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Question: If the line \(2x + 5y = k\)passes through origin then \(k = \) A. \(0\) B. \(1\) C. \[ -1\] ...

If the line 2x+5y=k2x + 5y = kpasses through origin then k=k =
A. 00
B. 11
C. 1 -1
D. 22

Explanation

Solution

Hint : If any general point say (a,b)(a,b)passes through any given straight line then that point satisfy the given straight line
Say: General equation of straight line is ax+by+c=0ax + by + c = 0
If (x,y)(x,y),point is passing through above straight line, then (x1,y1)({x_1},{y_1})will satisfy above straight line ax+by+c=0ax + by + c = 0
i.e. putting the value of (x1,y1)({x_1},{y_1})in straight line, we will get
ax1+by1+c=0a{x_1} + b{y_1} + c = 0

Complete step-by-step answer :
In general, therefore, the equation y = mx represents a straight line passing through the origin with gradient m. The equation of a straight line with gradient m passing through the origin is given by y = mx.
Here, given straight line is 2x+5y=K2x + 5y = K
It is given that the line2x+5y=K2x + 5y = Kpasses through origin,
It means line 2x+5y=K2x + 5y = K, will satisfy the origin point i.e. (0,0)(0,0)
So, putting (0,0)(0,0)in above line 2x+5y=K2x + 5y = K,
We will get
2(0)+5(0)=K2(0) + 5(0) = K
0+0=K\Rightarrow 0 + 0 = K
K=0\Rightarrow K = 0
Thus, the required value of KK for which the given line 2x+5y=K2x + 5y = K passes through origin is K=0K = 0
So, the correct answer is “Option A”.

Note : Any straight line say ax+by+c=0ax + by + c = 0 passes through origin will satisfy the point (0,0)(0,0)when (0,0)(0,0)is put into the line
i.e. a(0)+b(0)+c=0a(0) + b(0) + c = 0
C=0\Rightarrow C = 0. Hence the value of CCwill be zero.
So in case of competitive exams one can directly write the value of zero.