A square is formed by the lines x=0, y=0, x=1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The vertices of the square are $O(0,0), A(1,0), B(1,1),$ and $C(0,1)$.The diagonals are the line segments connecting opposite vertices,which are $

Vedclass · Accepted Answer

$y=x, x+y=1$

Vedclass · Answer

(C) The vertices of the square are $O(0,0), A(1,0), B(1,1),$ and $C(0,1)$.The diagonals are the line segments connecting opposite vertices,which are $OB$ and $AC$.$1$. Equation of diagonal $OB$ passing through $(0,0)$ and $(1,1)$:The slope $m = \frac{1-0}{1-0} = 1$.Using the point-slope form $y - y_1 = m(x - x_1)$:$y - 0 = 1(x - 0) \Rightarrow y = x$.$2$. Equation of diagonal $AC$ passing through $(1,0)$ and $(0,1)$:The slope $m = \frac{1-0}{0-1} = -1$.Using the point-slope form $y - 0 = -1(x - 1)$:$y = -x + 1 \Rightarrow x + y = 1$.Thus,the equations of the diagonals are $y = x$ and $x + y = 1$.

$A$ square is formed by the lines $x=0, y=0, x=1, y=1$. Then,the equations of its diagonals will be

Similar Questions

If the vertices of a parallelogram are $(0, 0)$,$(1, 0)$,$(2, 2)$,and $(1, 2)$,then the angle between its diagonals is:

$A$ rectangle $ABCD$ has its side parallel to the line $y=2x$ and vertices $A, B, D$ are on lines $y=1, x=1$ and $x=-1$ respectively. The coordinate of $C$ can be

In the given figure,$BC = AC$,$\angle AFD = 40^{\circ}$ and $CE = CD$. Then the value of $\angle BCE$ is equal to ......$^{\circ}$

If $(0, 1), (1, 1)$ and $(1, 0)$ are the midpoints of the sides of a triangle,find its incentre.

The set of values that $\beta$ can assume so that the point $(0, \beta)$ lies on or inside the triangle formed by the lines $3x+y+2=0$,$2x-3y+5=0$,and $x+4y-14=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module