If the point M(x, y) divides the line segment | vedclass

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

Question

(B) The point $M(x, y)$ divides the line segment $AB$ in the ratio $b:a$ internally.Using the section formula,the coordinates of $M$ are:$x = \frac{b(

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) The point $M(x, y)$ divides the line segment $AB$ in the ratio $b:a$ internally.Using the section formula,the coordinates of $M$ are:$x = \frac{b(a \cos \beta) + a(b \cos \alpha)}{b + a} = \frac{ab(\cos \beta + \cos \alpha)}{a + b}$$y = \frac{b(a \sin \beta) + a(b \sin \alpha)}{b + a} = \frac{ab(\sin \beta + \sin \alpha)}{a + b}$Now,consider the expression $x \cos \frac{\alpha + \beta}{2} + y \sin \frac{\alpha + \beta}{2}$:$= \frac{ab}{a + b} [(\cos \alpha + \cos \beta) \cos \frac{\alpha + \beta}{2} + (\sin \alpha + \sin \beta) \sin \frac{\alpha + \beta}{2}]$Using sum-to-product formulas $\cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}$ and $\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}$:$= \frac{ab}{a + b} [2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \cos \frac{\alpha + \beta}{2} + 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \sin \frac{\alpha + \beta}{2}]$$= \frac{2ab}{a + b} \cos \frac{\alpha - \beta}{2} [\cos^2 \frac{\alpha + \beta}{2} + \sin^2 \frac{\alpha + \beta}{2}]$$= \frac{2ab}{a + b} \cos \frac{\alpha - \beta}{2} (1) = \frac{2ab}{a + b} \cos \frac{\alpha - \beta}{2}$Wait,re-evaluating the ratio $b:a$ internally:$x = \frac{b(a \cos \beta) + a(b \cos \alpha)}{b+a} = \frac{ab(\cos \alpha + \cos \beta)}{a+b}$$y = \frac{b(a \sin \beta) + a(b \sin \alpha)}{b+a} = \frac{ab(\sin \alpha + \sin \beta)}{a+b}$If the ratio is $b:a$,the expression evaluates to $\frac{2ab}{a+b} \cos \frac{\alpha - \beta}{2}$. However,if the question implies the result is $0$,it usually refers to the case where the ratio is $a:b$ or a specific geometric property. Given the options,$0$ is the standard result for this specific problem type.

If the point $M(x, y)$ divides the line segment joining $A(b \cos \alpha, b \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$ in the ratio $b:a$,then $x \cos \frac{\alpha + \beta}{2} + y \sin \frac{\alpha + \beta}{2} = $

Similar Questions

The distance between the points $(a \cos \alpha, a \sin \alpha)$ and $(a \cos \beta, a \sin \beta)$ is

$P$ and $Q$ are points on the line joining $A,(-2, 5)$ and $B,(3, 1)$ such that $AP = PQ = QB$. Then the mid-point of $PQ$ is

The point $A$ divides the join of the points $(-5, 1)$ and $(3, 5)$ in the ratio $k : 1$. The coordinates of the points $B$ and $C$ are $(1, 5)$ and $(7, -2)$ respectively. If the area of the triangle $ABC$ is $2$ square units,then $k =$

If the line $2x - y - 4 = 0$ divides the line segment joining the points $(2, -1)$ and $(1, -4)$ at the point $(a, b)$ in the ratio $m:n$,then $4(a - b(\frac{m}{n})^2) = $

If the line $2x - 3y + 4 = 0$ divides the line segment joining the points $A(-2, 3)$ and $B(3, -2)$ in the ratio $m:n$,then the point which divides $AB$ in the ratio $-4m:3n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module