Question

#The extremities of the diagonal of a square are (1,1 )and (-2, -1).Obtain the two other vertices and the equation of a next diagonal.

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find the area of the triangle with vertices(1,2),(2,3),and(4,5).what inference can you draw about the points from your upshot?​

Answer 1

(i) (-3/2, 3/2) and (1/2, - 3/2)

6x + 4y = - 3

(ii) Points are col linear

Explanation:

Let the coordinates of the point(lower) be (a, b) and that of upper be (x, y).

Using mid point formula,

Mid point of diagonal(using the given points) is:

= ( (-2+1)/2 , (-1+1)/2) = (-½ , 0)

Using (a, b) and (x, y) :

= ( (a+x)/2 , (b+y)/2 ) = (-1/2, 0)

Which means, a+x= -1 & b + y = 0

Therefore, x = - 1 - a & b = - y

Using distance formula, diagonal = √(-2-1)² + (-1-1)² = √13

Knowing the relation in diagonal and side, side = diagonal/√2 = √13/√2 = √(13/2)

Again using distance formula, diagonal = √13

=> √(x - a)² + (y - b)² = √13

=> (-1-a -a)² + (-b - b)² = 13

=> a² + b² + a = 3 ... (1) [solved directly]

Length of side = √(13/2)

=> √(a-(-2))² + (b-(-1))² = √(13/2)

=> a² + b² + 4a + 2b = 3/2 ...(2)

On solving (1) and (2):

a = 1/2 or -3/2, but a lies in 4th quadrant so a > 0, thus, a = 1/2

b = -3/2 or 3/2, but b lies in 4th quadrant so b < 0, b = -3/2

Therefore,

x = - 1 - a = -1 - 1/2 = -3/2

y = - b = - (-3/2) = 3/2

Vertices are (x, y) = (-3/2, 3/2) and (a, b) = (1/2, -3/2)

Equation is just a relation in y and x, for a relation:

Subtract (1) from (2):

3a + 2b = -3/2 => 6a + 4b = - 3

By merely replacing a, b by x, y

6x + 4y = -3 is the required equation

(ii):

Ar. of ∆ = ½ | 1(3 - 5) + 2(5 - 2) + 4(2 - 3) |

Ar. of ∆ = ½ | -2 + 6 - 4 | = 0

As the area of the ∆ is 0, the given points don't form ∆, they are col linear.