# Straight line problem

1. Jan 21, 2014

### Dumbledore211

1. The problem statement, all variables and given/known data
Show that the point (-1/2, -2) is equidistant from the straight lines 2x-3y+4=0 and 6x+4y-7=0

2. Relevant equations

distance between the points= {(x1-x2)^2+(y1-y2)^2}^1/2

3. The attempt at a solution
All I know is that the given two straight lines are perpendicular to each other. It would be very helpful if anyone of you could drop a hint as to how I should use this fact in proving the fact that the given point is at an equal distance from the given straight lines

2. Jan 21, 2014

Did you draw any graphs?

3. Jan 21, 2014

### BvU

I don't see the relevance of your equation.
Do you know how to calculate the distance between two parallel lines ?

For example between 6x + 4y = 7 and 6x + 4y = -11 ?

(The latter happens to go through the point -1/2, -2 )

4. Jan 21, 2014

### Ray Vickson

I don't see the relevance of your questions. The two lines given by the OP are not parallel.

5. Jan 21, 2014

### Tanya Sharma

Just find the distance between the given point and the first line .Then calculate the distance between the point and the second line .You will get same distances in the two cases .

The distance between a point (p,q) and line Ax+By+C=0 is given by |Ap+Bq+C|/√(A2+B2)

6. Jan 21, 2014

### eXmag

sorry post error

7. Jan 22, 2014

### haruspex

No, but because the lines are perpendicular to each other, the 'dropped perpendicular' from the point to one line will be parallel to the other line. Whether that saves anything I'm not sure.

8. Jan 22, 2014

### BvU

Dear professor Dumbledore,
Do you recognize/understand the formula as spelled out by Tanya? If so, the exercise is straightforward as she says (work, yuch!).
In #1 (2) you made me a bit careful, since there are no two points in the OP. Yet in #1 (3) you quite correctly observe the two lines are perpendicular and you sniff that there might be a smart mathemagical trick to avoid superfluous work. Good attitude!