The area of the triangle formed by the line 5x-3y+10=0 with co-ordinate axis will be
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Hi ,
Find the intercepts made by the line
5x -3y + 10 = 0 with coordinate axes.
5x - 3y = - 10
Divide each term with - 10 , we get
( x / -2 ) + ( 3y / 10 ) = 1
x / ( -2 ) + y / ( 10/3 ) = 1
Therefore ,
X - intercept = a = -2
Y - intercept = b = 10/3
Required Area = 1/2 | ab |
A = 1/2 | ( -2 ) × ( 10/3 ) |
= 1/2 × ( 20/3 )
= 10/3 square units
I hope this helps you.
: )
Find the intercepts made by the line
5x -3y + 10 = 0 with coordinate axes.
5x - 3y = - 10
Divide each term with - 10 , we get
( x / -2 ) + ( 3y / 10 ) = 1
x / ( -2 ) + y / ( 10/3 ) = 1
Therefore ,
X - intercept = a = -2
Y - intercept = b = 10/3
Required Area = 1/2 | ab |
A = 1/2 | ( -2 ) × ( 10/3 ) |
= 1/2 × ( 20/3 )
= 10/3 square units
I hope this helps you.
: )
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