If (5,3),(4,2) and (1,-2) are the mid points of sides of triangle ABC, then the area of ABC is (

If (5,3),(4,2) and (1,-2) are the mid points of sides of...

If $(5,3),(4,2)$ and $(1,-2)$ are the mid points of sides of triangle $\mathrm{ABC}$, then the area of $\triangle \mathrm{ABC}$ is
(1) 2 sq. units
(2) $3 \mathrm{sq}$. units
(3) $1 \mathrm{sq}$. units
(4) $4 \mathrm{sq}$. units

Key Concepts

Medial Triangle

A medial triangle is formed by connecting the midpoints of the sides of a given triangle. This concept is key because the medial triangle is similar to the original triangle and its area is exactly one-fourth of the area of the original triangle.

Area Scaling Property

The area scaling concept in geometry indicates that if the lengths of a shape are scaled by a factor, the area scales by the square of that factor. In the case of the medial triangle, the sides are exactly one-half the length of the corresponding sides of the original triangle, so the area becomes (1/2)^2, or one-fourth that of the original triangle.

Coordinate Geometry

Coordinate geometry allows us to analyze geometric shapes using algebraic methods, particularly by handling vertices as coordinate pairs. The area of any triangle with given vertices can be computed using the determinant formula, which is especially useful when dealing with midpoints or other points calculated via coordinates.

Transcript

00:01 All those triangles are given here, we have the midpoints are given.

00:03 Our sides here we have, the midpoints here.

00:06 We find out the area of triangle a, b, z.

00:08 So we just now try to solve this.

00:11 So what can do here first find the coordinates of x1, y, 1, x2, y2, and x3, y3, we get the equation here in the midpoint formula.

00:17 That is x1 plus x2 over 2, x1, x1 plus x2 over 2, that equals 5.

00:27 Next we have x2 plus x3 over 2.

00:31 That equals 1.

00:32 Next we have x1 plus x3 over 2 that equals 4 to the midpoint formula here we have next coming out to be y 1 plus y2 over 2 that equals 3 next we have y 2 plus y 3 over 2 that equals 2 we have got this midpoint formula now we've got this equations from me here, you can solve this equation, three equations here, and you can solve this three equation here.

01:09 So we'll get x1 equals h, then x2 equals 2, x3 equals 0, then we get y1 equals 7, y2 equals negative 1 and y3 equals negative 3.

01:26 Now you'll find the area of triangle abc, so area is given as we can use determinant form, data of triangle a bc, that is given as determinant, x1, x2, x3 then y1, y2, y3, then one, one, one, one.

01:42 So it's coming out to be here, we are now.

01:45 Determine is coming out to be.

01:47 That is eight, two, and zero, seven, negative one, and negative three.

01:55 Then we have one, one, one.

01:57 We can now solve this eight times.

02:08 We have this coming out to be negative one, one, negative three, one.

02:12 Then we have negative 7, 2, 1, 0, 1.

02:18 Then we have positive 1 times 2, negative 1, 0, negative 3.

02:24 We're going to solve this here.

02:26 So, 8 times this is negative 1, positive 3...