Home › UP Board Class 10 › Mathematics › The midpoint of the line segment joining $(-6,10…
The midpoint of the line segment joining $(-6,10)$ and $(3,-8)$ is
A$\left(-\dfrac{3}{2},1\right)$
B$\left(-\dfrac{3}{2},2\right)$
C$( -3,1)$
D$\left(\dfrac{3}{2},1\right)$
Answer & Solution
Correct answer: A. $\left(-\dfrac{3}{2},1\right)$
The midpoint formula gives $\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$. So the midpoint is $\left(\dfrac{-6+3}{2},\dfrac{10+(-8)}{2}\right)=\left(-\dfrac{3}{2},1\right)$.
Related questions
What is the derivative of $x^x$ for $x>0$?If $y=x^{\tan^{-1}x}$, then $\dfrac{dy}{dx}$ equalsThe derivative of $ in^{-1}x$ isThe derivative of $\cos^{-1}x$ isWhat is $\dfrac{d}{dx}(\tan^{-1}x)$?If $y=\dfrac{1}{x^n}$, then $\dfrac{dy}{dx}$ isWhich identity is correct?For the expression $ qrt{1-x^2}$, a useful trigonometric substitution is