Triangle Geometry Problem Solving Find Equations Lengths And Intersections
Hey guys! Let's dive into a classic geometry problem involving triangles. We've got a triangle ABC with vertices A(7,1), B(9,2), and C(0,3). Our mission is to find several key properties: the equation and length of side AB, the equation and length of altitude CH, the equation and length of median AM, the intersection point N of AM and CH, and finally, the equation of a line passing through a specified point. Buckle up, because we're about to crunch some numbers and explore the fascinating world of coordinate geometry!
a) Equation and Length of Side AB
First, let's tackle finding the equation of the line AB and its length. This involves a couple of steps, but don't worry, we'll break it down. We'll first find the slope, then use the point-slope form to get the equation, and finally, calculate the distance using the distance formula.
Finding the Slope of AB
The slope (m) of a line passing through two points (x1, y1) and (x2, y2) is given by the formula:
m = (y2 - y1) / (x2 - x1)
In our case, A(7,1) and B(9,2) are our points. So, let's plug in the values:
m = (2 - 1) / (9 - 7) = 1 / 2
So, the slope of line AB is 1/2. Got it!
Determining the Equation of AB
Now that we have the slope, we can use the point-slope form of a line equation:
y - y1 = m(x - x1)
We can use either point A or B. Let's use point A(7,1) and our slope m = 1/2:
y - 1 = (1/2)(x - 7)
Let's simplify this equation to get it into slope-intercept form (y = mx + b):
y - 1 = (1/2)x - 7/2 y = (1/2)x - 7/2 + 1 y = (1/2)x - 5/2
To get rid of the fractions, we can multiply the entire equation by 2:
2y = x - 5
We can also write it in the standard form (Ax + By + C = 0):
x - 2y - 5 = 0
So, the equation of side AB is x - 2y - 5 = 0.
Calculating the Length of AB
To find the length of AB, we'll use the distance formula:
d = √((x2 - x1)² + (y2 - y1)²)
Using points A(7,1) and B(9,2):
d = √((9 - 7)² + (2 - 1)²) d = √(2² + 1²) d = √(4 + 1) d = √5
Therefore, the length of side AB is √5 units.
b) Equation and Length of Altitude CH
Next up, let's find the equation and length of altitude CH. Remember, an altitude is a perpendicular line from a vertex to the opposite side. So, CH is perpendicular to AB. This means we'll need to find the slope of CH first, using the fact that perpendicular lines have negative reciprocal slopes. Then, we'll use the point-slope form again to find the equation, and finally, calculate the length.
Finding the Slope of CH
Since CH is perpendicular to AB, its slope (mCH) is the negative reciprocal of the slope of AB (mAB). We found mAB to be 1/2. So:
mCH = -1 / mAB = -1 / (1/2) = -2
The slope of altitude CH is -2.
Determining the Equation of CH
Now we use the point-slope form again. We know the slope mCH = -2 and the altitude CH passes through point C(0,3):
y - y1 = m(x - x1) y - 3 = -2(x - 0) y - 3 = -2x y = -2x + 3
We can also write this in standard form:
2x + y - 3 = 0
So, the equation of altitude CH is 2x + y - 3 = 0.
Calculating the Length of CH
To find the length of CH, we first need to find the coordinates of the point H, which is the intersection of AB and CH. We have the equations for both lines:
AB: x - 2y - 5 = 0 CH: 2x + y - 3 = 0
Let's solve this system of equations. We can multiply the second equation by 2:
4x + 2y - 6 = 0
Now add this to the first equation:
(x - 2y - 5) + (4x + 2y - 6) = 0 5x - 11 = 0 5x = 11 x = 11/5
Now substitute x = 11/5 into the equation for CH:
2(11/5) + y - 3 = 0 22/5 + y - 15/5 = 0 y + 7/5 = 0 y = -7/5
So, the coordinates of point H are (11/5, -7/5).
