Find the area of the triangle formed by the lines y - x = 0, x + y =…

Question 54 of 70intermediate🔧 ApplyShort Answer4 marks

Find the area of the triangle formed by the lines $y - x = 0$ , $x + y = 0$ and $x - k = 0$ .

Correct Answer

The area of the triangle formed by the lines $y - x = 0$ , $x + y = 0$ and $x - k = 0$ is $k^2$ square units.

Exercise: Miscellaneous Exercise on Chapter 9 | Q: 7 | (Page 22)

For More Understanding

Explanation

This question tests the student's ability to find intersection points of lines and calculate the area of a triangle. The methodology is similar to Example 14 in the textbook context, which demonstrates finding vertices by solving pairs of linear equations. The three lines intersect to form a triangle with vertices at (0,0), (k,k), and (k,-k). The area formula for a triangle with given coordinates is then applied.

Solution Steps

Step 1: Find the intersection of $y - x = 0$ and $x + y = 0$ . Solving: $y = x$ and $y = -x$ , we get $x = 0, y = 0$ . First vertex is $(0, 0)$ .
Step 2: Find the intersection of $y - x = 0$ and $x - k = 0$ . From $x = k$ and $y = x$ , we get $y = k$ . Second vertex is $(k, k)$ .
Step 3: Find the intersection of $x + y = 0$ and $x - k = 0$ . From $x = k$ and $y = -x$ , we get $y = -k$ . Third vertex is $(k, -k)$ .
Step 4: Calculate area using formula: Area $= \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Step 5: Substituting vertices: Area $= \frac{1}{2}|0(k - (-k)) + k((-k) - 0) + k(0 - k)| = \frac{1}{2}|-k^2 - k^2| = \frac{1}{2}|-2k^2| = k^2$

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