Question 26858: Geometry. Floor plans for a building have the four corners of a room located at the points (2, 3), (11, 6), (-3, 18), and (8, 21). Determine whether the side through the points (2, 3) and (11, 6) is parallel to the side through the points (-3, 18) and (8, 21).
Geometry. For the floor plans given in exercise the first exersize determine whether the side through the points (2, 3) and (11, 6) is perpendicular to the side through the points(2, 3) and (-3, 18).
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! SEE THE FOLLOWING EXAMPLES WHICH ARE SAME AS YOURS AND TRY.IF STILL IN DIFFICULTY COME BACK..
Linear-equations/27633: Write an equation of a line that contains the given the given point and is parallel to the given line.
1.) (6,-2) 3x + 2y = 8
2.) (-1,7) 6x - 3y = 9
3.) (0,1) y = 3/7x - 8
Write an equation of a line that contains the given the given point and is perpendicular to the given line.
1.) (6,5) y = -1/2x + 1
2.) (9,-3) y = 3x + 8
3.) (0,4) y = -5/7x - 2
I'm desperate for some help!!!^^Thank you.
1 solutions
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Answer 15040 by venugopalramana(918) on 2006-02-17 00:22:12 (Show Source):
SEE THE FOLLOWING EXAMPLES WHICH ARE SAME AS YOURS AND TRY.IF STILL IN DIFFICULTY COME BACK..
Write an equation of a line that contains the given the given point and is parallel to the given line.
1.) (6,-2) 3x + 2y = 8
PARALLEL LINES HAVE EQUAL SLOPES.
SO PARALLEL LINE TO 3X+2Y=8 IS
3X+2Y=K....IT IS PASSING THROUGH (6,-2)..SO
3*6+2*-2=K=18-4=14
SO EQN.OF THE LINE IS
3X+2Y=14
THE OTHER PROBLEMS ARE SIMILAR
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1. Write in slope intercept the equation of theline passing through the two points.Show that the line is perpindicular to the given line.
(-2,-2), (1,3); y= 3x-1
EQN.OF LINE JOINING (X1,Y1) AND (X2,Y2) IS GIVEN BY
Y-Y1=(X-X1)*(Y2-Y1)/(X2-X1), WHERE (Y2-Y1)/(X2-X1) IS THE SLOPE OF THE LINE AND Y1-X1*(Y2-Y1)/(X2-X1)IS THE INTERCEPT.
HENCE EQN.OF LINE IS
Y+2=(X+2)*(3+2)/(1+2)=X(5/3)+2*5/3
Y=X(5/3)+10/3-2=X(5/3)+4/3.....HENCE SLOPE IS 5/3 AND INTERCEPT IS 4/3
SLOPE OF GIVEN LINE Y=3X-1 IS 3 ,,,FOR 2 LINES TO BE PERPENDICULAR,THE PRODUCT OF THEIR SLOPES SHOULD BE -1.HERE THE PRODUCT IS 3*5/3=5..HENCE THEY ARE NOT PERPENDICULAR TO EACH OTHER..CHECK BACK YOUR NUMBERS...COPY THE PROBLEM PROPERLY.AS GIVEN THE LINES ARE NOT PERPENDICULAR.
2. write in slope-intercept form the equation of the line passing through the given point and perpindicular to the given line.
(-4,-7), y=-4x-7
SLOPE OF Y=-4X-7 IS -4
AS GIVEN ABOVE FOR 2 LINES TO BE PERPENDICULAR,THE PRODUCT OF THEIR SLOPES SHOULD BE -1.HENCE THE SLOPE OF THE REQUIRED LINE IS -1/-4=1/4
EQN.OF REQD. LINE IS
Y+7=(1/4)(X+4)=X/4 + 1
Y= X/4 - 6
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