The above figure is a square. Find the value of $x_{1} + y_{1} + x_{2} + y_{2} + x_{3} + y_{3}$ .

Open in App

Solution

We know that,

Y coordinates below x-axis have negative value whereas above x-axis they have positive value.

So, B = (a, -a), C = (-a, -a) and D = (-a, a)
The value of $x_{1} + y_{1} + x_{2} + y_{2} + x_{3} + y_{3}$ = $a + (- a) + (- a) + (- a) + (- a) + a$ = $- 2 a$

Suggest Corrections

Similar questions

The above figure is a square. The value of $x_{1} + y_{1} + x_{2} + y_{2} + x_{3} + y_{3}$ is

Find the centroid of the triangle with vertices A $(x_{1}, y_{1})$ , B $(x_{2}, y_{2})$ and C $(x_{3}, y_{3})$ .

Q. If three points

(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

lie on the same line, prove that

\frac{y_{2} - y_{3}}{x_{2} x_{3}} + \frac{y_{3} - y_{1}}{x_{3} x_{1}} + \frac{y_{1} - y_{2}}{x_{1} x_{2}} = 0

ABCD is a parallelogram with vertices A (X₁, Y₁) , B(X₂, Y₂) and C (X₃, Y₃). Then the coordinates of the fourth vertex D in terms of the coordinates of A, B and C are

Q. If

A (x_{1}, y_{1}), B x_{2}, y_{2}), C (x_{3}, y_{3})

are the vertices of an equilateral triangle and

(x_{1} - 2)^{2} + (y_{1} - 3)^{2} = (x_{2} - 2)^{2} + (y_{2} - 3)^{2} = (x_{3} - 2)^{2} + (y_{3} - 3)^{2}

, then

Join BYJU'S Learning Program

The above figure is a square. Find the value of x1+y1+x2+y2+x3+y3 .

We know that, Y coordinates below x-axis have negative value whereas above x-axis they have positive value. So, B = (a, -a), C = (-a, -a) and D = (-a, a) The value of x1+y1+x2+y2+x3+y3 = a+(−a)+(−a)+(−a)+(−a)+a = −2a

The above figure is a square. Find the value of $x_{1} + y_{1} + x_{2} + y_{2} + x_{3} + y_{3}$ .

We know that,

Y coordinates below x-axis have negative value whereas above x-axis they have positive value.

So, B = (a, -a), C = (-a, -a) and D = (-a, a)
The value of $x_{1} + y_{1} + x_{2} + y_{2} + x_{3} + y_{3}$ = $a + (- a) + (- a) + (- a) + (- a) + a$ = $- 2 a$