JEE Straight Lines PYQs with Video Solutions, Download PDF

Rearrange the family of lines by grouping terms with $$\lambda$$:

$$(x + 2y - 5) + \lambda(3x + 7y - 17) = 0$$

This is in the form $$L_1 + \lambda L_2 = 0$$. To find the fixed point $$P$$, solve the system:

$$x + 2y = 5$$

$$3x + 7y = 17$$

Multiplying the first equation by 3 and subtracting gives $$y = 2$$, which means $$x = 1$$. So, $$P = (1, 2)$$.

The line from this family that is farthest from the origin is the line perpendicular to the segment $$OP$$ (where $$O$$ is the origin $$(0,0)$$).

Slope of $$OP = \frac{2 - 0}{1 - 0} = 2$$. Therefore, the slope of line $$L$$ is $$-\frac{1}{2}$$.

Equation of line $$L$$ passing through $$P(1, 2)$$:

$$y - 2 = -\frac{1}{2}(x - 1) \implies x + 2y - 5 = 0$$

$$d = \frac{|3 + 2(6) - 5|}{\sqrt{1^2 + 2^2}} = \frac{|3 + 12 - 5|}{\sqrt{5}} = \frac{10}{\sqrt{5}}$$

$$d^2 = \frac{100}{5} = 20$$

Solution :

For lines

$$a_1x+b_1y+c_1=0$$

and

$$a_2x+b_2y+c_2=0$$

to be perpendicular,

$$a_1a_2+b_1b_2=0$$

Here,

$$a_1=k-1,\quad b_1=1$$

and

$$a_2=1,\quad b_2=k+1$$

Therefore,

$$(k-1)(1)+(1)(k+1)=0$$

$$k-1+k+1=0$$

$$2k=0$$

$$k=0$$

Final Answer :

$$0$$

Let $$L_1: 3x+y-7=0$$ and the angle bisector be $$B: x-2y+5=0.$$

First, we find the point of intersection $$P$$ of the line $$L_1$$ and the bisector $$B$$. From the bisector equation, we can express $$x$$ as:

$$x = 2y - 5$$

Substituting this into the equation for $$L_1$$:

$$3(2y - 5) + y - 7 = 0 \implies 7y - 22 = 0 \implies y = \frac{22}{7}$$

Substituting $$y$$ back to find $$x$$:

$$x = 2\left(\frac{22}{7}\right) - 5 = \frac{9}{7}$$

Thus, both lines must intersect at the point $$P\left(\frac{9}{7}, \frac{22}{7}\right)$$.

Next, let the angle between $$L_1$$ and $$B$$ be $$\theta$$. The slopes are $$m_1 = -3$$ and $$m_B = \frac{1}{2}$$. Using the angle formula:

$$\tan\theta = \left|\frac{m_1 - m_B}{1 + m_1 m_B}\right| = \left|\frac{-3 - \frac{1}{2}}{1 + (-3)\left(\frac{1}{2}\right)}\right| = \left|\frac{-\frac{7}{2}}{-\frac{1}{2}}\right| = 7$$

Let the slope of the required line $$L_2$$ be $$m$$. Since $$B$$ is the angle bisector, the angle between $$L_2$$ and $$B$$ must also be $$\theta$$:

$$\left|\frac{m - \frac{1}{2}}{1 + \frac{m}{2}}\right| = 7 \implies \frac{2m - 1}{2 + m} = \pm 7$$

To find the line on the opposite side of the bisector, we solve the negative branch:

$$\frac{2m - 1}{2 + m} = -7 \implies 2m - 1 = -14 - 7m$$

$$9m = -13 \implies m = -\frac{13}{9}$$

Now, using the point-slope form for the line $$L_2$$ passing through $$P\left(\frac{9}{7}, \frac{22}{7}\right)$$ with slope $$m = -\frac{13}{9}$$:

$$y - \frac{22}{7} = -\frac{13}{9}\left(x - \frac{9}{7}\right)$$

Multiplying the entire equation by 63 to clear the denominators:

$$63y - 198 = -13(7x - 9)$$

$$63y - 198 = -91x + 117$$

$$91x + 63y - 315 = 0$$

Dividing by the common factor of 7 yields the final line equation:

$$13x + 9y - 45 = 0$$

To find the intercepts, put $$y=0$$ in

$$4x+3y=24$$

$$4x=24$$

$$x=6$$

So the x-intercept is

$$(6,0)$$

Now put

$$x=0$$

$$3y=24$$

$$y=8$$

So the y-intercept is

$$(0,8)$$

The triangle formed with the coordinate axes has base $$6$$ and height $$8$$

Therefore,

$$A=\frac12\times6\times8$$

$$=24$$

Hence,

$$\frac{A}{4}=\frac{24}{4}$$

$$=6$$

Final Answer :

$$6$$

By the geometric principles of reflection, the mirror image of the incident light source point across the reflecting line must lie directly on the straight line containing the reflected ray.

