Expert: Josh - 6/22/2009

Question
Find the equation of the line having intercept –3 and 7 on the co-ordinate axis
a)   7x-3y+21=0
b)   7x+3y+21=0
c)   7x+3y-21=0
d)   7x-3y-21=0
e)   None

2)If x+2y+2=0 and 2x+3y-4=0, x-4y+k=0 are concurrent then find k?


Answer
Shruti,

Giving you the right answer is not going to help you much if you do not explain what you are having difficulties with.

For Q1, it is unclear which one (-3 or 7) actually represents the intersection of the line with the y-axis. If the line crosses the x-axis at (-3,0) and the y-axis at (0,7), then it has a slope of (y2-y1)/(x2-x1)=(7-0)/(0--3)=7/3.

One way to proceed is to rearrange the equation into the y=mx+b form. Then, you can treat "b" as the y-intercept and the gradient "m" must agree with 7/3. Assuming the information was interpreted correctly (i.e., the y-intercept is (0,7) NOT (0,-3)), with b=7, m=7/3, we have y=(7/3)x+7 as the equation. Multiplying both sides of the equation by 3, this is equivalent to 3y=7x+21. Subtract 3y from both sides give 7x-3y+21=0.

Re Q2: By "concurrent", it means all three equations are simultaneously true.

x+2y+2=0  ..[1]
2x+3y-4=0 ..[2]
x-4y+k=0  ..[3]

First, we can eliminate x by taking 2[1]-[2].
This gives 2(x+2y+2)-(2x+3y-4)=0.
It reduces to y+8=0, which implies y=-8.

Second, we substitute this back into (say) equation [1].
x+2*(-8)+2=0
x-14=0
x=14.

At this point, we have used up the first two equations to determine the value of "x" and "y". It remains to find the value of k in [3]. i.e., we ask what value of "k" will make equation [3] true?

Knowing x=14 and y=-8, you can put these values into equation [3] to solve for k. I leave this as an exercise.
We can express x in terms of y from [1], to get x=