it's about analytic geometry...can you please help me?..i needed the answer tonight...thank you very much?

Question

1. find the distance bet. lines 3x + y -12=0 and 3x+y-4=0

2. find the distance bet. lines 7x-8y+12=0 and 7x-8y-8=0

3. A circle touches the lines 5x +2y-10=0. Find its area and the locus of its center

4. Two sides of a square are on the lines x+10y-5=0. Find the area of the square

5. The sides of a rectangle are on the lines x +5y-4=0, 5x-y+6=0, x+5y+8=0 and 5x-y+10=0. Find its area.

6. The sides of a parallelogram are on the lines x-3y
+20=0, x +y+6=0, x -3y-10=0 and x=y-2=0. Find its area.

7. The distance from a line to (6,2) is 3 units. The line is perpendicular to 4x +3y=8=0. Find its equation.

8. Find the equation of a line parallel to 5x =6y-10=0 at distance square root of 61 from (-3,7)

A moving point is always equidistant from (-3,2) and the line 2x-y-2=0. What is the equation of its locus?

10. find the equation of the bisector of the obtuse angle bet. the lines 4x+y6=0

find the equation of the bisector of the obtuse angle bet. the lines 4x+y+6=0 and 2x+8y-3=0.

Answer 1

1. assume x=1 in first equation, then
3+y-12=0
3+y=12
y=9 so (1,9) is a point on the line 1
assume x=1 in second equation, then
3+y-4=0
3+y=4
y=1 so (1,1) is a point on the line 2
use distance formula

d=sqrt{(x2-x1)^2 + (y2-y1)^2}
so sqrt{(1-1)^2 + ( 1-9)^2}
sqrt{0+ (-8)^2}
sqrt{64} =8

the same way is u do the questin 2
question 3 needs one more paramerter
question 4 requires the equation of another line

q5. using simultaneous mthod solve equation 1 & 2 get points of intersection, same way solve equations 2 a& 3 , 3&4 and 4 &1
with the 4 points, use the distance formula to find the distance 2 points, for all the 4 points.
then use the formula l*b to find the area.

q6. similar to q5, but apply the area of parellogram i.e
b*h

q7.as the given line is perpendicular to the other line, the product of their slope will be equal to -1
slope of the given line is
4x-3y=8
-3y=8-4x
y=-8/3+4/3x
so slope equals 4/3

therefore the slope of the required line is -3/4
so using ponit slope formula the equation of the line is
y-y1= m(x-x1)+c (where x1,y1 is 6,2 & c is the distance that is 3)
y-2 =-3/4(x-6)+3
y-2 =(-3x+18+12)/4
4y-8 = -3x +30
4y = -3x+30+8
4y=-3x+38 this is the reuired line

q8. similar to q.7, but the slope of 2 lines are equal when they are parallel, so the slope of this line is
5x-6y-10=0
5x-10=6y
5/6x- 10/6 =y
so the slope os 5/6, so go ahead as problem 7

i have no idea regarding q9 & 10

hope the answers help u/

Answer 2

For 1 and 2 use d=(y1-mx1-b)/sqrt(m^2+1).

3 and 4 cannot be answered since you only supplied 1 line.

5,6,7,8 use formula given above.

9 is a parabola.

10. tan x = (m2-m1)/(1+m1m2)

Answer 3

1)8 units
2)20 units
i don't know any of the others