c We want to find point M(x, y) on the line LD that is twice as far from point L as it is from point D. Let's denote the distance from M to D as d_(MD) and the distance from M to L as d_(ML). We want d_(ML) to be twice as long as d_(MD).
2 d_(MD) = d_(ML)
Our next step will be to find the expressions for d_(MD) and d_(ML). To do so, let's use the Distance Formula for the points M( x, y), L( -2, 1), and D( 4, -4).
d_(MD) = sqrt(( 4- x)^2 + ( -4- y)^2)
d_(ML) = sqrt(( -2- x)^2 + ( 1- y)^2)
Next, let's substitute those expressions into the previous equation.
2 d_(MD) &= d_(ML)
&⇓
2 sqrt((4 - x)^2 + (-4 - y)^2) &= sqrt((-2 - x)^2 + (1 - y)^2)
Notice that we have only one equation, but two variables. Therefore, we cannot solve it yet. However, we also know that point M must lie on the line LD. This means that the coordinates of M satisfy the line's equation.
y=m x+bHaving the equation of this line, we will be able to express the y -coordinate of the point M with respect to x. From Part A we know that the slope m is -56. This allows us to write an incomplete equation of LD.
y=-5/6x+b
To find the value of b, we will substitute the coordinates of either point L or D into the above equation. Let's choose point L(-2,1)
y= -5/6x + b
1=-5/6( -2) + b
1 = -5(-2)/6+b
1 = 10/6+b
1 = 5/3+b
3/3 = 5/3+b
-2/3 = b
b = -2/3
Now we can write the complete equation of the line.
y= -5/6x + -2/3
This means that (x, -56x + -23) are the coordinates of point M. Let's substitute -56x + -23 for y in the equation relating the two distances.
2 sqrt((4- x )^2 + ( -4 - ( -5/6x + -2/3 ) )^2) = sqrt((-2-x)^2 + (1 - ( -5/6x + -2/3 ))^2)
Finally, we have one equation with only one variable. To solve it for x, we will begin by raising both sides of the equation to the power of 2. Then we will continue simplifying the equation.
2 sqrt((4-x)^2 + (-4 - (-5/6x + -2/3 ))^2) = sqrt((-2-x)^2 + (1 - (-5/6x + -2/3 ))^2)
( 2 sqrt((4-x)^2 + (-4 - (-5/6x + -2/3 ))^2) )^2 = ( sqrt((-2-x)^2 + (1 - (-5/6x + -2/3 ))^2) )^2
2^2 (sqrt((4-x)^2 + (-4 - (-5/6x + -2/3 ))^2) )^2 = ( sqrt((-2-x)^2 + (1 - (-5/6x + -2/3 ))^2) )^2
4 (sqrt((4-x)^2 + (-4 - (-5/6x + -2/3 ))^2) )^2 = ( sqrt((-2-x)^2 + (1 - (-5/6x + -2/3 ))^2) )^2
4 ( (4-x)^2 + (-4 - (-5/6x + -2/3 ))^2 ) = (-2-x)^2 + (1 - (-5/6x + -2/3 ))^2
4 (4-x)^2 + 4(-4 - (-5/6x + -2/3 ))^2 = (-2-x)^2 + (1 - (-5/6x + -2/3 ))^2
4 (4^2-2(4)x + x^2 ) + 4((-4)^2 - 2(-4)(-5/6x + -2/3 ) + (-5/6x + -2/3 )^2) = (-2)^2 -2(-2)x +x^2 + 1^2 -2(1)(-5/6x + -2/3 ) + (-5/6x + -2/3 )^2
4 ( 16 -8x + x^2 ) + 4(16 + 8(-5/6x + -2/3 ) + (-5/6x + -2/3 )^2) = 4 +4x +x^2 + 1 -2(-5/6x + -2/3 ) + (-5/6x + -2/3 )^2
64 -32x + 4x^2 + 64 + 32(-5/6x + -2/3 ) + 4(-5/6x + -2/3 )^2 = 4 +4x +x^2 + 1 -2(-5/6x + -2/3 ) + (-5/6x + -2/3 )^2
4x^2 -32x + 128 + 32(-5/6x + -2/3 ) + 4(-5/6x + -2/3 )^2 = x^2 +4x + 5 -2(-5/6x + -2/3 ) + (-5/6x + -2/3 )^2
4x^2 -32x + 128 + -160/6x + -64/3 + 4(-5/6x + -2/3 )^2 = x^2 +4x + 5 -2(-5/6x + -2/3 ) + (-5/6x + -2/3 )^2
4x^2 -32x + 128 + -160/6x + -64/3 + 4(-5/6x + -2/3 )^2 = x^2 +4x + 5 +10/6x + 4/3 + (-5/6x + -2/3 )^2
