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Find the equation of the line that passes through the point (0 , -6) and has a slope equal to 5.
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Find the equation of the line that passes through the point (-2 , 3) and has a slope equal to -2.
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Is there a line that passes through all the following 3 points: A(2 , 3), B(4 , 7) and C(5 , 6).
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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).
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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)?
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Find the equation of the line through the point (-4 , 5) and is perpendicular to the x axis.
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Find the equation of the line through the point (6 , -1) and is perpendicular to the y axis.
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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?
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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.
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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.
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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
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given slope and y intercept, slope intercept form of the equation is: y = 5x - 6
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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
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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.
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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
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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
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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.
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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.
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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
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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.
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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)
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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
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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
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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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