Calculate the area of triangle QRS with altitude ST, given Q (0, 5), R (−5, 0), S (−3, 4), and T (−2, 3).
A- 6.2 square units B- 7 square units C- 5.9 square units D- 5 square units
r u familiar with the distance formula?
yeah
x2−x1)2+(y2−y1)2
@ml1042903
@Mertsj
using that find the lenghts of QR and ST. Area=(1/2)*QR*ST
Don't know sorry! :( Try @genius12
@britt5 did u find the lengths of QR and ST
i am working on it
oh k. carry on.:)
2
what's 2?
that what i got
for what did u get 2? for the length of QR?
@britt5
yes
it is nt right:( QR=sqrt((-5-0)^2+(0-5)^2) =sqrt(25+25) =sqrt(50) =5*sqrt(2)
i knew i looked wrong
@ajprincess so the answer is D- 5 square units
yup:)
Thanks can you help me on some more
ya post them:)
A segment with endpoints I (5, 2) and J (9, 10) is divided by a point K such that IK and IJ form a 2:3 ratio. Find the y value for K.
(A) 4.6 (B) 5.4 (C) 4.8 (D) 5.2
@ajprincess i think it is C 4.8
Write an equation of a line perpendicular to y = 1 over 2x − 5 in slope-intercept form that passes through the point (2, −7).
(A) y= 1 over 2x − 7 (B) y= 2x + 2 (C) y= −2x − 5 (D) y= −2x − 3
@Gabylovesyou
for the last question, the product of slopes of perpendicular lines is equal to -1. the slope of the given line is (1/2). let m be the slope of the required line. so m*(1/2)=-1 m=-2 slope intercept form of equation is y=mx+c y=-2x+c plugging the given point will give the value of c -7=-2*2+c -7=-4+c -7+4=c -3=c
@ajprincess So do yo think the other one is C 4.8 is the last one (D) y= −2x − 3
I am working on the second one. but last one is option D
Ok thanks on the second one i got 3.6 it is equivalent to 4.8
Line IJ has an equation of a line y = 3x − 8, and line KL has an equation of a line y = 3x + 4. These two lines are A) parallel because the slopes of the lines are the same B) perpendicular because the product of the slopes is −1 C.) both representations of the same line D) neither parallel nor perpendicular
C.) both representations of the same line
Write an equation of a line parallel to line EF below in slope-intercept form that passes through the point (2, 6). A) y = 2x + 3 B) y = 3x + 6 C.) y = negative two over 3x + twenty eight over three D) y = − two over threex + 6
@ajprincess
for the second one 3.6 or 4.8 is not right:( can u tell me how u workd it out?
may i knw y u say that both represent the same line
For the second one i used a ratio calculator
) perpendicular because the product of the slopes is −1
hmm. r u sure the product of the slopes is -1?
the second one would it be 4.6
nope.
B 5.4
say the total distance between I and J be 5d. 5d=sqrt((9-5)^2+(10-2)^2) =sqrt(4^2+8^2) =sqrt(16+64) =sqrt(80) =4*sqrt5 then d=(4*sqrt5)/5 the distance between I and K according to the given ratio 2d=(8*sqrt5)/5=sqrt((x-5)^2+(y-2)^2) the distance between K and J will be 3d=(12*sqrt5)/5=sqrt((9-x)^2+(10-y)^2) solve these two equations to get y.
5.2
yup
For the line segment whose endpoints are R (1, 2) and S (6, 7), find the y value for the point located 3 over 4 the distance from R to S. A) 4.25 B) 4.75 C.) 5.25 D) 5.75
before solving this one have u figured out the answer for 4th ques?
no
i only have that one and this one left
find the slopes of the two equations and see if they are equal or the product is -1.
and the question with the picture i still have
I am sry I'v gotta go nw. It's past midnight here.
y = 3x + 4 m=-3 y = 3x − 8 m=-3 A)slope are the same
m=3 nt -3 and yes slopes are same. If the slopes are same they are parallel.
Ok good night thank so much
ill guess on the other ones
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