4x^2 -32x + 128 + -160/6x + -64/3 + 4(( -5/6x)^2 + 2( -5/6x)( -2/3) + (-2/3 )^2) = x^2 +4x + 5 +10/6x + 4/3 + (-5/6x)^2 + 2(-5/6x)(-2/3 ) + (-2/3 )^2
4x^2 -32x + 128 + -160/6x + -64/3 + 4( 25/36x^2 + 20/18x + 4/9 ) = x^2 +4x + 5 +10/6x + 4/3 + 25/36x^2 + 20/18x+ 4/9
4x^2 -32x + 128 + -160/6x + -64/3 + 100/36x^2 + 80/18x + 16/9 = x^2 +4x + 5 +10/6x + 4/3 + 25/36x^2 + 20/18x+ 4/9
36 ( 4x^2 -32x + 128 + -160/6x + -64/3 + 100/36x^2 + 80/18x + 16/9 ) = 36( x^2 +4x + 5 +10/6x + 4/3 + 25/36x^2 + 20/18x+ 4/9 )
144x^2 - 1152 x + 4608 + -5760/6x + -2304/3 + 3600/36x^2 + 2880/18x + 576/9 = 36 x^2 +144x + 180 +360/6x + 144/3 + 900/36x^2 + 720/18x+ 144/9
144x^2 - 1152 x + 4608 + -5760/6x + (-768) + 3600/36x^2 + 2880/18x + 576/9 = 36 x^2 +144x + 180 +360/6x + 48 + 900/36x^2 + 720/18x+ 144/9
144x^2 - 1152 x + 4608 +(-960)x + (-768) + 3600/36x^2 + 2880/18x + 576/9 = 36 x^2 +144x + 180 + 60x + 48 + 900/36x^2 + 720/18x+ 144/9
144x^2 - 1152 x + 4608 +(-960)x + (-768) + 3600/36x^2 + 2880/18x + 64 = 36 x^2 +144x + 180 + 60x + 48 + 900/36x^2 + 720/18x+ 16
144x^2 - 1152 x + 4608 +(-960)x + (-768) + 3600/36x^2 + 160x + 64 = 36 x^2 +144x + 180 + 60x + 48 + 900/36x^2 + 40x+ 16
144x^2 - 1152 x + 4608 +(-960)x + (-768) + 100x^2 + 160x + 64 = 36 x^2 +144x + 180 + 60x + 48 + 25x^2 + 40x+ 16
144x^2 - 1152 x + 4608 -960x -768 + 100x^2 + 160x + 64 = 36 x^2 +144x + 180 + 60x + 48 + 25x^2 + 40x+ 16
244x^2 - 1952 x + 3904 = 61 x^2 +244x + 244
183x^2 - 1952 x + 3904 = +244x + 244
183x^2 - 2196 x + 3904 = + 244
183x^2 - 2196 x + 3660 =0
Now we have a which we can solve by . To do so, we will first factor out the of the coefficients. In this case it is 183.
183x^2 - 2196 x + 3660 =0
183 * x^2 - 183* 12 x + 183 * 20 =0
183 (x^2 - 12 x + 20) =0
x^2 - 12 x + 20 = 0
Next, we will rewrite the equation so all terms with x are on one side of the equation and all constants on the other side.
x^2 - 12 x + 20 = 0
⇕
x^2 - 12 x = -20
In a quadratic expression, b is the linear coefficient. For the equation above, we have that b=- 12. Let's now calculate ( b2 )^2.
Next, we will add ( b2 )^2=36 to both sides of our equation. Then, we will factor the on the left-hand side, and solve the equation.
x^2 - 12 x = -20
x^2 - 12 x + 36=- 20+ 36
(x-6)^2=- 20+36
(x-6)^2=16
sqrt((x-6)^2)=sqrt(16)
x-6=± 4
x=6 ± 4
Both x=6+4=10 and x=6-4=2 are solutions of the equation. This means that they both give us valid x-coordinates for point M. To get the corresponding y-coordinates, we will substitute 10 and 2 for x into the formula for LD and solve for y. Let's start with x = 10.
y = -5/6x+-2/3
y = -5/6( 10)+-2/3
y = -50/6+-2/3
y = -50/6+-4/6
y = -54/6
y= -9
Therefore, (10, -9) are valid coordinates for point M. To get the other ones, let's substitute x=2 and again solve for y.
y = -5/6x+-2/3
y = -5/6( 2)+-2/3
y = -10/6+-2/3
y = -10/6+-4/6
y = -14/6
y = -14/6
y = -7/3
The other pair of valid coordinates of M is (2, - 73). Let's show both of these points on a graph with L, D and a line passing through them.
As we can see, one of the points is between L and D, while the other, to the right of both L and D. However, since both of them satisfy our equation, each of them is twice as far from L as from D.
