Question 32248: Write an equation in slope-intercept form for each line described:
containing (2,-9) slope -1
containing (3,-4) slope 0
containing (3,1) and (6,-3)
containing (-2,7) and (-2,3) Please help!!!
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Graphs/30060: i need help with writing equasions in slope- intercept form and point slope equasion of a line and Slope intercept of a line
1 solutions
Answer 16809 by venugopalramana(1167) About Me on 2006-03-13 11:06:04 (Show Source):
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
Quadratic_Equations/30582: I have to know how to find X and y intercepts algebraically and graphically. I have to give 2 examples on how to do them. and explain it. Please help.
Thank you
1 solutions
Answer 17263 by venugopalramana(1167) About Me on 2006-03-18 04:44:48 (Show Source):
LET US TAKE EQN.AS
X+Y=1....
X INTERCEPT IS DISTANCE FROM ORIGIN TO POINT WHERE THE ABOVE LINE CUTS X AXIS.
IT IS OBTAINED BY PUTTING Y =0 AND SOLVING FOR X THE EQN.
HERE WE GET X+0=1...OR..X =1.....X INTERCEPT=1
Y INTERCEPT IS DISTANCE FROM ORIGIN TO POINT WHERE THE ABOVE LINE CUTS Y AXIS.
IT IS OBTAINED BY PUTTING X =0 AND SOLVING FOR Y THE EQN.
HERE WE GET 0+Y=1...OR..Y =1.....Y INTERCEPT=1
TO DO GRAPHICALLY PLOT THE GRAPH AND IND THE POINTS AS ABOVE AND THEIR DISTANCE FROM ORIGIN TO GET THE ANSWER
X=.........0..........1.........2..............3.........ETC...........
Y=1-X......1..........0.........-1.............-2.........ETC..........
GRAPH WILL LOOK LIKE THIS.....
graph( 600, 600, -5, 5, -5, 5, 1-x)
Graphs/24460: find the slope,m,and the y intercept,b,of the line equation 2y=28
1 solutions
Answer 13033 by venugopalramana(1088) About Me on 2006-01-15 10:53:09 (Show Source):
SEE THE FOLLOWING WHICH IS SIMILAR TO YOUR PROBLEM AND DO
GIVEN:
· There is a line (L1) that passes through the points
(8,-3) and (3,3/4).
· There is another line (L2) with slope m=2/3 that
intersects L1 at the point (-4,6).
· A third line (L3) is parallel to L2 that passes
through the (7,-13 1/2).
· Yet another line (L4) is perpendicular to L3, and
passes through the point (1/2,5 2/3).
· The fifth line (L5) has the equation
2/5y-6/10x=24/5.
Using whatever method, find the following:
2. The point of intersection of L1 and L3
3. The point of intersection of L1 and L4
4. The point of intersection of L1 and L5
5. The point of intersection of L2 and L3
6. The point of intersection of L2 and L4
7. The point of intersection of L2 and L5
8. The point of intersection of L3 and L4
9. The point of intersection of L3 and L5
10. The point of intersection of L4 and L5
PLEASE NOTE THE FOLLOWING FORMULAE FOR EQUATION OF A
STRAIGHT LINE:
slope(m)and intercept(c) form...y=mx+c
point (x1,y1) and slope (m) form...y-y1=m(x-x1)
two point (x1,y1)and(x2,y2)form.....................
y-y1=((y2-y1)/(x2-x1))*(x-x1)
standard linear form..ax+by+c=0..here by transforming
we get by=-ax-c..or y=(-a/b)x+(-c/b)..comparing with
slope intercept form we get ...slope = -a/b and
intercept = -c/b
*****************************************************
line (L1) that passes through the points (8,-3) and
(3,3/4).
eqn.of L1..y-(-3)=((3/4+3)/(3-8))*(x-8)
y+3=((15/4(-5)))(x-8)=(-3/4)(x-8)
or multiplying with 4 throughout
4y+12=-3x+24
3x+4y+12-24=0
3x+4y-12=0.........L1
There is another line (L2) with slope m=2/3 that
intersects L1 at the point (-4,6).
This means (-4,6)lies on both L1 and L2.(you can check
the eqn.of L1 we got by substituting this point in
equation of L1 and see whether it is satisfied).So
eqn.of L2...
y-6=(2/3)(x+4)..multiplying with 3 throughout..
3y-18=2x+8
-2x+3y-26=0.........L2
A third line (L3) is parallel to L2 that passes
through the (7,-13 1/2).
lines are parallel mean their slopes are same . so we
keep coefficients of x and y same for both parallel
lines and change the constant term only..
eqn.of L2 from above is ...-2x+3y-26=0.........L2
hence L3,its parallel will be ...-2x+3y+k=0..now it
passes through (7,-13 1/2)=(7,-13.5)......substituting
in L3..we get k
-2*7+3*(-13.5)+k=0...or k=14+40.5=54.5..hence eqn.of
L3 is........-2x+3y+54.5=0......................L3
Yet another line (L4) is perpendicular to L3, and
passes through the point (1/2,5 2/3).
lines are perpendicular when the product of their
slopes is equal to -1..so ,we interchange coefficients
of x and y from the first line and insert a negative
sign to one of them and then change the constant term.
L3 is........-2x+3y+54.5=0......................L3
hence L4,its perpendicular will be ..3x+2y+p=0...L4
this passes through (1/2,5 2/3)=(1/2,17/3).hence..
3*1/2+2*17/3+p=0..or ..p= -77/6.so eqn.of L4 is
3x+2y-77/6=0..............L4
The fifth line (L5) has the equation 2/5y-6/10x=24/5.
now to find a point of intersection means to find a
point say P(x,y) which lies on both the lines ..that
is, it satisfies both the equations..so we have to
simply solve the 2 equations of the 2 lines for x and
y to get their point of intersection.For example to
find the point of intersection of L1 and L3 we have to
solve for x and y the 2 equations....
3x+4y-12=0.........L1....(1) and
-2x+3y+40.5=0......L3.....(2)
I TRUST YOU CAN CONTINUE FROM HERE TO GET THE
ANSWERS.If you have any doubts or get into any
difficulty ,please ask me.
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