find joint equation of lines passing through point (3,2) one of which is parallel to line x+3y-1=0 and other is perpendicular
to to line y=2
Answers
Step-by-step explanation: Equation of the line parallel to x+3y-1 = 0 can be taken as x+3y+p =0. If thus passes through the point (-1,2) then -1+6+p = 0 => p=-5. So the equation of the parallel line is x+3y-5=0……(1).
The Equation of the line perpendicular to 2x-3y-1=0 is of the form 3x+2y+q=0. If this passes through the point (-1,2) then -3+4+q=0 which gives q=-1. So the equation of the perpendicular line is 3x+2y-1=0………………(2).
Hence the combined equation of (1) and (2) is (x+3y-5)(3x+2y-1) = 0 which when simplified yields 3x^2 + 11xy + 6y^2 – 16x - 15y + 5=0.
Answer:
2x+3y−5=0⟹
3y=−2x+5⟹
3y3=−2x3+53⟹
y=−23x+53
⟹slope of given line is
m=−23.
A line parallel to 2x+3y−5=0 has the same slope as this line.
To get the equation of the line that is parallel to 2x+3y−5=0 and that goes through (2, −1), use the point-slope formula:
y=m(x−x1)+y1
with m=−23
and (x1, y1) = (2, −1).
y=−23(x−2)+(−1)⟹
y=−23x−23(−2)−1⟹
y=−23x+43−33⟹
y=−23x+13.
This equation can be used for the answer.
You can multiply by 3 to clear it of fractions and get
3y=3(−23x)+3(13)⟹
3y=−2x+1⟹
2x+3y−1=0,
the equation of the line through (2, −1) and parallel to 2x+3y−5=0
in the form ax+by+c=0.