Question

The perpendiculars are drawn to lines $L_1$ and $L_2$ from the origin making an angle $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ respectively with the positive direction of X-axis. If both the lines are at unit distance from the origin, then their joint equation is

• A. $x^{2}-y^{2}+2\sqrt{2}y+2=0$
• B. $x^{2}-y^{2}-2\sqrt{2}y-2=0$
• C. $x^{2}-y^{2}+2\sqrt{2}y-2=0$
• D. $x^{2}-y^{2}-2\sqrt{2}y+2=0$

Hint:

Geometry Tip: When multiplying lines to find joint equations, always look for groupings that create a difference of squares. It reduces the chance of algebraic errors compared to distributing every single term manually.

Answer 1

The Correct Option is C

Concept: Coordinate Geometry - Normal Form of a Straight Line and Joint Equation of Lines.

Step 1: Recall the normal form of a line equation. If the perpendicular distance from the origin to a line is '$p$' and the angle this perpendicular makes with the positive x-axis is '$\alpha$', the equation of the line is given by $x\cos\alpha + y\sin\alpha = p$.

Step 2: Find the equation for line $L_{1}$. For $L_1$, we are given $p = 1$ (unit distance) and $\alpha = \frac{\pi}{4}$. Substitute these into the normal form: $x\cos\left(\frac{\pi}{4}\right) + y\sin\left(\frac{\pi}{4}\right) = 1$. Since $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$, the equation becomes $\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 1$. Multiply by $\sqrt{2}$ to get standard form: $x + y - \sqrt{2} = 0$.

Step 3: Find the equation for line $L_{2}$. For $L_2$, we are given $p = 1$ and $\alpha = \frac{3\pi}{4}$. Substitute these into the normal form: $x\cos\left(\frac{3\pi}{4}\right) + y\sin\left(\frac{3\pi}{4}\right) = 1$.
Since $\cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}$ (2nd quadrant) and $\sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}$, the equation becomes $-\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 1$.
Multiply by $\sqrt{2}$ to get standard form: $-x + y - \sqrt{2} = 0$, which can also be written as $x - y + \sqrt{2} = 0$ (multiplying by -1). Let's use $-x + y - \sqrt{2} = 0$ to easily match difference of squares. Actually, matching the solution format, we can write it as $x - y + \sqrt{2} = 0$. Wait, let's look at the given solution steps.
They used $-x/\sqrt{2} + y/\sqrt{2} = 1 \implies x - y + \sqrt{2} = 0$. Let's use that.

Step 4: Set up the joint equation. The joint equation of two lines $u=0$ and $v=0$ is given by their product: $u \cdot v = 0$. Multiply the equations of $L_1$ and $L_2$: $(x + y - \sqrt{2})(x - y + \sqrt{2}) = 0$.

Step 5: Expand and simplify the joint equation. Group the terms to utilize the difference of squares formula $(A+B)(A-B) = A^2 - B^2$. Notice that $(x + y - \sqrt{2})(x - (y - \sqrt{2})) = 0$. Let $A = x$ and $B = (y - \sqrt{2})$. The expansion is $x^2 - (y - \sqrt{2})^2 = 0$. Expand the squared term: $x^2 - (y^2 - 2\sqrt{2}y + 2) = 0$. Distribute the negative sign: $x^2 - y^2 + 2\sqrt{2}y - 2 = 0$. This perfectly matches option C. $$ \therefore \text{The joint equation is } x^{2}-y^{2}+2\sqrt{2}y-2=0. $$