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satishinamdar
08-26-2005, 02:42 PM
find the equations of the two straight lines passing through (4,5) and making equal angles with 3x=4y+7 & 5y=12x+6.

pka
08-26-2005, 04:41 PM
The acute angle between two non-perpendicular lines with slopes m<SUB>1</SUB> & m<SUB>2</SUB> is given by
arctan[|(m<SUB>2</SUB>− m<SUB>1</SUB>)/(1+m<SUB>2</SUB>m<SUB>1</SUB>)|]. (Note the absolute value.)

Thus you want to solve |[m−(3/4)]/(1+3m/4)|= |[m−(12/5)]/(1+12m/5)| for m.

Using m as the slope and the point (4,5) write the equation of the line.

wjm11
08-26-2005, 05:20 PM
find the equations of the two straight lines passing through (4,5) and making equal angles with 3x=4y+7 & 5y=12x+6.

Hi, Satishinamdar,

I also used trig to find one line. Perhaps someone else can solve using geom.:

y-5 = -(7/9)*(x-4)

Having one line, it can be shown using geometry that the other line is perpendicular to it. Thus, the second line would be:

y-5 = (9/7)*(x-4)

Can anyone else help out here? Is there a way to solve this using only geometric principles?

mathmike
08-27-2005, 02:17 AM
find the equations of the two straight lines passing through (4,5) and making equal angles with 3x=4y+7 & 5y=12x+6.
I think, that one of the lines in question has to pass through the common point of the two given ones to bisect angle between them and another has to be perpendicular to it. So, if we solve two given equations for x & y we'll find coordinates of another point on the the line in question. Given coordinates of two points it is standard procedure to find equation for a line as well as equation of the perpendicular one that has to go through (4, 5) too.

wjm11
08-27-2005, 03:46 AM
I think, that one of the lines in question has to pass through the common point of the two given ones to bisect angle between them

That would only be true in the specialized case where the point in question happens to lie on the angle bisector.

mathmike
08-27-2005, 11:02 AM
That would only be true in the specialized case where the point in question happens to lie on the angle bisector.
Right, my bad- it is enough that line is parallel to the bisector or perpendicular to it. So, if we take average slope from initial equations it would give us the bisector slope and negative reciprocal would give a slope for a perpendicular. Then given point can be used to determine intersepts of these two lines.