Two sides of a rhombus are along the lines x- | vedclass

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

Question

(C) The equations of the two sides of the rhombus are $L_1: x-y+1=0$ and $L_2: 7x-y-5=0$. Solving these equations,we find the vertex $V_1$ where these

Vedclass · Accepted Answer

$\left(\frac{1}{3}, -\frac{8}{3}\right)$

Vedclass · Answer

(C) The equations of the two sides of the rhombus are $L_1: x-y+1=0$ and $L_2: 7x-y-5=0$. Solving these equations,we find the vertex $V_1$ where these sides intersect: $x-y = -1$ and $7x-y = 5$. Subtracting the first from the second gives $6x = 6$,so $x=1$. Substituting $x=1$ into $x-y+1=0$ gives $y=2$. Thus,$V_1 = (1,2)$. The diagonals of a rhombus bisect each other at the intersection point $P(-1,-2)$. Since $P$ is the midpoint of the diagonal connecting $V_1(1,2)$ and its opposite vertex $V_3(x_3, y_3)$,we have: $-1 = \frac{1+x_3}{2} \Rightarrow x_3 = -3$ and $-2 = \frac{2+y_3}{2} \Rightarrow y_3 = -6$. So,$V_3 = (-3,-6)$. The other diagonal passes through $P(-1,-2)$ and is perpendicular to the diagonal $V_1V_3$. The slope of $V_1V_3$ is $m = \frac{-6-2}{-3-1} = \frac{-8}{-4} = 2$. The slope of the other diagonal is $m' = -\frac{1}{2}$. The equation of the other diagonal is $y - (-2) = -\frac{1}{2}(x - (-1))$,which simplifies to $2y + 4 = -x - 1$,or $x + 2y + 5 = 0$. The other two vertices lie on this diagonal. Checking the options,for $C \left(\frac{1}{3}, -\frac{8}{3}\right)$: $\frac{1}{3} + 2\left(-\frac{8}{3}\right) + 5 = \frac{1-16+15}{3} = 0$. Thus,$\left(\frac{1}{3}, -\frac{8}{3}\right)$ is a vertex.

Two sides of a rhombus are along the lines $x-y+1=0$ and $7x-y-5=0$. If its diagonals intersect at $(-1,-2)$,then one of the vertices of this rhombus is

Similar Questions

If two opposite vertices of a square are $(5, -4)$ and $(-3, 2)$,find its area.

The opposite angular points of a square are $(3, 4)$ and $(1, -1)$. Then the coordinates of the other two points are

The perimeter of a triangle is $16 \text{ cm}$,one of the sides is of length $6 \text{ cm}$. If the area of the triangle is $12 \text{ cm}^2$,then the triangle is:

Let $ABC$ be an isosceles triangle in which $A$ is at $(-1, 0)$,$\angle A = \frac{2\pi}{3}$,$AB = AC$ and $B$ is on the positive $x$-axis. If $BC = 4\sqrt{3}$ and the line $BC$ intersects the line $y = x + 3$ at $(\alpha, \beta)$,then $\frac{\beta^4}{\alpha^2}$ is :

For what value of $k$ are the points $(k, 2 - 2k)$,$(1 - k, 2k)$,and $(-4 - k, 6 - 2k)$ collinear?

Vedclass Test Series

Exam Paper Generator

Online Exam Module