Point A = (3,5) , P = (1,3), and Q = (2,5).
If line AB is parallel to line PQ, find the equation of line AB.

Open in App

Solution

The equation of a line with slope m and passing through the point $(x_{1}, y_{1})$ is given by
$y - y_{1} = m \times (x - x_{1})$
Now on line AB we already know the point $(x_{1}, y_{1})$ = (3,5)
To find slope we can use the property of paralllel lines that is Slope of parallel lines are always the same i.e.
Slope of AB = Slope of PQ
Slope of a line = ( $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ )
$Let, x_{1} = 1, x_{2} = 2 y_{1} = 3, y_{2} = 5$
Slope of PQ = slope of AB = ( $\frac{5 - 3}{2 - 1}$ )
= 2
So equation of line AB = $y - 5 = 2 \times (x - 3)$
= $y - 5 = 2 x - 6$
= $y - 2 x + 1 = 0$

Suggest Corrections

Similar questions

Q. The equation of line

A B

\frac{x}{2} = \frac{y}{- 3} = \frac{z}{6}

. Through a point

P (1, 2, 5)

, line

P N

is drawn perpendicular to

A B

and line

P Q

is drawn parallel to the plane

3 x + 4 x + 5 z = 0

to meet

A B

Q

. Then

Q. Find the equation of the line passing through the point (−3, 5) and perpendicular to the line joining (2, 5) and (−3, 6).

AB is a line segment which is inclined to line XY, such that they intersect at point O. A line PQ is drawn which intersects line XY at point P at the same angle of inclination as AB. Line PQ is __ to line segment AB.

Q. Draw a line AB=6cm. Mark a point P anywhere outside the line AB. Through the point P, construct a line parallel to AB.

Q. The equations of line AB and line PQ are y =

- \frac{1}{2} x

and y=2x respectively. Find the measure of angle

∠

BOQ which is formed by intersection of line AB and line PQ. (Point P and point A are in first and second quadrant respectively)

Join BYJU'S Learning Program

Point A = (3,5) , P = (1,3), and Q = (2,5). If line AB is parallel to line PQ, find the equation of line AB.

Point A = (3,5) , P = (1,3), and Q = (2,5).
If line AB is parallel to line PQ, find the equation of line AB.