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3 years ago

The perimeter of the triangle

Mathematics

1 answer:

Illusion [34]3 years ago

8 0

Area=228 feet sqa.
Length=12 ft.
Breadth=(228/12) ft.
=19 ft.
Perimeter=2(12+19) ft.
=2×31 ft.
=62 ft.

HOPE IT HELPS UH!!☺️☺️

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Let D be the mid point of side BC, [B(2, - 1), C(5, 2)].

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4 (a)

Equation of line AB[A(2, 1), B(-2, - 11)] in two point form is given as:

$\frac{y-y_1}{y_1-y_2} =\frac{x-x_1}{x_1 - x_2} \\\\\therefore \frac{y-1}{1-(-11)} =\frac{x-2}{2 - (-2) } \\\\\therefore \frac{y-1}{1+11} =\frac{x-2}{2 +2} \\\\\therefore \frac{y-1}{12} =\frac{x-2}{4} \\\\\therefore \frac{y-1}{3} =\frac{x-2}{1} \\\\\therefore y-1= 3(x - 2)\\\\\therefore y= 3x - 6+1\\\\\therefore y= 3x - 5\\\\ \huge \purple {\boxed {\therefore 3x - y-5=0}} \\$

is the equation of line AB.

Now we have to check whether C(4, 7) lie on line AB or not.

Let us substitute x = 4 & y = 7 on the Left hand side of equation of line AB and if it gives us 0, then C lies on the line.

$LHS = 3x - y-5\\=3\times 4-7-5\\= 12-12\\=0\\= RHS$

Hence, point C (4, 7) lie on the straight line AB.

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Like we did in 4(a), first find the equation of line AB and then substitute the coordinates of point C in equation and if they satisfy the equation, then all the three points lie on the straight line.

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