Question

distance of the origin from the line (1+√3) y+(1-√3) x=10 along the line y=√3 x+k
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Answer 1

Answer:

5 units

Step-by-step explanation:

Equation of line 1

L1 : (1+√3) y+(1-√3) x=10

Equation of line 2

L2 : y=√3 x+k

Slope of line L2 = √3

We have to find a line along this slope which passes through the origin (O)

and then the point of intersection (lets say P) of this line and line L1. OP will be the distance of line L1 from the origin along line L2

equation of line passing through origin and having slope √3 is

y = √3x

Putting this value of y in the equation of line L1

(1+√3)√3x + (1-√3) x=10

⇒ √3x + 3x + x - √3x = 10

⇒ 4x = 10

⇒ x = 5/2

Therefore, y = 5√3/2

So the coordinate of point P = (5/2, 5√3/2)

Distance OP

$=\sqrt{(\frac{5}{2}-0)^2+(\frac{5\sqrt{3}}{2}-0)^2}$

$=\sqrt{\frac{25}{4}+\frac{75}{4}}$

$=\sqrt{\frac{25}{4}+\frac{75}{4}}$

$=\sqrt{\frac{100}{4}}$

$=\frac{10}{2}$

$=5$

Thus, the distance of the origin from the line (1+√3) y+(1-√3) x=10 along the line y=√3 x+k is 5