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Draw the graphs of the equations $x - y + 1 = 0$ and $3 x + 2 y - 12 = 0$ . Determine the coordinates of the vertices of the triangle formed by these lines and the $x$ -axis.

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(1, 3), (- 1, 6)

and

(4, 0)

(2, 5), (- 1, 6)

and

(2, 0)

(- 2, 5), (- 1, 6)

and

(2, 0)

(2, 3), (- 1, 0)

and

(4, 0)

answer is D.

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Detailed Solution

Given that,

x - y + 1 = 0

and

3 x + 2 y - 12 = 0

.
For equation

x–y + 1 = 0

x = y - 1

, we have following points which lie on the line.
Putting

y = 0

x = y - 1

we get:

x = 0 - 1

\Rightarrow x = - 1

Putting

y = 1

x = y - 1

we get:

x = 1 - 1

\Rightarrow x = 0

Putting

y = 2

x = y - 1

we get:

x = 2 - 1

\Rightarrow x = 1

Therefore, the required points are

(- 1, 0), (0, 1)

, (1,2).
For equation

3 x + 2 y– 12 = 0

y = \frac{12 - 3 x}{2}

, we have following points which lie on the line.
Putting

x = 0

y = \frac{12 - 3 x}{2}

we get,

y = \frac{12 - 0}{2}

\Rightarrow y = 6

Putting

x = 4

y = \frac{12 - 3 x}{2}

we get,

y = \frac{12 - 12}{2}

\Rightarrow y = 0

Putting

x = 2

y = \frac{12 - 3 x}{2}

we get,

y = \frac{12 - 6}{2}

\Rightarrow y = 3

Therefore, the required points are

(0, 6), (4, 0) and (2, 3)

.
Plot both the lines using the points.
Question Image

We can see from the graphs that points of intersection of the lines with the x–axis and their intersection are

(- 1, 0), (2, 3)

and

(4, 0)

.
Therefore, option (4) is correct.

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Draw the graphs of the equations x-y+1=0and 3x+2y-12=0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis.

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Draw the graphs of the equations $x - y + 1 = 0$ and $3 x + 2 y - 12 = 0$ . Determine the coordinates of the vertices of the triangle formed by these lines and the $x$ -axis.