Mathematics

Grade-8

Easy

Question

# Identify the equation which matches the following graph

- Y = 2x – 1
- X
^{2} + y^{2} = 1 - Y = –2x – 1
- Y = x
^{2} - 1

^{2}+ y^{2}= 1^{2}- 1Hint:

### 1. An equation refers to the relationship between 2 expressions represented with an equal to sign i.e. '='.

2. A point is said to be on a given line when the coordinates of that point when substituted in the given equation, satisfies the given equation.

## The correct answer is: Y = –2x – 1

### GIVEN-

The given line passes through the points (-1,1) & (0,-1) as shown in the given graph.

TO FIND-

Equation of the given line.

SOLUTION-

From the given graph, we observe that-

(-1,1) & (0,-1) are 2 points on the line.

We know that when a line passes through a given point, the coordinates of that point satisfies the equation of the said line.

Hence, we substitute the coordiantes of both the points in the given options to find the correct equation, for which LHS = RHS.

a. y = 2x - 1

We substitute (-1,1) in the given equation-

y = 2x - 1

∴ 1 = 2 (-1) - 1

∴ 1 = -2 - 1

∴ 1 ≠ -3

∴ LHS ≠ RHS

Since (-1,1) does not satisfy the given equation, y = 2x - 1 is not the equation of the given line.

b. x^{2} + y^{2} = 1

We substitute (-1,1) in the given equation-

x^{2} + y^{2} = 1

∴ (-1)^{2} + (1)^{2} = 1

∴ 1 + 1 = 1

∴ 2 ≠ 1

∴ LHS ≠ RHS

Since, (-1,1) does not satisfy the given equation, x2 + y2 = 1 is not the equation of the given line.

c. y = -2x - 1

We substitute (-1,1) in the given equation-

y = -2x - 1

∴ 1 = -2 (-1) - 1

∴ 1 = 2 - 1

∴ 1 = 1

∴ LHS = RHS

We substitute (0,-1) in the given equation-

y = -2x - 1

∴ -1 = -2 (0) - 1

∴ -1 = 0 - 1

∴ -1 = -1

∴ LHS = RHS

Since both (-1,1) and (0,-1) satisfy the given equation, y = -2 x - 1 is the equation of the given line.

FINAl ANSWER-

Option 'c' i.e. 'y = -2x - 1' is the correct answer to the given question.

**Alternatively, we can use the 2 point formula to find the equation of the given line-****When we have 2 points that lie on a given line then we can find the equation of the said line by using the 2-point formula-**** (y-y1) = (y2-y1) * (x-x1)**** (x2-x1)****In the given question, x1 = -1, y1 = 1, x2 = 0 & y2 = -1****∴ (y-y1) = (y2-y1) * (x-x1)**** (x2-x1)****∴ (y - 1) = (-1 - 1) * [x - (-1)]**** [0 - (-1)]****∴ (y - 1) = -2 * (x + 1)**** 0 + 1****∴ (y - 1) = -2/1 * (x + 1)****∴ y - 1 = -2x - 2****∴ y = -2x - 2 + 1****∴ y = -2x - 1****Final Answer remains the same.**