Question

The straight line passing through the points $(0,0),( - 1,1){\text{ and (1, - 1)}}$ has equation:
$A.{\text{ }}2 - x = y \\\ B.{\text{ }}y = x \\\ C.{\text{ }}2x - y = 0 \\\ D.{\text{ }}x + y = 0 \\\$

Answer 1

Hint- Sometimes there is no need to evaluate the equation with the help of formulas whether we can check it with the help of given options. Just use the simple property of the line that any point lying on the line satisfies its equation.

Complete step-by-step solution -
We have to check which straight line given in option, passes through the points $(0,0),( - 1,1){\text{ and (1, - 1)}}$
Any point lying on the line satisfies its equation.
Let us take option A, $2 - x = y$
For the point (0,0), LHS =2-0=2
RHS=$3(0) = 0$
Since $LHS \ne RHS$
Hence, the line $2 - x = y$ does not pass through (0,0). So, no need to check whether other points lie on the line or not.
Next taking option B, $y = x$
For the point $(0,0),y = x$ holds.
For the point $( - 1,1)$ , LHS=1,RHS=-1
Since $LHS \ne RHS$
Hence, line does not pass through $(1, - 1)$
Next we take option C, $2x - y = 0$
For the point $(0,0)$ LHS $= 2(0) - 0 = 0,{\text{ RHS = 0}}$
LHS=RHS
For the point $( - 1,1)$ , LHS=-3,RHS=0
Since $LHS \ne RHS$
Hence, the line does not pass through (1,-1).
Finally taking option D, $x + y = 0$
For point (0, 0), LHS$= 0 + 0 = 0 = {\text{RHS}}$
For point ${\text{(-1, 1),LHS = - 1 + 1 = 0 = RHS}}$
For point $(1, - 1),{\text{LHS = 1 - 1 = 0 = RHS}}$
Hence, the line $x + y = 0$ passes through all the points $(0,0),( - 1,1){\text{ and (1, - 1)}}$
So,Option (D) is the answer.

Note-To solve this type of question using formula is also possible but it is less time consuming and easier to check which point satisfies the equation. This question can also be solved by a straight line equation formula. Remember all types of straight line equation formulas. It is helpful when you have to form the equation of a line.