Slopes of Perpendicular Lines

Which of the following lines are perpendicular?

\[ \text{a) } y = -2x - 3 \quad \text{b) } -4y + 5x = 4 \quad \text{c) } \frac{1}{2}x - y = 2 \quad \text{d) } -6y - 3x = 9 \]

Solution

\[ \begin{aligned} \text{a)}\;& y = -2x - 3 \implies m = -2 \\ \text{b)}\;& -4y + 5x = 4 \implies 4y = 5x - 4 \implies y = \frac{5}{4}x - 1 \implies m = \frac{5}{4} \\ \text{c)}\;& \frac{1}{2}x - y = 2 \implies y = \frac{1}{2}x - 2 \implies m = \frac{1}{2} \\ \text{d)}\;& -6y - 3x = 9 \implies 6y = -3x - 9 \implies y = -\frac{1}{2}x - \frac{3}{2} \implies m = -\frac{1}{2} \end{aligned} \]

Checking the product of slopes for lines (a) and (c): \[ (-2)\left(\frac{1}{2}\right) = -1 \] Therefore, the lines in parts a and c are perpendicular.

Find \(k\) so that the line through \((1,-2)\) and \((0,3)\) is perpendicular to the line through \((3,-k)\) and \((2,-5)\).

Solution

\[ m_1 = \frac{3 - (-2)}{0 - 1} = \frac{5}{-1} = -5 \] \[ m_2 = \frac{-5 - (-k)}{2 - 3} = \frac{-5 + k}{-1} = 5 - k \]

Since the lines are perpendicular, \(m_1 m_2 = -1\): \[ -5(5 - k) = -1 \] \[ -25 + 5k = -1 \implies 5k = 24 \implies k = \frac{24}{5} \]

Find the value of \(k\) such that the lines with equations \[ - kx + 2x - y = -3 \quad \text{and} \quad -2x + y = 7 \] are perpendicular.

Solution

First, write each equation in slope–intercept form \(y = mx + b\).

\[ - kx + 2x - y = -3 \quad \implies \quad y = (-k + 2)x + 3 \] \[ \text{Slope of first line: } m_1 = -k + 2 \]

\[ -2x + y = 7 \quad \implies \quad y = 2x + 7 \] \[ \text{Slope of second line: } m_2 = 2 \]

Since the lines are perpendicular, the product of their slopes is \(-1\): \[ m_1 m_2 = -1 \] \[ 2(-k + 2) = -1 \] \[ -2k + 4 = -1 \implies -2k = -5 \implies k = \frac{5}{2} \]

Find the coordinates of point \(C\) such that the points \[ A(0,2), \quad B(2,6), \quad C(2k, k) \] are the vertices of a right-angled triangle with hypotenuse \(AC\).

Solution

Since \(AC\) is the hypotenuse, segments \(AB\) and \(BC\) must be perpendicular at vertex \(B\).

Slope of segment \(AB\): \[ m_{AB} = \frac{6 - 2}{2 - 0} = \frac{4}{2} = 2 \]

Slope of segment \(BC\): \[ m_{BC} = \frac{k - 6}{2k - 2} \]

Because the segments are perpendicular, the product of their slopes is \(-1\): \[ 2 \cdot \frac{k - 6}{2k - 2} = -1 \] \[ \frac{2(k - 6)}{2(k - 1)} = -1 \implies \frac{k - 6}{k - 1} = -1 \] \[ k - 6 = -(k - 1) \implies k - 6 = -k + 1 \implies 2k = 7 \implies k = \frac{7}{2} \]

Therefore, the coordinates of point \(C\) are: \[ C(2k, k) = \left(2\left(\frac{7}{2}\right), \frac{7}{2}\right) = \left(7, \frac{7}{2}\right) \]

Find the equation of a line \(L_1\) that passes through the point \((3,1)\) and is perpendicular to the line \(L_2\) with equation \[ - x + 2y - 6 = 0 \]

Solution

First, find the slope of line \(L_2\) by writing its equation in slope–intercept form.

\[ - x + 2y - 6 = 0 \quad \implies \quad 2y = x + 6 \implies y = \frac{1}{2}x + 3 \] \[ m_2 = \frac{1}{2} \]

Since the lines are perpendicular, the slope \(m_1\) of \(L_1\) satisfies: \[ m_1 \cdot \frac{1}{2} = -1 \implies m_1 = -2 \]

Using point–slope form with point \((3,1)\): \[ y - 1 = -2(x - 3) \] \[ y - 1 = -2x + 6 \implies y = -2x + 7 \]

Find the equation of a line \(L_1\) that passes through the point \((-2,-1)\) and is perpendicular to the line \(L_2\) passing through \((0,-2)\) and \((2,4)\).

Solution

First, find the slope of line \(L_2\):

\[ m_2 = \frac{4 - (-2)}{2 - 0} = \frac{6}{2} = 3 \]

Since the lines are perpendicular: \[ m_1 \cdot 3 = -1 \implies m_1 = -\frac{1}{3} \]

Using point–slope form with point \((-2,-1)\): \[ y - (-1) = -\frac{1}{3}(x - (-2)) \] \[ y + 1 = -\frac{1}{3}(x + 2) \implies y + 1 = -\frac{1}{3}x - \frac{2}{3} \] \[ y = -\frac{1}{3}x - \frac{2}{3} - 1 \implies y = -\frac{1}{3}x - \frac{5}{3} \]