Now we can use the distance formula to find the length of CH, using points C(0,3) and H(11/5, -7/5):
d = √((x2 - x1)² + (y2 - y1)²) d = √((11/5 - 0)² + (-7/5 - 3)²) d = √((11/5)² + (-7/5 - 15/5)²) d = √((11/5)² + (-22/5)²) d = √(121/25 + 484/25) d = √(605/25) d = √(121/5) d = 11/√5
To rationalize the denominator, we multiply the numerator and denominator by √5:
d = (11√5) / 5
Therefore, the length of altitude CH is (11√5) / 5 units.
c) Equation and Length of Median AM
Let's switch gears and find the equation and length of median AM. A median connects a vertex to the midpoint of the opposite side. So, AM connects vertex A to the midpoint M of BC. This means we need to find the midpoint M first, then find the equation of the line AM, and finally calculate the length.
Finding the Midpoint M of BC
The midpoint M of a line segment with endpoints (x1, y1) and (x2, y2) is given by the midpoint formula:
M = ((x1 + x2) / 2, (y1 + y2) / 2)
For BC, B(9,2) and C(0,3):
M = ((9 + 0) / 2, (2 + 3) / 2) M = (9/2, 5/2)
So, the coordinates of midpoint M are (9/2, 5/2).
Determining the Equation of AM
Now we'll find the equation of the line AM. We have two points: A(7,1) and M(9/2, 5/2). First, let's find the slope:
m = (y2 - y1) / (x2 - x1) m = (5/2 - 1) / (9/2 - 7) m = (5/2 - 2/2) / (9/2 - 14/2) m = (3/2) / (-5/2) m = -3/5
So, the slope of median AM is -3/5.
Now, use the point-slope form with point A(7,1):
y - 1 = (-3/5)(x - 7) y - 1 = (-3/5)x + 21/5 y = (-3/5)x + 21/5 + 1 y = (-3/5)x + 21/5 + 5/5 y = (-3/5)x + 26/5
To get rid of fractions, multiply the entire equation by 5:
5y = -3x + 26
In standard form:
3x + 5y - 26 = 0
Therefore, the equation of median AM is 3x + 5y - 26 = 0.
Calculating the Length of AM
Now, let's find the length of AM using the distance formula with points A(7,1) and M(9/2, 5/2):
d = √((x2 - x1)² + (y2 - y1)²) d = √((9/2 - 7)² + (5/2 - 1)²) d = √((9/2 - 14/2)² + (5/2 - 2/2)²) d = √((-5/2)² + (3/2)²) d = √(25/4 + 9/4) d = √(34/4) d = √(17/2) d = (√34) / √2
To rationalize the denominator, multiply the numerator and denominator by √2:
d = (√34 * √2) / 2 d = √(68) / 2 d = √(4 * 17) / 2 d = 2√17 / 2 d = √17
So, the length of median AM is √17 units.
d) Point N of Intersection of AM and CH
Now, we need to find the coordinates of point N, the intersection of median AM and altitude CH. We already have the equations for both lines:
AM: 3x + 5y - 26 = 0 CH: 2x + y - 3 = 0
Let's solve this system of equations. We can multiply the second equation by -5:
-10x - 5y + 15 = 0
Now add this to the first equation:
(3x + 5y - 26) + (-10x - 5y + 15) = 0 -7x - 11 = 0 -7x = 11 x = -11/7
Substitute x = -11/7 into the equation for CH:
2(-11/7) + y - 3 = 0 -22/7 + y - 21/7 = 0 y - 43/7 = 0 y = 43/7
Therefore, the coordinates of point N are (-11/7, 43/7).
e) Equation of the Line Passing Through...
Oops! It seems like the original problem statement is incomplete. We need a point and either another point or a slope to define a line. However, we've covered a ton of ground already! We've found equations of lines, lengths of sides and altitudes, and even the intersection point of two important lines within the triangle. If we had a specific point and condition (like parallel to another line or perpendicular to another line), we could easily find the equation of that line using the methods we've already discussed.
Conclusion
We've successfully tackled a comprehensive triangle geometry problem! We found equations and lengths of sides, altitudes, and medians. We even pinpointed the intersection of the median and altitude. Geometry problems like these are a fantastic way to sharpen your skills in algebra and coordinate geometry. Keep practicing, and you'll become a geometry whiz in no time! Remember, the key is to break down the problem into smaller, manageable steps, and use the right formulas and concepts. You got this!
