Master Intermediate Algebra: Equations of Lines
Practice Problems & Detailed Solutions

Given slope of \( 5 \) and the y intercept at \( (0 , -6) \) , the slope intercept form of the equation is: \[ y = 5x - 6 \]

Given a slope of \( -2 \) and the point \( (-2 , 3) \), the point slope form of the equation is: \[ y - 3 = -2(x - (- 2)) \] \[ y - 3 = -2 (x+2) \] 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 \[ m_{AB} = \dfrac{7 - 3}{4 - 2} = 2 \] and slope through BC \[ m_{BC} = \dfrac{6 - 7}{5 - 4} = -1 \]

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 \((4, 5)\). \[ m = \dfrac{5 - 1}{4 - 2} = 2 \]

The slope \( s \) of the line perpendicular to a line of slope \( 2 \) is found by solving: \[ m \times s = -1 \] Hence, \[ s = -\dfrac{1}{2} \]

The point slope form of the equation of the line through the point \( (3 , 0) \) and perpendicular to the line through the points \( (2 , 1) \; \text{and} \; (4 , 5) \) is given by \[ y - 0 = -\dfrac{1}{2}(x - 3) \] or equivalently \[ y = -\dfrac{1}{2} x + \dfrac{3}{2} \]

The line with equation \[ y - mx = w \] passes through the point \( (0, 4) \). Substitute \( x = 0 \) and \( y = 4 \) into the equation to obtain: \[ 4 - m \cdot 0 = w \] Solve for \( w \): \[ w = 4 \] The line with equation \[ y - mx = w \] also passes through the point \( (2, 0) \). Substitute \( x = 2 \) and \( y = 0 \) into the equation to obtain: \[ 0 - 2m = w \] Solve for \( m \): \[ m = \dfrac{w}{-2} = \dfrac{4}{-2} = -2 \]

The equation of a line perpendicular to the x-axis is parallel to the y-axis and is of the form: \[ x = \text{constant} \].

The equation of a line parallel to the y-axis and passing through the point \((-4,\, 5)\) is given by: \[ x = -4 \].

The equation of a line perpendicular to the \( y \)-axis is parallel to the \( x \)-axis and is of the form: \[ y = \text{constant} \,.\]

The equation of a line parallel to the \( x \)-axis and passing through the point \( (6, -1) \) is: \[ y = -1 \,.\]

The function \( f \) is linear and passes through the origin. It has the form: \[ f(x) = m x \] where \( m \) is a constant representing the slope of the line as shown on the graph. To find the slope \( m \), we use the points \( (0, 0) \) and \( (4, 1) \) from the graph: \[ m = \dfrac{1 - 0}{4 - 0} = \dfrac{1}{4} \] Now we calculate the value of \( f(120) \): \[ f(120) = \dfrac{1}{4} \cdot 120 = 30 \]

Write the equation \( 0.5y + 0.5x = 5 \) in slope-intercept form and find its slope. \[ 0.5y = -0.5x + 5 \Rightarrow y = -x + 10 \] slope \[ m = -1 \].

The slope of the line given by its graph is \( \dfrac{1}{4} \).

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 \( \dfrac{1}{4} \) is not equal to \( -1 \) and therefore the line with equation \( 0.5y + 0.5x = 5 \) 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 equations: \[ \begin{cases} 2x - 3y = 5 \\ -4x + 5y = 7 \end{cases} \]

Multiply all terms of the equation \( 2x - 3y = 5 \) by 2 to obtain: \[ 4x - 6y = 10 \]

Now add it to the equation \( -4x + 5y = 7 \): \[ (4x - 6y) + (-4x + 5y) = 10 + 7 \] \[ -y = 17 \quad \Rightarrow \quad y = -17 \]

Substitute \( y = -17 \) into one of the original equations, say \( 2x - 3y = 5 \): \[ 2x - 3(-17) = 5 \quad \Rightarrow \quad 2x + 51 = 5 \quad \Rightarrow \quad 2x = -46 \quad \Rightarrow \quad x = -23 \]

The two lines intersect at the point: \[ (-23, -17) \]

The line through the points \( (2, -3) \) and the origin \( (0, 0) \) has slope \[ m = \dfrac{-3}{2} \]

Its equation is given by: \[ y = \dfrac{-3}{2}x \]

The second line is perpendicular to the line with slope \( -\dfrac{3}{2} \). Its slope \( s \) is found by solving the equation:

The second line has slope \( \dfrac{2}{3} \) and passes through the point \( (2, -3) \). Its equation is given by:

We are given a circle with center at \( (2, 4) \) and radius \( 1 \). We need to find the equations of all lines that:

• Pass through the origin \((0, 0)\), and
• Are tangent to the circle.

Step 1: Equation of the Line

A line passing through the origin has the form: \[ y = kx \]

Step 2: Equation of the Circle

The equation of the circle with center \( (2, 4) \) and radius \( 1 \) is: \[ (x - 2)^2 + (y - 4)^2 = 1 \]

Step 3: Substitute \( y = kx \) into the Circle

\[ (x - 2)^2 + (kx - 4)^2 = 1 \] Expand both terms: \[ x^2 - 4x + 4 + k^2x^2 - 8kx + 16 = 1 \] Combine like terms: \[ (1 + k^2)x^2 + (-4 - 8k)x + 19 = 0 \]

Step 4: Tangent Condition

The line is tangent to the circle if this quadratic equation has exactly one real solution, so the discriminant must be zero: \[ \Delta = (-4 - 8k)^2 - 4(1 + k^2)(19) \] Simplify: \[ \Delta = -12k^2 + 64k - 60 \] Set \( \Delta = 0 \): \[ -12k^2 + 64k - 60 = 0 \]

Step 5: Solve the Quadratic in \( k \)

Divide the equation by 4: \[ -3k^2 + 16k - 15 = 0 \] Use the quadratic formula: \[ k = \dfrac{-16 \pm \sqrt{16^2 - 4(-3)(-15)}}{2(-3)} = \dfrac{-16 \pm \sqrt{256 - 180}}{-6} \] \[ k = \dfrac{-16 \pm \sqrt{76}}{-6} = \dfrac{-16 \pm 2\sqrt{19}}{-6} \] \[ k = \dfrac{16 \mp 2\sqrt{19}}{6} = \dfrac{8 \mp \sqrt{19}}{3} \]

Step 6: Final Line Equations

Therefore, the two lines through the origin that are tangent to the circle are: \[ y = \left( \dfrac{8 + \sqrt{19}}{3} \right)x \quad \text{and} \quad y = \left( \dfrac{8 - \sqrt{19}}{3} \right)x \]

Final Answer

\[ \boxed{ \begin{aligned} y &= \left( \dfrac{8 + \sqrt{19}}{3} \right)x \\ y &= \left( \dfrac{8 - \sqrt{19}}{3} \right)x \end{aligned} } \]

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 \(y \) coordinates:

Check point: \( A(3, 0) \) : \[ -3(0) + 2(3) - 6 = 0 \] Equation is satisfied, point \( A \) lies on the line.

Check point: \( B(-2, -5) \) : \[ -3(-5) + 2(-2) - 6 = 15 - 4 - 6 = 5 \] Equation is not satisfied, point \( B \) does not lie on the line.

Check point: \( C(0, 2) \) : \[ -3(2) + 2(0) - 6 = -6 - 0 - 6 = -12 \] Equation is not satisfied, point \( C \) does not lie on the line.