Let a neq 0, b neq 0, c be three real numbers | vedclass

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

Question

(B) Given $L(p, q) = \frac{ap + bq + c}{\sqrt{a^2 + b^2}}$.Substituting the given values:$L\left(\frac{2}{3}, \frac{1}{3}\right) = \frac{a(\frac{2}{3}

Vedclass · Accepted Answer

$(1, 1)$

Vedclass · Answer

(B) Given $L(p, q) = \frac{ap + bq + c}{\sqrt{a^2 + b^2}}$.Substituting the given values:$L\left(\frac{2}{3}, \frac{1}{3}\right) = \frac{a(\frac{2}{3}) + b(\frac{1}{3}) + c}{\sqrt{a^2 + b^2}} = \frac{2a + b + 3c}{3\sqrt{a^2 + b^2}}$$L\left(\frac{1}{3}, \frac{2}{3}\right) = \frac{a(\frac{1}{3}) + b(\frac{2}{3}) + c}{\sqrt{a^2 + b^2}} = \frac{a + 2b + 3c}{3\sqrt{a^2 + b^2}}$$L(2, 2) = \frac{a(2) + b(2) + c}{\sqrt{a^2 + b^2}} = \frac{2a + 2b + c}{\sqrt{a^2 + b^2}} = \frac{6a + 6b + 3c}{3\sqrt{a^2 + b^2}}$Summing these values:$\frac{2a + b + 3c + a + 2b + 3c + 6a + 6b + 3c}{3\sqrt{a^2 + b^2}} = 0$$\frac{9a + 9b + 9c}{3\sqrt{a^2 + b^2}} = 0$$9(a + b + c) = 0 \implies a + b + c = 0$This implies that the line $ax + by + c = 0$ satisfies the condition $a(1) + b(1) + c = 0$ for the point $(1, 1)$.Thus,the line always passes through $(1, 1)$.

Similar Questions

$A(-4,0)$ and $B(4,0)$ are two fixed points. $C$ and $D$ are two points on the $Y$-axis such that $CD=4$ and $C$ is a point below $D$. Then the locus of the point of intersection of the lines $AC$ and $BD$ is

$A (a, 0)$ and $B (-a, 0)$ are two fixed points of $\Delta ABC$. If the vertex $C$ moves such that $\cot A + \cot B = \lambda$,where $\lambda$ is a constant,then what is the locus of point $C$?

$A$ line segment joining a point $A$ on $x$-axis to a point $B$ on $y$-axis is such that $AB=15$. If $P$ is a point on $AB$ such that $\frac{AP}{PB}=\frac{2}{3}$,then the locus of $P$ is:

$A(a, 0)$ and $B(-a, 0)$ are two fixed points. If $\angle A - \angle B = \theta$,what is the locus of point $C$ of triangle $ABC$?

The locus of a point which moves such that the area of the triangle formed by it with the vertices $(1, 2)$ and $(-2, 5)$ is $8$ sq. units is/are

Vedclass Test Series

Exam Paper Generator

Online Exam Module