y axis is parallel to the x axis and is of the form: x = constant. The equation of line parallel to the x axis and through the point (6 , 1 ) has the equation: x = 6.

Function f is linear and passes through the origin has the form: f(x) = m x , where m is a constant equal to the slope of the line given by its graph. We find the slope m using the points (0 , 0) and (4 , 1) from the graph.
m = (1  0) / (4  0) = 1/4
f(120) = (1/4)*120 = 30

Write the equation 0.5 y + 0.5 x = 5 in slope intercept form and find its slope.
0.5 y =  0.5 x + 5 or y =  x + 10 , slope m = 1
The slope of the line given by its graph is 1/4
For two lines to be perpendicular, the product of their slopes must be equal to  1. It is clear that the product of  1 and 1/4 is not equal to  1 and therefore the line with equation 0.5 y + 0.5 x = 5 is not perpendicular to the line given by its graph.

The point of intersection (if any) of the two lines is the solution to the system of the two equations: 2 x  3 y = 5 and  4 x + 5 y = 7
Multiply all terms of the equation 2 x  3 y = 5 by 2 to obtain: 4 x  6 y = 10 and add it to the equation  4 x + 5 y = 7 which gives  y = 17 or y =  17
Substitute y by  17 and find x using one of the equations: 2 x  3 (17) = 5 gives x =  23.
The two lines intersect at the point: (23 , 17)

The line through (2 , 3) and the origin (0 , 0) has slope
m =  3/2
Its equation is given by: y =  3/2 x
The second line is perpendicular to the line with slope 3/2. It slope s is found by solving the equation
(3/2) s =  1 which give s = 2/3
The second line has slope 2/3 and passes through the point (2 , 3); its equation is given by
y  (3) = 2/3(x  2) or y = (2/3) x  13 / 3

The equation of the tangent through the origin has the form: y = k x. The equation of the circle with center at (2 , 4) and radius equal to 1 is written as:
(x  2)^{ 2} + (y  4)^{ 2} = 1^{ 2}
The point of tangency of the line and the circle is a point of intersection. The point of intersection of the line and the circle is obtained by solving the system of equations of the line and the circle which are:
y = k x and (x  2)^{ 2} + (y  4)^{ 2} = 1
Substitute y by k x in the equation of the circle to eliminate y.
(x  2)^{ 2} + (k x  4)^{ 2} = 1
Expand the squares
x^{ 2}  4 x + 4 + (k x)^{ 2}  8 k x + 16 = 1
Rewrite the above quadratic equation (in x) in standard form
(1 + k^{ 2}) x^{ 2} + ( 4  8 k) x + 19 = 0
Find discriminant
Δ = ( 4  8 k)^{ 2}  4(1 + k^{ 2})*19
Expand the expression of Δ
Δ = 12 k^{ 2} + 64 k  60
The x coordinate of the point of tangency may be considered as a unique solution to the quadratic equation in x obtained above. Hence for the line y = k x to be tangent to the circle Δ must be equal to zero (case of a unique solution to a quadratic equation). Hence Δ = 0 gives the equation in k:
12 k^{ 2} + 64 k  60 = 0
Solve the above quadratic equation for k to obtain two solutions
k = (8 + √19) / 3 and k

For a point to lie on the graph of a line, its x and y coordinates must satisfy the equation of the line.
Check each point by substituting the x and y coordinate:
Check point: A(3 , 0) :  3(0) + 2(3)  6 = 0 , equation is satisfied , point A lies on the line.
Check point: B(  2 ,  5) :  3(5) + 2(2)  6 = 5 , equation is not satisfied , point B does not lie on the line.
Check point: C( 0 , 2) :  3(2) + 2(0)  6 =  12 , equation is not satisfied , point C does not lie on the line.
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