Find the area of a triangle with vertices (1, 5), (1, -2), (3,-2).

Mathematics

1 answer:

ollegr [7]3 years ago

3 0

Answer:

Step-by-step explanation:

A(3, - 2), B(1, 5), C(1, - 2)

ΔABC is right angled Δ

AC and BC are legs (cathetus)

$A_{ABC}$ = (AC)(BC) ÷ 2 = $\frac{2*7}{2}$ = 7 units²

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Really need help guys!<br> Love you if you can solve them wiz steps!!!!

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Step-by-step explanation:

$x^2+y^2-8x+4y+4=0\\\\a)\\\\x^2-8x+y^2+4y+4=0\\\\x^2-2*x*4+(y^2+2*y*2+2^2)=0\\\\x^2-2*x*4+4^2+(y+2)^2=4^2\\\\(x-4)^2+(y+2)^2=4^2\\$

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$b)\\\\y=0\\x^2+0^2-8x+4*0+4=0\\x^2-8x+4=0\\a=1\ \ \ \ b=-8\ \ \ \ c=4\\D=(-8)^2-4*1*4\\D=64-14\\D=48\\\sqrt{D}=\sqrt{48} \\ \sqrt{D}=\sqrt{16*3} \\\sqrt{D}=\sqrt{4^2*3} \\\sqrt{D}=4\sqrt{3} \\\displaystyle\\x=\frac{-(-8)б4\sqrt{3} }{2*1} \\\\x=\frac{8б4\sqrt{3} }{2} \\x_1=4-2\sqrt{3} \\x_2=4+2\sqrt{3}$

$c)\\\\A(6,2\sqrt{3}-2)\\ (x-4)^2+(y+2)^2=4^2\\(6-4)^2+(2\sqrt{3} -2+2)^2=16\\2^2+(2\sqrt{3} )^2=16\\4+2^2*(\sqrt{3})^2=16\\ 4+4*3=16\\4+12=16\\16\equiv16$