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Based on the calculations, the coordinates of the point of congruency of the altitudes (H) are (-160/11, 40/11).
A triangle can be defined as a two-dimensional geometric shape that comprises three (3) sides, three (3) vertices and three (3) angles only.
A slope is also referred to as gradient and it's typically used to describe both the ratio, direction and steepness of the function of a straight line.
Mathematically, the slope of a straight line can be calculated by using this formula;
[tex]Slope, m = \frac{Change\;in\;y\;axis}{Change\;in\;x\;axis}\\\\Slope, m = \frac{y_2\;-\;y_1}{x_2\;-\;x_1}[/tex]
Assuming the following parameters for triangle ABC:
For the slope of BC, we have:
Slope of BC = (2 - 8)/(4 - 0)
Slope of BC = -6/4
Slope of BC = -3/2.
For the slope of CA, we have:
Slope of CA = (2 - 0)/(4 - (-2))
Slope of CA = 2/6
Slope of CA = 1/3.
For the slope of AB, we have:
Slope of AB = (8 - 0)/(0 - (-2))
Slope of AB = 8/2
Slope of AB = 4.
Note: The point of concurrency of three altitudes in a triangle is referred to as orthocenter.
Since side AB is perpendicular to side QC, we have:
m₁ × m₂ = -1
Slope of AB × Slope of QC = -1
Slope of QC = (k - 4)/(h - 2)
4 × (k - 4)/(h - 2) = -1
(4k - 16)/(h - 2) = -1
4k - 16 = -h + 2
4k + h = 18 .......equation 1.
Similarly, we have the following:
Slope of BC × Slope of AH = -1
-3/2 × (k)/(h + 2) = -1
3k/(2h + 4) = 1
3k = 2h + 4
3k - 2h = 4 .......equation 2.
Solving eqn. 1 and eqn. 2 simultaneously, we have:
8k + 2h = 36
3k - 2h = 4
11k = 40
k = 40/11.
For the value of h, we have:
h = -4k
h = -4 × (40/11)
h = -160/11
Therefore, the coordinates of the point of congruency of the altitudes (H) are (-160/11, 40/11).
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