Question

the vertices of Δ ABC are A ( 0, 6 ), B ( 8, 0 ) and C ( 5, 8 ).

If CD ⊥ AB, then find the length of altitude CD.​

Answer 1

Answer:

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Answer 2

The length of altitude CD = 4.6 unit

Given :

• The vertices of Δ ABC are A(0,6) , B(8,0) and C(5,8)

• CD ⊥ AB

To find :

The length of altitude CD

Formula Used :

1. For the given two points ( x₁ , y₁) & (x₂ , y₂)

The equation of the line passing through the points is

$\displaystyle \sf{ \frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1} }$

2. The perpendicular distance from a point ( x₁ , y₁) to the line ax + by + c = 0 is

$\displaystyle \sf = \bigg| \frac{ax_1 + by_1 + c}{ \sqrt{ {a}^{2} + {b}^{2} } } \bigg| \: \: \: unit$

Solution :

Step 1 of 2 :

Find the equation of the line AB

Here the given points are A(0,6) , B(8,0)

So the equation of the line AB passing through the points A(0,6) , B(8,0) is

$\displaystyle \sf \frac{y - 6}{x - 0} = \frac{0 - 6}{8 - 0}$

$\displaystyle \sf{ \implies }\frac{y - 6}{x } = \frac{ - 6}{8 }$

$\displaystyle \sf{ \implies }\frac{y - 6}{x } = - \frac{3}{4 }$

$\displaystyle \sf{ \implies } - 3x = 4y - 24$

$\displaystyle \sf{ \implies }3x + 4y - 24 = 0$

Step 2 of 2 :

Find length of altitude CD

The coordinates of the point C is (5,8)

Hence the length of altitude CD

$\displaystyle \sf = \bigg| \frac{(3 \times 5)+(4 \times 8) - 24}{ \sqrt{ {3}^{2} + {4}^{2} } } \bigg| \: \: \: unit$

$\displaystyle \sf = \bigg| \frac{15+32 - 24}{ \sqrt{ 9 + 16} } \bigg| \: \: \: unit$

$\displaystyle \sf = \bigg| \frac{23}{ \sqrt{ 25} } \bigg| \: \: \: unit$

$\displaystyle \sf = \bigg| \frac{23}{ 5 } \bigg| \: \: \: unit$

$\displaystyle \sf = 4.6\: \: \: unit$

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