Find the equation of a line that is parallel to line g that contains (P, Q).
the coordinate plane has a line g that passes through the points (-3,2) and (0,5).

3x − y = 3P − Q
3x + y = Q − 3P
x − y = P − Q
x + y = Q − P

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danielmaduroh danielmaduroh · Answer 1 · 2020-02-11T15:03:34+01:00

Correct option:

[tex]\boxed{x-y=P-Q}[/tex]

Given that the line we are looking for is parallel to g, then the slope of that line and g is the same, therefore:

[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ (x_{1},y_{1})=(-3,2) \\ \\ (x_{2},y_{2})=(0,5) \\ \\ \\ m=\frac{5-2}{0-(-3)}=1[/tex]

So we can write the point-slope form of the equation of the line as follows:

[tex]y-y_{0}=m(x-x_{0}) \\ \\ (x_{0},y_{0})=(P,Q) \\ \\ \\ y-Q=1(x-P) \\ \\ y-Q=x-P \\ \\ \\ Arranging: \\ \\ P-Q=x-y \\ \\ \boxed{x-y=P-Q}[/tex]

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raphealnwobi raphealnwobi · Answer 2 · 2022-01-18T11:42:12+01:00

A linear equation is in the form:

y = mx + b

where y, x are variables, m is the slope and b is the y intercept.

Line g passes through (-3,2) and (0,5). hence:

Two lines are parallel if they have the same slope. Hence:

Line parallel to line g has a slope of 1. Since it passes through (P, Q), hence:

[tex]y-y_1=m(x-x_1)\\\\y-Q=1(x-P)\\\\y-Q=x-P\\\\x-y=P-Q[/tex]

The equation of a line that is parallel to line g that contains (P, Q) is x - y = P - Q

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