Co-ordinate Geometry questions (1 Viewer)

[Follz21]

Any help with these questions will be very much appreciated.

Q1] Find d if the line passing through (-2,-1) and (3,2) is perpendicular to the line dx + 2y - 3 = 0

(I know the gradient of the line is 3/5, and therefore the gradient of the equation dx + 2y - 3 = 0 must be -5/3, but I can't figure out how to find d through algebraic terms... or am I wrong about the gradient?)

Q2] The lines L1, L2 and L3 are concurrent.

L1: 2x - 3y + 6 = 0
L2: x - y - 2 = 0
L3: 2x - 3y +6 + k(x - y - 2) = 0

a) At what point do these lines intersect?
b) If L3 passes through the point (-7,5) find k. Hence find the equation of L3.


Q3] Find the equation of the locus of a point P which is equidistant from the y-axis and the point (3,-1).


Thanks for taking the time

 

[ninetypercent]

1) gradient. (2 + 1)/(3+2) = 3/5
so the line dx + 2y - 3 has a gradient of -5/3

2y = -dx + 3
y = (-dx + 3)/2
-d/2 = -5/3
-3d = 10
d = -10/3

 

[MATHSCAPE]

1) gradient. (2 + 1)/(3+2) = 3/5
so the line dx + 2y - 3 has a gradient of -5/3

2y = -dx + 3
y = (-dx + 3)/2
-d/2 = -5/3
-3d = 10
d = -10/3

damn it, beat me to it.

 

[ninetypercent]

Q2] The lines L1, L2 and L3 are concurrent.

L1: 2x - 3y + 6 = 0
L2: x - y - 2 = 0
L3: 2x - 3y +6 + k(x - y - 2) = 0

a) At what point do these lines intersect?

solve simultaneously

Spoiler

2x - 3y + 6 = 0 (1)
x - y - 2 = 0 (2)

(2) x 2
2x - 2y - 4 = 0 (3)

(1) - (3)
-y + 10 = 0
y = 10

x = y + 2
= 10 + 2 = 12

(12,10)

b) If L3 passes through the point (-7,5) find k. Hence find the equation of L3.

L3: 2x - 3y +6 + k(x - y - 2) = 0

passes through (-7,5)
sub it in

Spoiler

2(-7) - 3(5) + 6 + k(-7 - 5 - 2) = 0
-23 + k(-14) = 0
-14k = 23
k = -23/14

substituting back in
2x - 3y +6 + k(x - y - 2) = 0
2x - 3y + 6 + -23/14(x-y-2) = 0
28x - 42y + 84 - 23x + 23y + 46 = 0
5x -19y + 130 = 0

 

[ninetypercent]

Q3] Find the equation of the locus of a point P which is equidistant from the y-axis and the point (3,-1).

recall focus-directrix definition of parabola. For a parabola,

PM = PS where M is the directrix, S is the focus and P is any point on the parabola

Here you have the y axis as the directrix, and (3,-1) as the focus.

then find the equation of the parabola. try it yourself

Spoiler

focal length = 3/2
therefore, the vertex is equal to (3/2, -1)

using the equation.. (y-k)^2 = 4a(x-h) where (h,k) is the vertex, and a is the focal length

the locus is the parabola ( y +1)^2 = 6(x- 1.5)

I think...

 

[Follz21]

Thanks a lot ninetypercent, you were very helpful!

Cheers