Two consecutive sides of a parallelogram are | vedclass

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

Question

(C) Let the parallelogram be $OABC$ with vertex $O$ at the origin $(0,0)$ because both side equations pass through the origin.Let the sides be $OA: 4x

Vedclass · Accepted Answer

$x - y = 0$

Vedclass · Answer

(C) Let the parallelogram be $OABC$ with vertex $O$ at the origin $(0,0)$ because both side equations pass through the origin.Let the sides be $OA: 4x + 5y = 0$ and $OC: 7x + 2y = 0$.The diagonal $AC$ is given by $11x + 7y = 9$.To find vertex $A$,solve $OA$ and $AC$:$4x + 5y = 0 \implies x = -\frac{5}{4}y$.Substitute into $11x + 7y = 9$: $11(-\frac{5}{4}y) + 7y = 9 \implies -\frac{55}{4}y + \frac{28}{4}y = 9 \implies -\frac{27}{4}y = 9 \implies y = -\frac{36}{27} = -\frac{4}{3}$.Then $x = -\frac{5}{4}(-\frac{4}{3}) = \frac{5}{3}$. So $A = (\frac{5}{3}, -\frac{4}{3})$.To find vertex $C$,solve $OC$ and $AC$:$7x + 2y = 0 \implies y = -\frac{7}{2}x$.Substitute into $11x + 7y = 9$: $11x + 7(-\frac{7}{2}x) = 9 \implies \frac{22x - 49x}{2} = 9 \implies -27x = 18 \implies x = -\frac{18}{27} = -\frac{2}{3}$.Then $y = -\frac{7}{2}(-\frac{2}{3}) = \frac{7}{3}$. So $C = (-\frac{2}{3}, \frac{7}{3})$.The midpoint $M$ of diagonal $AC$ is $(\frac{5/3 - 2/3}{2}, \frac{-4/3 + 7/3}{2}) = (\frac{1/3}{2}, \frac{3/3}{2}) = (\frac{1}{6}, \frac{1}{2})$.The other diagonal passes through the origin $O(0,0)$ and $M(\frac{1}{6}, \frac{1}{2})$.The equation is $y - 0 = \frac{1/2 - 0}{1/6 - 0}(x - 0) \implies y = \frac{1/2}{1/6}x \implies y = 3x \implies 3x - y = 0$. Wait,re-evaluating the midpoint calculation: $M = (\frac{1}{2}, \frac{1}{2})$.Equation: $y - 0 = \frac{1/2 - 0}{1/2 - 0}(x - 0) \implies y = x \implies x - y = 0$.

Two consecutive sides of a parallelogram are $4x + 5y = 0$ and $7x + 2y = 0$. If the equation to one diagonal is $11x + 7y = 9$,then the equation to the other diagonal is:

Similar Questions

If the lines $3x + y - 4 = 0$,$x - \alpha y + 10 = 0$,$\beta x + 2y + 4 = 0$,and $3x + y + k = 0$ represent the sides of a square,then $\alpha \beta (k + 4)^2 = $

The area enclosed by the graphs of $|x + y| = 2$ and $|x| = 1$ is

The coordinates of the feet of the perpendiculars from the vertices of a triangle to the opposite sides are $(20, 25)$,$(8, 16)$,and $(8, 9)$. Find the orthocenter of the triangle.

If $ABCD$ is a quadrilateral,and the midpoints of consecutive sides $AB, BC, CD$,and $DA$ are joined by straight lines to form a quadrilateral $PQRS$,then $PQRS$ is always:

The straight lines $x+y=0$, $5x+y=4$, and $x+5y=4$ form

Vedclass Test Series

Exam Paper Generator

Online Exam Module