The incident ray originates from $$A(1, 4)$$ and reflects off the horizontal line mirror given by the equation $$y = 2$$.
Let $$A'(x_1, y_1)$$ be the mirror image of point $$A(1, 4)$$ with respect to the horizontal line $$y = 2$$. Since the line is horizontal, the x-coordinate remains unchanged, and the line midway between $$y_1$$ and $$4$$ must equal $$2$$:
$$x_1 = 1$$
$$\frac{y_1 + 4}{2} = 2 \implies y_1 + 4 = 4 \implies y_1 = 0$$

Thus, the image point is $$A'(1, 0)$$. The reflected ray behaves as if it originates from this virtual image point $$A'(1, 0)$$ and passes through the given path point $$B(5, 4)$$.
We find the equation of the straight line passing through $$A'(1, 0)$$ and $$B(5, 4)$$. First, compute its slope ($$m$$):
$$m = \frac{4 - 0}{5 - 1} = \frac{4}{4} = 1$$

Using the point-slope form of a straight line equation with point $$(1, 0)$$:
$$y - 0 = 1(x - 1)$$
$$y = x - 1 \quad \text{--- (Equation of Reflected Ray)}$$

The point of incidence $$P$$ is the intersection point of the reflected ray line with the actual mirror line $$y = 2$$. Substitute $$y = 2$$ into the equation of the reflected ray:
$$2 = x - 1 \implies x = 3$$

Since $$P$$ lies on the line mirror $$y = 2$$, its coordinates are given by:
$$P = (3, 2)$$

To find the distance between two parallel lines, their coefficients of $$x$$ and $$y$$ must be identical. Let's rewrite the given equations:
$$\text{Line 1: } 3x - 4y + \sqrt{2}\sin\theta = 0$$
$$\text{Line 2: } 6x - 8y - 3\sqrt{2}\cos\theta = 0 \implies 3x - 4y - \frac{3\sqrt{2}}{2}\cos\theta = 0$$

Now, both lines match the standard form $$Ax + By + C = 0$$, where:
$$A = 3, \quad B = -4$$
$$C_1 = \sqrt{2}\sin\theta, \quad C_2 = -\frac{3\sqrt{2}}{2}\cos\theta$$

$$ \textbf{Step 2: Apply the parallel line distance formula} $$
The formula for the distance $d$ between two parallel lines is:
$$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$$

Substitute the values into the formula:
$$d = \frac{\left| \sqrt{2}\sin\theta - \left(-\frac{3\sqrt{2}}{2}\cos\theta\right) \right|}{\sqrt{3^2 + (-4)^2}}$$
$$d = \frac{\left| \sqrt{2}\sin\theta + \frac{3\sqrt{2}}{2}\cos\theta \right|}{\sqrt{9 + 16}}$$
$$d = \frac{\sqrt{2}}{5} \left| \sin\theta + \frac{3}{2}\cos\theta \right|$$

$$ \textbf{Step 3: Maximize the trigonometric expression} $$
We know that the maximum value of any expression in the form $a\sin\theta + b\cos\theta$ is given by $$\sqrt{a^2 + b^2}$$.

Here, $$a = 1$$ and $$b = \frac{3}{2}$$. Let's calculate the maximum value:
$$\text{Maximum value of } \left( \sin\theta + \frac{3}{2}\cos\theta \right) = \sqrt{1^2 + \left(\frac{3}{2}\right)^2} = \sqrt{1 + \frac{9}{4}} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2}$$

$$\textbf{Step 4: Calculate the maximum distance} $$
Substitute this maximum value back into the distance equation:
$$d_{\max} = \frac{\sqrt{2}}{5} \times \frac{\sqrt{13}}{2} = \frac{\sqrt{26}}{10}$$

Step 1: Find the acute and obtuse angle bisectors of L2 and L3
The equations of the angle bisectors between $$L_2: y = 0$$ and $$L_3: 3x - 4y = 0$$ are given by:
$$\frac{3x - 4y}{\sqrt{3^2 + (-4)^2}} = \pm \frac{y}{\sqrt{0^2 + 1^2}}$$
$$\frac{3x - 4y}{5} = \pm y$$

This yields two separate lines:
1) Taking the positive sign:
$$3x - 4y = 5y \implies 3x - 9y = 0 \implies x - 3y = 0$$

2) Taking the negative sign:
$$3x - 4y = -5y \implies 3x + y = 0$$

To determine which one is the acute bisector, we can check the angle formed with $$L_2: y = 0$$ (the $$x$$-axis). The line $$x - 3y = 0$$ has a slope of $$\frac{1}{3}$$, making an angle less than $$\frac{\pi}{4}$$ with the $$x$$-axis. Thus, the acute angle bisector is $$B_{\text{acute}}: x - 3y = 0$$ and the obtuse angle bisector is $$B_{\text{obtuse}}: 3x + y = 0$$.

Step 2: Determine the point of incidence
The point of incidence $$I$$ is the intersection point of the original line $$L_1: x - y - 2 = 0$$ and the reflecting mirror line $$B_{\text{acute}}: x - 3y = 0$$.

Substituting $$x = 3y$$ from the bisector equation into $$L_1$$:
$$3y - y - 2 = 0 \implies 2y = 2 \implies y = 1$$
$$x = 3(1) = 3$$

Thus, the point of incidence is $$I(3, 1)$$.

