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# Find the coorddinates of the point of intersection of the straight lines 2x-3y=1 and 5y-x=3 and determine the angle between them

2080 Points
2 years ago
to find the point of intersection simply solve the two eqns together. 2x-3y=1 and 5y-x=3 gives x= 2 and y= 1.
so coordinates of the point of intersection= (2, 1)
now, the formula for angle between 2 lines is
theta= arctan[|(m1 – m2)/(1+m1*m2)|], where m1 and m2 are the slopes of the lines.
here, m1= 2/3 for 2x-3y=1 and m2= 1/5 for 5y-x=3.
so, theta= arctan[|(m1 – m2)/(1+m1*m2)|]= arctan(7/17) which is approx 22.38 degrees.
kindly approve :)
Vikas TU
14149 Points
2 years ago
Dear student
in equation 1 , a1 = 2 , b1 = -3 ,
in equation 2 a2 = 5 , b2 = -1
a1/a2 is not equal to b1 / b2
so, this is the condition of line to intersect ,
and angle between them = 22.38 degree.
Pawan Kumar Karela
14 Points
2 years ago
Given equation    2x-3y=1   (1)
5y-x=3     (2)
we use the elimination method to get the intersection point.
therefore multiply equation (2) by 2 and then add it to equation (1) to eliminate variable y.
2x-3y=1     (3)
-2x+10y=6  (4)
adding (3) and (4) we get
7y=7
y=1      (*)
use value of y to get x
2x-3*1=1
2x=1+3
x=2
therefore point of intersection is (2,1)

to find angle between these two lines we use formula
$tan\Theta =\left | \frac{m1-m2}{1+m1*m2} \right |$
from equation (1)  2x-3y=1...... implies y=(2/3)x-1
therefore m1=2/3
similarly for equation (2)  y=1/5+3/5
therefore m2=1/5
hence $tan\Theta =\left | \frac{(2/3)-(1/5)}{1+(2/3)*(1/5)} \right |$
therefore on solving we get
$tan\Theta =\left | \frac{7}{17}\right |$
$\Theta =tan^-1\left | \frac{7}{17}\right |$