SOLUTION: Find the slope of any line perpendicular to the line through points (0, 5) and (-3, -4).

Question 26857: Find the slope of any line perpendicular to the line through points (0, 5) and
(-3, -4).
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING EXAMPLES 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(921) 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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