Step 3: Find the equation of the reflected line
Since $$I(3, 1)$$ lies on the mirror line, it remains invariant. To find the path of the reflected line, let us pick another convenient point on $$L_1$$, say $$Q(2, 0)$$, and reflect it across $$B_{\text{acute}}: x - 3y = 0$$:
$$\frac{x_Q' - 2}{1} = \frac{y_Q' - 0}{-3} = -2 \frac{1(2) - 3(0)}{1^2 + (-3)^2}$$
$$\frac{x_Q' - 2}{1} = \frac{y_Q'}{-3} = -2 \frac{2}{10} = -\frac{2}{5}$$

Solving for the coordinates of the reflected point $$Q'$$:
$$x_Q' = 2 - \frac{2}{5} = \frac{8}{5}$$
$$y_Q' = \frac{6}{5}$$

The reflected line passes through $$I(3, 1)$$ and $$Q'\left(\frac{8}{5}, \frac{6}{5}\right)$$. Let us find its slope $$m$$:
$$m = \frac{1 - \frac{6}{5}}{3 - \frac{8}{5}} = \frac{-\frac{1}{5}}{\frac{7}{5}} = -\frac{1}{7}$$

The equation of the reflected line is:
$$y - 1 = -\frac{1}{7}(x - 3) \implies x + 7y - 10 = 0$$

Step 4: Locate point P on the reflected line
Point $$P$$ lies on $$x + 7y - 10 = 0$$ at a distance of $$k = 5\sqrt{2} = \sqrt{50}$$ units from the point of incidence $$I(3, 1)$$. Using the parametric form of a straight line with slope angle $$\alpha$$ where $$\tan\alpha = -\frac{1}{7}$$:
$$\cos\alpha = \frac{7}{\sqrt{50}}, \quad \sin\alpha = -\frac{1}{\sqrt{50}}$$

The coordinates of $$P$$ are given by:
$$x = 3 + k\cos\alpha = 3 + \sqrt{50}\left(\frac{7}{\sqrt{50}}\right) = 3 + 7 = 10$$
$$y = 1 + k\sin\alpha = 1 + \sqrt{50}\left(-\frac{1}{\sqrt{50}}\right) = 1 - 1 = 0$$

Since the question specifies that $$P$$ has a positive $$x$$-coordinate, $$P(10, 0)$$ is the correct point.

Step 5: Find the final image of P across the obtuse bisector
Now, we reflect $$P(10, 0)$$ across the obtuse angle bisector $$B_{\text{obtuse}}: 3x + y = 0$$:
$$\frac{x' - 10}{3} = \frac{y' - 0}{1} = -2 \frac{3(10) + 1(0)}{3^2 + 1^2}$$
$$\frac{x' - 10}{3} = \frac{y'}{1} = -2 \frac{30}{10} = -6$$

Solving for the final image coordinates:
$$x' - 10 = 3(-6) = -18 \implies x' = -8$$
$$y' = 1(-6) = -6$$

Thus, the final image of point $$P$$ is $$\begin{bmatrix} -8 & -6 \end{bmatrix}$$.

Let the equation of the line be

$$\frac{x}{a}+\frac{y}{b}=1$$

where

$$a=OA$$

and

$$b=OB$$

Since the line passes through

$$(4,9)$$

we have

$$\frac{4}{a}+\frac{9}{b}=1$$

We need to minimize

$$a+b$$

Now,

$$a+b=(a+b)\left(\frac{4}{a}+\frac{9}{b}\right)$$

Expanding,

$$a+b=\frac{4a}{a}+\frac{9a}{b}+\frac{4b}{a}+\frac{9b}{b}$$

$$=13+\frac{9a}{b}+\frac{4b}{a}$$

Applying AM-GM,

$$\frac{\frac{9a}{b}+\frac{4b}{a}}{2}\ge\sqrt{\frac{9a}{b}\cdot\frac{4b}{a}}$$

$$=\sqrt{36}$$

$$=6$$

Therefore,

$$\frac{9a}{b}+\frac{4b}{a}\ge12$$

Hence,

$$a+b\ge13+12$$

$$=25$$

Equality holds when

$$\frac{9a}{b}=\frac{4b}{a}$$

$$9a^2=4b^2$$

$$3a=2b$$

Thus the minimum value of

$$a+b$$

is

$$25$$

Given, $$2a+3b-c=0$$

Therefore,

$$c=2a+3b$$ Substitute in the line

$$ax+by+c=0$$

$$ax+by+(2a+3b)=0$$

Group the terms containing $$a$$ and $$b$$

$$a(x+2)+b(y+3)=0$$

Since $$a$$ and $$b$$ are arbitrary non-zero numbers satisfying the relation, the above equation must hold for all such values.

Hence,$$x+2=0$$ and $$y+3=0$$

Therefore the fixed point is $$P(-2,-3)$$

Thus,

$$h=-2,\quad k=-3$$

Hence,

$$h^2+k^2=(-2)^2+(-3)^2$$

$$=4+9$$

$$=13$$

Final Answer :

$$13$$