If the graph of the equation 2x-7y =28 cuts the coordinate axes at P and Q ,then find the length of hypotenuse of right triangle POQ ( Where O is the origin)
Answers
Answér :
2√53 units
Note :
★ Equation of a straight line in intercept form is given as : x/a + y/b = 1
Where a = x-intercept and b = y-intercept
★ The length of the hypotenuse of the right angled triangle formed by the the straight line and the coordinate axes is given as ; h = √(a² + b²)
Solution :
Here ,
The given equation of straight line is ;
2x - 7y = 28 ---------(1)
The given equation can be rewritten as ;
=> 2x - 7y = 28
=> (2x - 7y) / 28 = 1
=> 2x/28 - 7y/28 = 1
=> x/14 - y/4 = 1
=> x/14 + y/-4 = 1 ----------(2)
Clearly ,
Eq-(2) is in intercept form , where
x-intercept , a = OP = 14
y-intercept , b = OQ = -4
Also ,
Point of interception on x-axis is P(14,0)
Point of interception on y-axis is Q(0,-4)
Now ,
The length of the hypotenuse of the right angled ∆POQ will be ;
=> h = PQ
=> h = √(a² + b²)
=> h = √(14² + (-4)²)
=> h = √(196 + 16)
=> h = √212
=> h = 2√53
Hence ,
Required answer is 2√53 units .
Step-by-step explanation:
2x+7y=28
(2x-7y)/28=1 __1
2x/28-7y/28=1
x/14-y/4=1
x/14+y/-4=1