Straight Line equation in 4th quadrant of cartesian coordinate system

Hi all,

Thanks for the attention. I want to know the staight line equation for 4th quadrant of cartesian coordinate system. Also i need to find out the slope of the straight line and Y intercept or X intercept constant, which ever is applicable to the staight line equation.

Thanks,
drv

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Originally Posted by drv123

Hi all,

Thanks for the attention. I want to know the staight line equation for 4th quadrant of cartesian coordinate system. Also i need to find out the slope of the straight line and Y intercept or X intercept constant, which ever is applicable to the staight line equation.

Thanks,
drv

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Hi drv123, welcome to MHF.
When the straight line is in the fourth quadrant, intercepts on the x and y axis will be either x1 and y1 or -x1 and -y1. And the angle made by the line with x -axis is obtuse. So the slope is negative. One point on the x-axis will be either (x1,0) or (-x1,0)
Then the equations of line will be
y = -(y1/x1)*(x-x1) or
y = -(y1/x1)*(x+x1)

Hi Sa-re-ga-ma,
Thank you so much for your reply.
Please let me know what can be the slope calculation for the same line in 4th quadrant?
For a straight line, ((x1,y1) , (x2,y2)), can we take
Slope in 4th quadrant = (y2-y1)/(x2-x1) as we take in 1st quadrant?
Does it depend on the angle it makes.

thanks,
drv

If x1 and y1 are the intercepts on the co-ordinate axis, the points on the axis are (x1,0) and (0,y1) or (-x1,0) and (0,-y1). In both the cases slope is (0-y1)/(x1-0) = -(y1/x1)