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

The problem asks for the area bounded by $y = x^2$ (an upward-opening parabola with vertex at the origin) and $y = 16$ (a horizontal line). The key idea: area between two curves is found by integrating the top function minus the bottom function over the interval where they intersect.

First, find where the curves meet. Set $x^2 = 16$, so $x = \pm 4$. The region is symmetric about the $y$-axis, so we can integrate from $x = -4$ to $x = 4$, or double the area from $0$ to $4$.

Over this interval, the line $y = 16$ is above the parabola $y = x^2$. The vertical height between them at any $x$ is $16 - x^2$.

1. Set up the integral

   Area $= \int_{-4}^{4} (16 - x^2) \, dx$.

2. Use symmetry

   Since the integrand is even ($16 - x^2$ is symmetric),

   Area $= 2 \int_{0}^{4} (16 - x^2) \, dx$.

3. Integrate

   $\int (16 - x^2) \, dx = 16x - \frac{x^3}{3}$.

   Evaluate from $0$ to $4$:

   $\left[16(4) - \frac{4^3}{3}\right] - \left[16(0) - \frac{0^3}{3}\right] = 64 - \frac{64}{3} = \frac{192 - 64}{3} = \frac{128}{3}$.

4. Double for full area …

The problem asks for the area bounded by a straight line and two vertical lines. This is a classic application of definite integration: the area between a curve $y = f(x)$ and the $x$-axis, from $x = a$ to $x = b$, is $\int_a^b f(x) \, dx$. Here $f(x) = x+1$ is linear, so the region is a trapezoid — but integration handles it cleanly.

1. Set up the integral. The region is bounded by $y = x+1$, $x = 3$, $x = 5$, and the $x$-axis (since no other lower bound is given, the $x$-axis is the natural boundary). The area is

2. Integrate. The antiderivative of $x+1$ is $\frac{x^2}{2} + x$. So

3. Evaluate at the limits. At $x = 5$: $\frac{25}{2} + 5 = \frac{25}{2} + \frac{10}{2} = \frac{35}{2}$. At $x = 3$: $\frac{9}{2} + 3 = \frac{9}{2} + \frac{6}{2} = \frac{15}{2}$. Subtract: $A = \frac{35}{2} - \frac{15}{2} = \frac{20}{2} = 10.$ …

The key idea here is that the area of a triangle can be computed directly from the coordinates of its vertices using the determinant formula. This formula gives the area as half the absolute value of a certain expression — and that absolute value is what introduces the possibility of two signs for $k$.

Let’s see why this works. The three points are $(-3,0)$, $(3,0)$, and $(0,k)$. Notice that the first two lie on the x-axis, symmetric about the origin. So the base of the triangle is the segment joining them, which has length $6$. The third vertex is on the y-axis, so the height of the triangle is simply the vertical distance from $(0,k)$ to the x-axis, which is $|k|$. Area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$, so we get $\frac{1}{2} \times 6 \times |k| = 3|k|$. Setting this equal to $9$ gives $|k| = 3$, so $k = 3$ or $k = -3$. Only $3$ is among the options.

Now let’s do it step by step using the coordinate formula, which is the standard method for any three points.

1. Write the area formula. For vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, the area is

2. Plug in the given points. Here $(x_1, y_1) = (-3, 0)$, $(x_2, y_2) = (3, 0)$, $(x_3, y_3) = (0, k)$. Then

3. Simplify inside the absolute value. $(-3)(-k) + 3k + 0 = 3k + 3k = 6k.$ …

Step 1 — Find the points of intersection.

Substitute $y = 2x$ into $y^2 = 8x$:

Corresponding $y$-values from $y = 2x$: $y = 0$ and $y = 4$.

So the region is bounded between $(0,0)$ and $(2,4)$.

Step 2 — Choose the integration variable.

Integrating with respect to $y$ is cleaner here, because both curves give $x$ as a single-valued function of $y$:

• Line: $y = 2x \Rightarrow x_{\text{line}} = \dfrac{y}{2}$
• Parabola: $y^2 = 8x \Rightarrow x_{\text{par}} = \dfrac{y^{2}}{8}$

Step 3 — Decide which curve is on the right.

Test a value inside the range, say $y = 2$:

So the line lies to the right of the parabola. The horizontal width of a strip is $\left(\dfrac{y}{2} - \dfrac{y^{2}}{8}\right)$, which is positive throughout $0<y<4$. $\checkmark$

Step 4 — Integrate from $y = 0$ to $y = 4$. …

The key idea: area bounded by a curve and the $x$-axis between two vertical lines is $\int_{a}^{b} |y| \, dx$. The line $y = 2x + 1$ crosses the $x$-axis where $2x + 1 = 0$, i.e. at $x = -\frac{1}{2}$. Since this lies between $x = -1$ and $x = 1$, the function changes sign. Area is always positive, so we must integrate the absolute value — splitting the integral at the root.

