Worked examples · Sheet 1

Lines through two points, parallel and perpendicular lines

Theory used: Slope, lines through a point, parallelism and perpendicularity Equation of a line Distance from a point to a line

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Write the equation of the line passing through $A (- 1, 3)$ and $B (2, - 3)$ .

The slope is

m = \frac{y _{B} - y _{A}}{x _{B} - x _{A}} = \frac{- 3 - 3}{2 - ( - 1 )} = \frac{- 6}{3} = - 2.

The line through $A$ with slope $- 2$ is

y - 3 = - 2 (x + 1), y = - 2 x + 1.

Result: $y = - 2 x + 1$ .

The abscissas of the two points are different, so the line is not vertical and has a slope, which is computed as the ratio between the change in $y$ and the change in $x$ in going from $A$ to $B$ . Once $m$ is found, one imposes the passage through one of the two points with the point-slope formula; the choice of point does not change the result, because both belong to the line. A check with the other point: for $x = 2$ one obtains $y = - 4 + 1 = - 3$ , which is the ordinate of $B$ , therefore the line also passes through $B$ .

Exercise 2

Write the equations of the line through $P (1, 2)$ parallel to the line $y = 3 x - 5$ and of the line through $P$ perpendicular to the same line.

The given line has slope $m = 3$ . The parallel line through $P$ has the same slope:

y - 2 = 3 (x - 1), y = 3 x - 1.

The perpendicular line through $P$ has slope $m^{'} = - \frac{1}{3}$ , since perpendicularity requires $m m^{'} = - 1$ :

y - 2 = - \frac{1}{3} (x - 1), y = - \frac{1}{3} x + \frac{7}{3} .

Result: parallel $y = 3 x - 1$ , perpendicular $y = - \frac{1}{3} x + \frac{7}{3}$ .

Parallelism and perpendicularity fix the slope of the required line, therefore the passage through $P$ determines everything else. For the parallel line one copies $m = 3$ ; for the perpendicular line one imposes $m m^{'} = - 1$ , so $m^{'}$ is the negative reciprocal of $3$ . In both cases one uses the point-slope formula with the slope found and brings the equation to slope-intercept form. A check: for $x = 1$ the parallel line gives $y = 3 - 1 = 2$ and the perpendicular line gives $y = - \frac{1}{3} + \frac{7}{3} = 2$ , therefore both pass through $P (1, 2)$ .

Exercise 3

Compute the distance from the point $P (2, 1)$ to the line $r$ of equation

y = \frac{3}{4} x + 1.

Bring the line to general form by multiplying by $4$ and rearranging:

4 y = 3 x + 4, 3 x - 4 y + 4 = 0.

The distance formula gives

d (P, r) = \frac{∣3 \cdot 2 - 4 \cdot 1 + 4∣}{3 ^{2} + ( - 4 ) ^{2}} = \frac{∣6 - 4 + 4∣}{25} = \frac{6}{5} .

Result: $d (P, r) = \frac{6}{5}$ .

The point-line distance formula requires the general form $a x + b y + c = 0$ , so the first operation moves all the terms to the left and eliminates the fraction. In the numerator one substitutes the coordinates of $P$ for $x$ and $y$ and takes the absolute value, because a distance is not negative; the denominator contains the square root of the sum of the squares of the coefficients of $x$ and $y$ , which here equals $9 + 16 = 5$ .