Form an equation of a parallel or perpendicular line

Line g

\(\displaystyle y=4x+1\)

Point T on h

\(\displaystyle (8,3)\)

Required relation

\(\displaystyle h\perp g\)

Which equation describes line h?

Change the gradient of g. Compare its direction with h and watch the product of their gradients

Which relationship remains true at every slider position?

Find the direction, then use the point

The relation determines the direction. The supplied point determines the position of the new line

1. 1Parallel Nonvertical parallel lines have the same gradient \(\displaystyle m_h=m_g\)
2. 2Perpendicular For two oblique perpendicular lines, use the negative reciprocal. Do not just change the sign \(\displaystyle m_h=-\frac{1}{m_g},\qquad m_gm_h=-1\)
3. 3Gradient from points Use the same point order in both coordinate differences \(\displaystyle m_g=\frac{y_Q-y_P}{x_Q-x_P}\)
4. 4Use the new point Substitute its two coordinates with the target gradient, then solve for the intercept \(\displaystyle \begin{aligned}y_R&=m_hx_R+c\\c&=y_R-m_hx_R\\h:\quad y&=m_hx+c\end{aligned}\)

Point A on k

\(\displaystyle (-2,7)\)

Point B on k

\(\displaystyle (4,3)\)

Point P on the new line

\(\displaystyle (6,-5)\)

Find the vertical-axis intercept of the line through P that is parallel to k

Find the given gradient from A and B, keep it for the parallel line, then use P

Point A on k

\(\displaystyle (-2,7)\)

Point B on k

\(\displaystyle (4,3)\)

Point P on the new line

\(\displaystyle (6,-5)\)

Find the vertical-axis intercept of the line through P that is perpendicular to k

Find the given gradient, take its negative reciprocal, then substitute P

Check both intercepts

The original line k passes through A = (−2, 7) and B = (4, 3). Both new lines pass through P = (6, −5). The two applications use different target directions

1. 1Original gradient Calculate the two changes in the same point order \(\displaystyle m_k=\frac{3-7}{4-(-2)}=-\frac23\)
2. 2Parallel line Keep the gradient and substitute the new point \(\displaystyle \begin{aligned}-5&=-\frac23(6)+c\\-5&=-4+c\\c&=-1\\y&=-\frac23x-1\end{aligned}\)
3. 3Perpendicular line Use the negative reciprocal before substituting the point \(\displaystyle \begin{aligned}m_h&=\frac32\\-5&=\frac32(6)+c\\-5&=9+c\\c&=-14\\y&=\frac32x-14\end{aligned}\)

General form and horizontal or vertical lines

Use the same direction-and-point method. A vertical line needs its fixed horizontal coordinate rather than a gradient

1. 1Read a general equation If the coefficient of the vertical variable is nonzero, rearrange to find the gradient. A zero vertical coefficient gives a vertical line \(\displaystyle \begin{gathered}Ax+By+C=0\\y=-\frac ABx-\frac CB\\(B\ne0)\end{gathered}\)
2. 2Write a general equation This line has gradient negative three-halves and passes through the given point. Clear the fraction and move all terms to the left \(\displaystyle \begin{aligned}R&=(2,5)\\y-5&=-\frac32(x-2)\\2y-10&=-3x+6\\3x+2y-16&=0\end{aligned}\)
3. 3Coefficient convention When asked for primitive general form, use integer coefficients with greatest common divisor one and the first nonzero coefficient positive \(\displaystyle \begin{gathered}-6x-4y+32=0\\\Downarrow\\3x+2y-16=0\end{gathered}\)
4. 4Horizontal and vertical Horizontal and vertical lines are perpendicular. Through this point, use the appropriate fixed coordinate. Parallel lines keep their horizontal or vertical direction \(\displaystyle \begin{gathered}P=(3,7)\\x=3\text{ is vertical}\\y=7\text{ is horizontal}\end{gathered}\)

Given line

\(\displaystyle g:\ y=3x-2\)

Point on h

\(\displaystyle (6,1)\)

Required relation

\(\displaystyle h\perp g\)

Where is the first error in this student's construction of h?

Line g

\(\displaystyle y=-4\)

Point P

\(\displaystyle (3,7)\)

Which line through P is perpendicular to g?