what is equation of the line pass through the origin & perpendicular to Line whose X&Y intercept are 5 &-3/2?

"Point-Point form" (y-y1)/(x-x1)=(y2-y1)/(x2-x1)

In your known line you have two points.

The x intercept (5,0), and the y intercept (0,-3/2)

Plug these values into the "Point-Point" form and solve for y in terms of x, that is y= mx + b

When you get that, the slope of your perpendicular will be the negative reciprocal of m... (-1/m)

Now, you know a point on that perpendicular (0,0).

Use

(y-y1)/(x-x1) = -1/m

====== OR =====

You could do it all at once by using

(x1,y1) = (0, -3/2)

(x2,y2) = (5, 0)

and

(x3,y3) = (0, 0)

and use

(y-y3)/(x-x3)= -(y2-y1)/(x2-x1)

and solve for y in terms of x.

The equation of any line perpendicular to the line whose x and y intercepts are 5 an - 3/2 is of the form

x / (3/2) + y / 5 = k

As (0, 0) lies on it k = 0.

=> eqn. of the line is 10x + 3y = 0.

By reading the problem, we can obtain two points that is in the line: (5, 0) and (0,-3/2). We will use these points to obtain the slope of the line.

m = (y2 - y1) / (x2 - x1)

m = (-3/2 - 0) / (0 - 5)

m = 3 / 10

PARALLEL LINES ( Two lines are parallel if their slopes are equal)

m = 3 / 10 with point (0,0) ==> origin.

POINT - SLOPE FORM:

(y-y1) = m (x-x1)

(y - 0) = 3/10 (x - 0)

10y = 3x

[ 3x - 10y = 0 ] == Line parallel to the given line

PERPENDICULAR LINES ( Two lines are perpendicular if their slopes are negative reciprocals of each other)

m = -10 / 3 with point (0,0) ==> origin.

POINT - SLOPE FORM:

(y-y1) = m (x-x1)

(y - 0) = -10/3 (x - 0)

3y = -10x

[ 10x +3y = 0 ] == Line perpendicular to the given line

Let a (x1,y1) be (5,0) Let b (x2,y2) be (0,-3/2)

Slope of the line

through the two points = (y2-y1)/(x2-x1).

=(-3/2-0)/(0-5) = +3/10

Therefore slope of per = -10/3

Also line required goes throught the origin, therefore it is:

y = -(10/3)x

y=3.5/5x