Equations of lines through a point

Three quick questions from earlier lessons.

A line has slope 2 and passes through $(3, 1)$ . Does it pass through $(4, 3)$ ?

A point and a slope fix a line.

Point-slope form:

y = y_1 + m(x - x_1)

. To get

y = mx + b

, substitute the point to find

b

. A line parallel (perpendicular) to a given line uses the same (negative reciprocal) slope, with its own intercept.

Through $(2, 5)$ , parallel to $y = 3x - 4$ .

m = 3

Same slope.

b = -1

5 = 3·2 + b.

y = 3x - 1

The line.

Your turn: slope $-2$ through $(1, 4)$ . The intercept $b$ .

b = 6

4 = −2 + b.

Write the line with slope $3$ through $(4, -1)$ : enter the right-hand side of $y = \ldots$
Write the line with slope $1$ through $(2, -1)$ : enter the right-hand side of $y = \ldots$

Find the $y$ -intercept of the line parallel to $y = -3x + -1$ through $(-3, -5)$ .
Find the $y$ -intercept of the line parallel to $y = 1x + -1$ through $(-1, 2)$ .

Find the $y$ -intercept of the line perpendicular to $y = 4x$ through $(-3, -3)$ .

Boss problem. $A(-2, -2)$ and $B(2, 0)$ . Find the perpendicular bisector of $AB$ .

Three questions to check you have got it.

Write the line with slope $3$ through $(4, -1)$ : enter the right-hand side of $y = \ldots$
Write the line with slope $1$ through $(2, -1)$ : enter the right-hand side of $y = \ldots$
Write the line with slope $5$ through $(4, -1)$ : enter the right-hand side of $y = \ldots$

Equations of lines

$y = y_1 + m(x - x_1)$ ; find $b$ by substituting the point.

Watch out: parallel lines share the slope, not the intercept.

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