

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
? = ( 4  8 k)^{ 2}  4(1 + k^{ 2})*19
Expand the expression of ?
? = 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 ? must be equal to zero (case of a unique solution to a quadratic equation). Hence ? = 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 = (8  √19) / 3
Two lines through the origin are tangent to the given circle:
y = ((8 + √19) / 3) x and y = ((8  √19) / 3) x

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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