The midpoint of the diagonal of a rectangle f | vedclass

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

Question

(C) The given equations are $x^2+5x-6=0$ and $y^2-8y-20=0$.Solving for $x$:$x^2+6x-x-6=0$ $\Rightarrow (x+6)(x-1)=0$ $\Rightarrow x_1=-6, x_2=1$.Solvi

Vedclass · Accepted Answer

$\left(\frac{-5}{2}, 4\right)$

Vedclass · Answer

(C) The given equations are $x^2+5x-6=0$ and $y^2-8y-20=0$.Solving for $x$:$x^2+6x-x-6=0$ $\Rightarrow (x+6)(x-1)=0$ $\Rightarrow x_1=-6, x_2=1$.Solving for $y$:$y^2-10y+2y-20=0$ $\Rightarrow (y-10)(y+2)=0$ $\Rightarrow y_1=10, y_2=-2$.The vertices of the rectangle are formed by the intersection of these lines: $(-6, 10), (1, 10), (1, -2), (-6, -2)$.The midpoint of the diagonal is the midpoint of the segment connecting opposite vertices,such as $(-6, 10)$ and $(1, -2)$.Midpoint $= \left(\frac{-6+1}{2}, \frac{10-2}{2}\right) = \left(\frac{-5}{2}, \frac{8}{2}\right) = \left(\frac{-5}{2}, 4\right)$.

The midpoint of the diagonal of a rectangle formed by $x^2+5x-6=0$ and $y^2-8y-20=0$ is

Similar Questions

If $p$ and $q$ are distinct prime numbers and the equation $x^2 - px + q = 0$ has positive integers as its roots,then the roots of the equation are:

The equation $4^{(x^2 + 2)} - 9 \cdot 2^{(x^2 + 2)} + 8 = 0$ has the solution:

If $\frac{k}{kx+3}+\frac{3}{3x-k}=\frac{12x+5}{(kx+3)(3x-k)}$ for all $x \in R - \{-\frac{3}{k}, \frac{k}{3}\}$,then both the roots of the equation $kx^2-7x+3=0$ are

If the product of roots of the equation ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$ is $7$,then its roots will be real when:

If the roots of the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ are equal in magnitude but opposite in sign,then $r = ......$

Vedclass Test Series

Exam Paper Generator

Online Exam Module