Given two points A(3,1) and B(-12,-1), determine a third point P(x,y) such that the slopes of AP and BP are -2/3 and 2/3 respectively.
Thank you! Now I've got (-3,5)
To find the coordinates of point P (x, y) such that the slopes of AP and BP are -2/3 and 2/3 respectively, follow these steps:
Step 1: Find the slope of line AB
The slope is given by the formula:
slope = (y2 - y1) / (x2 - x1)
Let A(3, 1) be (x1, y1) and B(-12, -1) be (x2, y2):
slope_AB = (y2 - y1) / (x2 - x1)
= (-1 - 1) / (-12 - 3)
= -2 / -15
= 2 / 15
Step 2: Find the equation of line AB using the point-slope form
The equation of a line is given by the formula:
y - y1 = m(x - x1)
Using point A(3, 1) and slope 2/15, the equation of line AB is:
y - 1 = (2/15)(x - 3)
Step 3: Determine the equations of lines AP and BP
Since the slopes of AP and BP are -2/3 and 2/3 respectively, the equations of these lines can be written as follows:
AP: y - 1 = (-2/3)(x - 3)
BP: y - 1 = (2/3)(x - 3)
Step 4: Solve the system of equations by setting the two equations equal to each other
(-2/3)(x - 3) = (2/3)(x - 3)
Step 5: Solve for x
Multiply both sides of the equation by 3 to eliminate the fractions:
-2(x - 3) = 2(x - 3)
Expand and simplify:
-2x + 6 = 2x - 6
Add 2x to both sides:
6x + 6 = 6
Subtract 6 from both sides:
6x = 0
Divide by 6:
x = 0
Step 6: Find the value of y
Substitute the value of x = 0 in any of the two equations of AP or BP:
Let's use AP:
y - 1 = (-2/3)(0 - 3)
y - 1 = (-2/3)(-3)
y - 1 = 2
Add 1 to both sides:
y = 2 + 1
y = 3
The coordinates of point P (x, y) are (0, 3).
To find the coordinates of point P(x, y), we can use the concept of slope.
The slope of a line passing through two points (x1, y1) and (x2, y2) is given by the formula:
slope = (y2 - y1) / (x2 - x1)
Let's first find the slopes of line AP and line BP:
Slope of AP = -2/3
Slope of BP = 2/3
Now, let's find the slope of AB:
Slope of AB = (y2 - y1) / (x2 - x1)
= (-1 - 1) / (-12 - 3)
= -2 / -15
= 2/15
Since line AP has a slope of -2/3, we have:
-2/3 = (y - 1) / (x - 3)
Cross-multiplying:
-2(x - 3) = 3(y - 1)
-2x + 6 = 3y - 3
3y = -2x + 9
y = (-2/3)x + 3
Similarly, since line BP has a slope of 2/3, we have:
2/3 = (y - (-1)) / (x - (-12))
Cross-multiplying:
2(x + 12) = 3(y + 1)
2x + 24 = 3y + 3
3y = 2x + 21
y = (2/3)x + 7
To find the common point P(x, y) for both equations, we can set them equal to each other:
(-2/3)x + 3 = (2/3)x + 7
Adding (2/3)x to both sides:
(4/3)x + (-2/3)x = 7 - 3
(2/3)x = 4
x = (3/2) * 4
x = 6
Substituting x = 6 into either equation (let's use (-2/3)x + 3 = (2/3)x + 7), we can find y:
(-2/3) * 6 + 3 = (2/3) * 6 + 7
-4 + 3 = 4 + 7
-1 = 11
As -1 does not equal 11, we can conclude that there is no solution to this system of equations.
slope AP = (y-1)/(x-3) = -2/3
3y-3 = -2x + 6
2x + 3y = 9
slope BP = (y+1)/(x+12) = 2/3
2x + 24 = 3y + 3
2x - 3y = -21
Solve the two equations, adding them would be the simplest way.