Question

6) The perpendicular from the origin intersects the line√3/2x+1/2y=5
at the coordinates​

Answer 1

The perpendicular from the origin intersects the line√3/2x+1/2y=5

at the coordinates​

Answer:

The point at where perpendicular from the origin intersect the given line is $(\frac{5}{2}\sqrt{3},\frac{5}{2} )$

Step-by-step explanation:

Since we are provide with the equation

$(\frac{\sqrt{3}}{2})x+\frac{1}{2} y=5$    …..(1)

which can be written as

$y=-\sqrt{3} x+10$

The given line has slope m = $-\sqrt{3}$

And line perpendicular to given ln will be -1/m which is  $1/\sqrt{3}$

So line perpendicular to given line with y- intercept equal to zero is

$y = (1/\sqrt{3})x$   ….(2)

Now we need to solve the equations simultaneously to find the common points of line which will give us pair of point which are required here

Putting equation (2) in equation (1) we get

$\frac{1}{\sqrt{3} }x =-\sqrt{3} x +10\\=> x = -3x+10\sqrt{3\\}\\=> 4x = 10\sqrt{3}\\=> x = \frac{5\sqrt{3}}{2}$

Noe putting the value of x in equation (2) we get

$y = 5/2$

So the point at where perpendicular from the origin intersect the given line is $(\frac{5}{2}\sqrt{3},\frac{5}{2} )$

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Answer 2

Given :   (√3 / 2) x + (1/2)y = 5

To Find : perpendicular from the origin intersects the line at the coordinates​

Solution:

(√3 / 2) x + (1/2)y = 5

=> (√3) x + y = 10

=> y = -√3  x  + 10

=> Slope = - √3

Hence Slope of perpendicular must be:   1/√3

Perpendicular from origin

Hence y =  x/√3

(√3 / 2) x + (1/2)y = 5

=>  (√3 / 2) x + (1/2) x/√3 = 5

=> 3x + x = 10√3

=> x = 5√3/2

  y = 5/2

The perpendicular from the origin intersects the line√3/2x+1/2y=5

at the coordinates​ ( 5√3/2 ,  5/2)

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