im so sad on this i cannot make a move im so so confused pls guide me on this what i should do
find the equation of the line passing through the point of intersection of the lines x - y = 0 and 3x - 2y = 2 and satisfying the condition indicated: (without finding the intersection of the two lines)
a. containing the point ( - 1, 3)
b. having x - intercept of 4
c. parallel to line x + y = 2
d. cutting from the first quadrant a triangle whose area is 9
im so super confused on these... i can some simple finding equation when slopes or points are given ... but these things are quite confusing
thanks
because if i did... in the statement above is (without finding the intersection of the two lines)... that is y if solved it ... i was thinking that it not the possible answer
i just solved it sir: the intersection is ( 2, 2) how are they related to the problem below like a.) b.) c.) .... does it need a family of lines procedure? the one that has a very long solution, sir?
thank you
A (non-vertical) straight line can always be written in the form y= Ax+ B for some numbers A and B.
Saying That it passes through the intersection of x- y= 0, which is the same as y= x, means Ax+ B= x for some x. Saying that it passes through the intersection of 3x- 2y= 2, which is the same as y= (3/2)x- 1, means Ax+ B= (3/2)x- 1. Saying that it passes through the intersection of those two lines means those are true for the [b]same x. From Ax+ B= x, we get (1- A)x= B or x= B/(1- A).
a) Saying the line passes through (-1, 3) means that 3= -A+ B
So you have Ax+ B= (3/2)x- 1, x= B/(1- A), and 3= -A+ B. Replace the value of x in the first equation by B/(1-A) and you have two equations to solve for A and B.
b) Saying that the line has an x intercept -4 means it passes through (-4, 0) so you can now do the same as in (a).
c) Saying the the line is parallel to x+ y= 2, which is the same as y= -x+ 2, means that it has slope A= -1. You still have the two equations Ax+ B= (3/2)x- 1 and x= B/(1- A) which becomes -x+ B= (3/2)x- 1 and x= B/(-2). Eliminate x as I suggested before and solve for B.
d) cutting from the first quadrant a triangle whose area is 9. If the two points at which the line cuts the axes are (X, 0) and (0, Y), the area of the triangle formed is (1/2)XY= 9 so XY= 18. Using y= Ax+ B, Y= B and X= -B/A so that equation becomes . Again, you can eliminate x form the two equations I cited above to give you two equations for A, and B.