About NUR ARINA BASILAH BINTI KAMISAN Expertise i can answer derivatives and integral questions but not so expert in answering limits,vector and infinite series.
Experience i teached calculus for foundation students at university malaya and i also teached tutorial class in university for subject differential equation when i was doing my degree.currently i'm teaching algebra for foundation students.
Education/Credentials i have degree in mathematics with education and now i'm futhering my study in master of statistics at university malaya.
Expert: NUR ARINA BASILAH BINTI KAMISAN - 8/13/2008
Question I have a summer packet and i can't find any website that tells me how to do any of this correctly lol . here are a few questions i am completely lsot on:
Find the points of intersection of the two graphs (if any exist):
1. 3x-4y=8 and x+y=5
2. y=2(xsquared)+3 and 4y-x=12
Find the equation of the lines passing through _-2, 4) and having the following characteristics:
1. slope of 7/16
2. parallel to the line 5x-3y=3
3. perpendicular to the line x+y=0
4. passing through the origin
Find t such that all three points are colinear:
1. (-2,5), (0, t) and (1,1)
2. (-3,3), (t, -1) and (8,6)
thank you =]
Answer dear jac,
for your first question,if they want intersection point all you have to do is compare the two equation together.
3x-4y=8
3x-8=4y
y=(3x-8)/4--->(1)
and
x+y=5
y=5-x---->(2)
now substitute y=5-x into (1)
5-x=(3x-8)/4
solve the equation above
7x/4=7
7x=28
x=4
substitute x=4 into (2)
y=5-4
y=1
so the intersection point is (4,1)
also the same for your second question
y=2x^2+3--->(1)
4y-x=12
y=(x+12)/4--->(2)
substitute (2) into (1)
(x+12)/4=2x^2+3
x+12=8x^2+12
8x^2-x=0
x(8x-1)=0
so x=0 or x=1/8
for x=0,y=3 (by substitute x=0 into (2))
for x=1/8,y=97/32
so the intersection points are (0,3) and (1/8,97/32)
for 2nd question:
for the first characteristics:
general equation is y=mx+c
the slope is 7/16 which is also equal to m so now the equation will become y=7/16x+c
we need to find the value of c so substitute the point that they have gave u which is (-2,4)
4=7/16(-2)+c
c=39/8
so the equation will become y=7/16x+39/8
when they said parallel that means it has same slope
so first change the equation into general form:
5x-3y=3
y=(-5x+3)/-3 or
y=5x/3-1
so the slope is 5/3
the general equation will be
y=5x/3+c
to find c we must substitute (-2,4)
4=5(-2)+c
c=14
so the equation is y=5x/3+14
for perpendicular dat means the slope is -ve from the equation that is given
x+y=0
y=-x
so the slope is -1
and we have rules that said when 2 equation that are perpendicular,when we multiply the slope together the answer will be -1.so let m1 as the first equation slope and m2 as the slope for the 2nd equation.
m1=-1
m1*m2=-1
-1*m2=-1
m2=1
so the slope for second equation is 1
y=x+c
substitute (-2,4)
4=-2+c
c=-2
so the equation is y=x-2
passing through the origin means we have two point here (0,0) and (-2,4).from here we can calculate the slope by this formula (y2-y1)/(x2-x1)=m
so m=(4-0)/(-2-0)
m=4/-2
m=-2
and because it mention there that it passing through origin dat means the c=0
so the equation will become y=-2x
when they said collinear that means the points are in a straight line and we have this formula:
x1(y2-y3)+x2(y3-y1)+x3(y1-y2)=0
so just substitute the points into the formula above
-2(t-1)+0(1-5)+1(5-t)=0
-2t+2+5-t=0
-3t+7=0
t=7/3
for second question
-3(-1-6)+t(6-3)+8(3+1)=0
21+3t+32=0
3t+53=0
t=-53/3