1. Find the $x$-intercept Set $y = 0$:

This point lies inside $[-1, 1]$, so the line is below the $x$-axis for $x < -\frac{1}{2}$ and above it for $x > -\frac{1}{2}$.

2. Set up the area as two integrals For $x \in [-1, -\frac{1}{2}]$, $y$ is negative, so $|y| = -(2x+1)$. For $x \in [-\frac{1}{2}, 1]$, $y$ is positive, so $|y| = 2x+1$. Hence

3. Evaluate the first integral

At $x = -\frac{1}{2}$: $\frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$

At $x = -1$: $1 - 1 = 0$

So $-\left( -\frac{1}{4} - 0 \right) = \frac{1}{4}$

4. Evaluate the second integral

At $x = 1$: $1 + 1 = 2$

At $x = -\frac{1}{2}$: $\frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$ …

The key idea here is the relationship between intercepts and slope. When a line cuts off equal intercepts on both axes, it means the distances from the origin to where the line meets the X-axis and Y-axis are the same in magnitude. But the sign matters — if both intercepts are positive, the line slopes downward; if both are negative, it also slopes downward. Either way, the slope becomes $-1$.

The angle a line makes with the positive X-axis is found from its slope: $\text{slope} = \tan \theta$, where $\theta$ is that angle (measured anticlockwise from the positive X-axis). So we need the angle whose tangent is $-1$.

Let’s go through it step by step.

1. Write the equation in intercept form. If a line cuts intercepts $a$ on the X-axis and $b$ on the Y-axis, its equation is

Here, "equal intercepts" means $a = b$ (in magnitude). But intercepts can be positive or negative depending on which side of the origin the line lies. The condition "cuts off equal intercepts" usually means $|a| = |b|$, but in coordinate geometry problems, it often implies $a = b$ (both positive) or $a = -b$ (one positive, one negative). Let’s examine both.

2. Case 1: $a = b$ (both positive or both negative). Then the equation becomes

Rewrite in slope-intercept form: $y = -x + a$.

Slope $m = -1$.

3. Case 2: $a = -b$ (one positive, one negative). Then

So $y = x - a$, slope $m = 1$.

But does this line cut off equal intercepts? Yes, in magnitude: intercepts are $a$ and $-a$, so lengths are $|a|$ and $|a|$. However, the phrase "cuts off equal intercepts" in most exam contexts means the signed intercepts are equal (both $a$), not just their absolute values. The NCERT and typical board problems treat "equal intercepts" as $a = b$, not $a = -b$. So the intended interpretation is $a = b$. …

The key is to see the geometry clearly. The parabola $y^2 = x$ opens to the right, with its vertex at the origin. The circle $x^2 + y^2 = 2x$ can be rewritten by completing the square:

$x^2 - 2x + y^2 = 0 \implies (x-1)^2 + y^2 = 1$.

So it is a circle of radius 1 centred at $(1,0)$. The region “above the X‑axis” means we only consider $y \ge 0$. The two curves intersect where both equations hold — that will give the limits for integration.

1. Find the intersection points Substitute $y^2 = x$ into the circle’s equation:

So $x=0$ or $x=1$.

• For $x=0$, $y^2=0 \Rightarrow y=0$.
• For $x=1$, $y^2=1 \Rightarrow y = \pm 1$. Above the X‑axis, we take $y=1$. Thus the two intersection points in the upper half‑plane are $(0,0)$ and $(1,1)$.

2. Identify which curve is on top

   For $0 \le x \le 1$, compare the upper‑half functions.

   From the parabola: $y = \sqrt{x}$.

   From the circle: $(x-1)^2 + y^2 = 1 \Rightarrow y^2 = 1 - (x-1)^2 = 2x - x^2$, so $y = \sqrt{2x - x^2}$.

   At $x=0$, $\sqrt{0}=0$ and $\sqrt{0}=0$ — they meet. At $x=0.5$, $\sqrt{0.5} \approx 0.707$ and $\sqrt{2(0.5)-0.25} = \sqrt{0.75} \approx 0.866$. So the circle’s upper half lies above the parabola’s upper half for $0 < x < 1$.

   Therefore the area is the integral of (circle top minus parabola top) from $x=0$ to $x=1$.

3. Set up the area integral

4. Evaluate $\int_{0}^{1} \sqrt{2x - x^2} \, dx$ Complete the square inside the root: $2x - x^2 = 1 - (x-1)^2$. So

Let $u = x-1$, then $du = dx$, and when $x=0$, $u=-1$; when $x=1$, $u=0$.

$\int_{-1}^{0} \sqrt{1 - u^2} \, du.$ …

Concept: area determinant. The area of a triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is

The column of $1$s is what turns coordinates into an area, so $|D| = 2\Delta$.

1. Relate the given determinant to $D$. The determinant in the question,

differs from $D$ only in its third column ($4$ instead of $1$). A determinant is linear in each column, so pulling the common factor $4$ out of that column gives

2. Square it.

The coefficient $64$ (from $16 \times 4$) is what fixes the answer.

3. Identify the option. …