A triangle has corners A, B, and C located at #(1 ,7 )#, #(7 ,4 )#, and #(2 , 8 )#, respectively. What are the endpoints and length of the altitude going through corner C?

Answer 1

The end points are (2, 8) and (4.6, 5.2), The length is #a ~~ 3.82#

Going from point A to B we go right 6 and down 3, therefore, the slope is #-1/2#

The line connecting points A and B has the following equation:

#y - 7 = -1/2(x -1)#
#y - 14/2 = -1/2x + 1/2#
#y = 15/2 - 1/2x#
The altitude through point C will perpendicular to the above line, thereby, making its slope the negative reciprocal of #-1/2#. The slope is 2.

The line through point C has the following equation:

#y - 8 = 2(x - 2)#
#y - 8 = 2x - 4#
#y = 2x + 4#

After equating the two equations' right sides, find the x coordinate where they intersect:

#15/2 - 1/2x = 2x + 4#
#15/2 - 1/2x = 4/2x + 8/2#
#23/2 = 5/2x#
#x = 23/5#
#x = 4.6#
#y = 15/2 - 1/2(4.6)#
#y = 5.2#

The altitude's length, a, is:

#a = sqrt((4.6 - 2)² + (5.2 - 8)²)#
#a ~~ 3.82#
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Answer 2

The altitude of a triangle going through corner C is perpendicular to the side AB. To find the endpoints of this altitude, you can find the equation of the line passing through A and B (the base of the triangle) and then find the intersection point with the line perpendicular to AB passing through C.

  1. Find the slope of the line passing through A and B. [ \text{Slope} = \frac{y_B - y_A}{x_B - x_A} = \frac{4 - 7}{7 - 1} = -\frac{3}{6} = -\frac{1}{2} ]

  2. The slope of the altitude is the negative reciprocal of the slope of AB, which is ( \frac{1}{2} ).

  3. Using the point-slope form of a line with point C, find the equation of the altitude. [ y - y_C = \frac{1}{2}(x - x_C) ] Substituting ( (x_C, y_C) = (2, 8) ): [ y - 8 = \frac{1}{2}(x - 2) ] [ y = \frac{1}{2}x + 7 ]

  4. To find the endpoints of the altitude, you need to find where it intersects with the line AB. Set the equations of the altitude and AB equal to each other and solve for ( x ). [ \frac{1}{2}x + 7 = -\frac{1}{2}x + b ] Substitute the coordinates of point A: [ \frac{1}{2}(1) + 7 = -\frac{1}{2}(1) + b ] [ \frac{1}{2} + 7 = -\frac{1}{2} + b ] [ b = \frac{15}{2} ]

So, the equation of the altitude is ( y = \frac{1}{2}x + \frac{15}{2} ). To find the endpoints, you can set ( y ) equal to 0 (since the altitude intersects the x-axis). [ 0 = \frac{1}{2}x + \frac{15}{2} ] [ x = -15 ]

Plug ( x = -15 ) into the equation of the altitude to find ( y ). [ y = \frac{1}{2}(-15) + \frac{15}{2} ] [ y = -\frac{15}{2} + \frac{15}{2} ] [ y = 0 ]

So, one endpoint is ( (-15, 0) ). Now, find the other endpoint by setting ( y = 0 ) and solving for ( x ). [ 0 = \frac{1}{2}x + \frac{15}{2} ] [ x = -15 ]

The other endpoint is ( (-15, 0) ).

To find the length of the altitude, you can use the distance formula between the two endpoints: [ \text{Length} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ] [ \text{Length} = \sqrt{((-15) - (-15))^2 + (0 - 0)^2} ] [ \text{Length} = \sqrt{0 + 0} ] [ \text{Length} = \sqrt{0} ] [ \text{Length} = 0 ]

So, the length of the altitude going through corner C is 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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