```Question 22537
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peter wanted to draw a line parallel to the line of a window with equation y=x+2.if the line is supposed to pass through point (2,3).what is the equation of the line desired by peter.
1 solutions
y=x+2....line parallel to this is given by y=x+k..it is passing through (2,3)
so substituting...3=2+k..or k=3-2=1
hence the eqn.of the parallel line is y=x+1
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write an equation for the line, in point slope form, that passes through the points (-5,7) and (-4,-3)
1 solutions
write an equation for the line, in point slope form, that passes through the points (-5,7) and (-4,-3)
formula for eqn.of line connecting 2 points x1,y1 and x2,y2 is given by
y-y1=(x-x1)*{(y2-y1)/(x2-x1)}....and slope of the line =(y2-y1)/(x2-x1)
so eqn.of line is y-(-3)=(x-(-4))*{(-3-(7))/(-4-(-5))
y+3=(x+4)*(-10/1)=-10(x+4)=-10x-40
y=-10x-40-3
y=-10x-43
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i.. i need help with writing and solving an equation in standard form. for instance.. y- intercept=8 and the slope is 3. does that mean the equation in standard form is 3x+y=-8?? I also need help with.. if the slope = 3/4 and y-intercept= -2 what's the equation in standard form and how can you get 3/4 to not be a fraction or negative? same with slope= 3/5 and passes through 0, -6.. i think that means the y-int. is -6 but how do you graph slopes that are fractions and write them in an equation. sorry it's such a long question but please help me.. thanks!!
1 solutions
hi.. i need help with writing and solving an equation in standard form. for instance.. y- intercept=8 and the slope is 3. does that mean the equation in standard form is 3x+y=-8?? I also need help with.. if the slope = 3/4 and y-intercept= -2 what's the equation in standard form and how can you get 3/4 to not be a fraction or negative? same with slope= 3/5 and passes through 0, -6.. i think that means the y-int. is -6 but how do you graph slopes that are fractions and write them in an equation. sorry it's such a long question but please help me.. thanks!!
The slope intercept form of eqn.of a st.line is y=mx+c..where m is the slope and c is the y intercept.so your answer should be
..y=3x+8 and not y+3x=8
now slope=3/4.. we are given that y intercept is 2.. so the eqn.is..
..y=(3/4)x-2.to remove the fraction multiply both sides of the eqn.with 4 we get
4y=(3*4/4)x-2*4or 4y=3x-8if you so desire you put all of them on one side as
4y-3x+8=0
now slope is 3/5 and passes through 0, -6.. i think that means the y-int. is -6 YES CORRECT YOU ARE ..so proceed as above to get the eqn.and remove fraction in slope as is convenient.the resulting eqn.you should be able to graph easily by giving different values for x say 0,1,2,3..etc and finding corresponding values of y and plotting.
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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)
What is the point of interectstion of line L1 and L2
1 solutions
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
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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