# Math Help - need help in lines...!!

1. ## need help in lines...!!

hi guys!! i really need help in my assignment...can you help me solve this?:
a.) What are the equations of the lines through (-5,-3) and passing at distance two square root of five from (5,7)?
b.)A point moves so that the ratio of its distances from 3x+4y+8=0 and 4x+3y-6=0 is 2. find the equation of its locus.

2. Originally Posted by april29
hi guys!! i really need help in my assignment...can you help me solve this?:
a.) What are the equations of the lines through (-5,-3) and passing at distance two square root of five from (5,7)?
[snip]
I assume you mean the perpendicular distance?

A rough diagram will help to visualise the following situation.

Let P(a, b) be the point on the line such that the distance between P and (5, 7) is 2sqrt{5}. Then:

(a - 5)^2 + (b - 7)^2 = 20 .... (1)

The lines obviously have the form y + 3 = m(x + 5).

Also (a, b) is on the line and so:

b + 3 = m(a + 5) .... (2)

The line segment joining P and (5, 7) is perpendicular to the line and so:

(b - 7)/(a - 5) = -1/m .... (3)

Solve simultaneously to get the value of m.

I get m = 2 (for P(1,9)) or m = 1/2 (for P(7,3)) but I sometimes slip up .......

Off-hand, I don't see a more elegant way.

3. Originally Posted by april29
hi guys!! i really need help in my assignment...can you help me solve this?:
b.)A point moves so that the ratio of its distances from 3x+4y+8=0 and 4x+3y-6=0 is 2. find the equation of its locus.
I assume you mean distance from 3x+4y+8=0 is twice distance from 4x+3y-6=0.

Here are two hints to get you started:

1. The locus passess through the point of intersection of 3x+4y+8=0 and 4x+3y-6=0 (do you see why?)

2. The locus consists of TWO lines (do you see why?) and they both pass through the aforementioned intersection point.

4. Originally Posted by april29
hi guys!! i really need help in my assignment...can you help me solve this?:
a.) What are the equations of the lines through (-5,-3) and passing at distance two square root of five from (5,7)?
[snip]
Originally Posted by mr fantastic
[snip]
Off-hand, I don't see a more elegant way.
But now I've just thought of one ......

Find the values of m such that the line y + 3 = m(x + 5) <=> y = mx + 5m - 3 is tangent to the circle with centre (5, 7) and radius 2 sqrt{5}, that is, the circle $(x - 5)^2 + (y - 7)^2 = 20$. (What circle geometry theorem does this approach exploit?)

To do this, set up the equation that gives the x-coordinates of the intersection points of the line with the circle by substituting y = mx + 5m - 3 into $(x - 5)^2 + (y - 7)^2 = 20$.

You get a quadratic in x.

Set the discriminant (which will be a function of m) of this quadratic equal to zero and solve for m.