Name: Condition for Concurrency of Three Lines
Uploaded: 2019-07-19
Duration: 10 min 26 s
Description: Well, this video beautifully explains the conditions for concurrency of three lines with some interesting examples. Come, let's learn and watch the video...!!

Question

A

is a variable point on

x

-axis and

B (0, b)

is a fixed point. An equilateral triangle

A B C

is completed with vertex

C

away from origin. If the locus of the point

C

is

α x + β y = b,

then

α^{2} + β^{2}

is

Accepted Answer

Let coordinates of point $x$ on $B (0, b)$ axis be $(a, 0)$ and $B (0, b)$ is a fixed point.

Given that an equilateral triangle $A B C$ is completed with vertex $C$ away from origin, so let coordinates of point $C$ be $(x, y)$ .

Now, $A B = \sqrt{a^{2} + b^{2}}$

Let $D$ be the midpoint of $A B$ , then $D (\frac{a}{2}, \frac{b}{2})$ .

$A D = \sqrt{{(a - \frac{a}{2})}^{2} + {(0 - \frac{b}{2})}^{2}} = \frac{1}{2} \sqrt{a^{2} + b^{2}}$

Since $A B C$ is an equilateral triangle, then $A B = A C = B C$

In $A$ ,

${(A C)}^{2} = {(A D)}^{2} + {(C D)}^{2}$

$\Rightarrow {(C D)}^{2} = {(A C)}^{2} - {(A D)}^{2} = {(A B)}^{2} - {(A D)}^{2} = a^{2} + b^{2} - (\frac{a^{2} + b^{2}}{4}) = \frac{3}{4} (a^{2} + b^{2})$

$\Rightarrow C D = \frac{\sqrt{3}}{2} \sqrt{a^{2} + b^{2}}$

Let slope of line $A B$ $= m_{1} = \frac{0 - b}{a - 0} = - \frac{b}{a}$

So slope of line $C D$ $= - \frac{1}{m_{1}} = \frac{a}{b} = \tan θ$

Let $C D = r$ , $\cos θ = \frac{b}{\sqrt{a^{2} + b^{2}}} & \sin θ = \frac{a}{\sqrt{a^{2} + b^{2}}}$

So, coordinates of point $C$ is given by equation of straight line in parametric form :

$\frac{x - \frac{a}{2}}{\cos θ} = \frac{y - \frac{b}{2}}{\sin θ} = r$

$\Rightarrow x = \frac{a}{2} + r \cos θ & y = \frac{b}{2} + r \sin θ$

$\Rightarrow x = \frac{a}{2} + \frac{\sqrt{3}}{2} \sqrt{a^{2} + b^{2}} (\frac{b}{\sqrt{a^{2} + b^{2}}}) = \frac{a + b \sqrt{3}}{2}$

and $y = \frac{b}{2} + (\frac{\sqrt{3}}{2} \sqrt{a^{2} + b^{2}}) (\frac{a}{\sqrt{a^{2} + b^{2}}}) = \frac{b + a \sqrt{3}}{2}$

$\Rightarrow b = 2 y - a \sqrt{3}$

$∴ x = \frac{a + \sqrt{3} (2 y - a \sqrt{3})}{2} \Rightarrow x - \sqrt{3} y = - a$

Since it is given that $α x + β y = b$

$\Rightarrow α = 1, β = - \sqrt{3}$

So, $α^{2} + β {^{2} = (1)}^{2} + {(- \sqrt{3})}^{2} = 4$

$A$ is a variable point on $x$ -axis and $B (0, b)$ is a fixed point. An equilateral triangle $A B C$ is completed with vertex $C$ away from origin. If the locus of the point $C$ is $α x + β y = b,$ then $α^{2} + β^{2}$ is

Important Questions on Point and Straight Line

Learn and Prepare for any exam you want !

A is a variable point on x-axis and B0, b is a fixed point. An equilateral triangle ABC is completed with vertex C away from origin. If the locus of the point C is αx+βy=b, then α2+β2 is

Important Questions on Point and Straight Line

Learn and Prepare for any exam you want !

$A$ is a variable point on $x$ -axis and $B (0, b)$ is a fixed point. An equilateral triangle $A B C$ is completed with vertex $C$ away from origin. If the locus of the point $C$ is $α x + β y = b,$ then $α^{2} + β^{2}$ is