Question

Find the length of the altitude drawn to the longest side of the triangle formed by the vertices A(4, -6) , B(3, -2), C(5, 2)

Answer 1

12 /√65 = 1.488 is the length of the altitude drawn to the longest side of the triangle formed by the vertices A(4, -6) , B(3, -2), C(5, 2)

Step-by-step explanation:

A(4, -6) , B(3, -2), C(5, 2)

AB = √(3 - 4)² + (-2 -(-6))² = √1 + 36 = √37

AC =√(5 - 4)² + (2 -(-6))² = √1 + 64 = √65

BC = √(3 - 5)² + (-2 -2)² = √4 + 16 = √20

Longest side = AC

Area of Triangle

A(4, -6) , B(3, -2), C(5, 2)

= (1/2) |  4 ( -2 - 2)  + 3(2 -(-6)) + 5 (-6 -(-2)) |

= (1/2) |  -16 + 24 - 20 |

= (1/2) | -12 |

= 12/2

= 6

Area of triangle = (1/2) * AC * Altitude  = 6

=> Altitude  = 12 /√65

= 1.488

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Answer 2

The length of the altitude drawn to the longest side of the triangle is 1.49

Step-by-step explanation:

The vertices of a triangle ABC are

A(4, -6) , B(3, -2), C(5, 2)

The length of altitude on longest side of triangle.

First find the length of each side of triangle using distance formula.

• Distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AB=\sqrt{(4-3)^2+(-6+2)^2}$

$AB=\sqrt{1+16}$

$AB=\sqrt{17}$

$BC=\sqrt{(5-3)^2+(2+2)^2}$

$BC=\sqrt{4+16}$

$BC=\sqrt{20}$

$AC=\sqrt{(5-4)^2+(2+6)^2}$

$AC=\sqrt{1+64}$

$AC=\sqrt{65}$

Longest side is AC.

So, the altitude of a triangle from B on AC

First find the equation of line AC using two point formula of equation of line.

A(4, -6) and C(5, 2)

• Equation of line AC:-

$y-2=\dfrac{-6-2}{4-5}(x-5)$

$y-2=8(x-5)$

$8x-y-38=0$

Now find distance of point B from line AC

using formula point line distance formula

$d=\dfrac{ax_0+by_0+c}{\sqrt{a^2+b^2}}$

where, $B(x_0,y_0)$ and ax+by+c=0 is equation of line AC

$d=\dfrac{8\cdot 3-1\cdot -2-38}{\sqrt{8^2+1^2}}$

$d=\dfrac{24+2-38}{\sqrt{65}}$

$d=\dfrac{12}{\sqrt{65}}$

hence, the length of altitude on longest side of triangle ABC is $\dfrac{12}{\sqrt{65}}\approx 1.49$

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