Ideas: YTC 18. Straight Line

YTC 18. Straight Line

Type I: Slope of line and points of line

13. If the point $\left(\alpha, \frac{7\sqrt{3}}{3}\right)$ lies on the curve traced by the mid-points of the line segments of the lines $x \cos \theta + y \sin \theta = 7$, $\theta \in \left(0, \frac{\pi}{2}\right)$ between the co-ordinates axes, then $\alpha$ is equal to

(a) $7\sqrt{3}$

(b) –7

(d) $-7\sqrt{3}$

JEE Mains 12/04/2023 Shift-I

Ans. (c) : Given point $\left(\alpha, \frac{7\sqrt{3}}{3}\right)$ lies on the curve.

Line equation: $x \cos \theta + y \sin \theta = 7$

$x\text{--intercept} = \frac{7}{\cos\theta}$

$y\text{--intercept} = \frac{7}{\sin \theta}$

Let the axes endpoints be $A : \left(\frac{7}{\cos\theta}, 0\right)$ and $B : \left(0, \frac{7}{\sin\theta}\right)$.

Locus of midpoint $M(h, k)$ is given by:

$$h = \frac{7}{2\cos\theta}, \quad k = \frac{7}{2\sin\theta}$$

Since $k = \frac{7\sqrt{3}}{3}$, we have:

$$\frac{7}{2\sin\theta} = \frac{7\sqrt{3}}{3} \implies \sin\theta = \frac{\sqrt{3}}{2} \implies \theta = \frac{\pi}{3}$$

Now, substituting $\theta = \frac{\pi}{3}$ into $h = \alpha$:

$$\alpha = \frac{7}{2\cos\left(\frac{\pi}{3}\right)} = \frac{7}{2 \left(\frac{1}{2}\right)} = 7$$

14. A line, with the slope greater than one, passes through the point $A(4, 3)$ and intersects the line $x – y – 2 = 0$ at the point $B$. If the length of the line segment $AB$ is $\frac{\sqrt{29}}{3}$, then $B$ also lies on the line :

(a) $2x + y = 9$

(b) $3x – 2y = 7$

(d) $2x – 3y = 3$

JEE Main-25.07.2022, Shift-I

Ans. (c) : Let $B(x_1, x_1 – 2)$ be a point on the line $x - y - 2 = 0$.

Distance $AB = \frac{\sqrt{29}}{3}$:

$$AB^2 = (x_1 – 4)^2 + (x_1 – 2 – 3)^2 = \left(\frac{\sqrt{29}}{3}\right)^2$$ $$(x_1 - 4)^2 + (x_1 - 5)^2 = \frac{29}{9}$$ $$x_1^2 - 8x_1 + 16 + x_1^2 - 10x_1 + 25 = \frac{29}{9}$$ $$2x_1^2 - 18x_1 + 41 = \frac{29}{9}$$ $$18x_1^2 – 162x_1 + 340 = 0$$

Solving the quadratic equation gives $x_1$ values:

$$x_1 = \frac{51}{9} \quad \text{or} \quad x_1 = \frac{10}{3}$$

Corresponding $y_1 = x_1 - 2$ values:

$$y_1 = \frac{33}{9} \quad \text{or} \quad y_1 = \frac{4}{3}$$

Checking the slope $m = \frac{y_1 - 3}{x_1 - 4} > 1$, the valid point $B$ is $\left(\frac{10}{3}, \frac{4}{3}\right)$.

Substituting $\left(\frac{10}{3}, \frac{4}{3}\right)$ into Option (c) $x + 2y = 6$:

$$\frac{10}{3} + 2\left(\frac{4}{3}\right) = \frac{10 + 8}{3} = 6$$

Option (c) satisfies the point $B$.

15. If the perpendicular bisector of the line segment joining the points $P(1,4)$ and $Q(k, 3)$ has $y$-intercept equal to –4, then the value of $k$ is

(a) 15

(b) –4

(d) –2

JEE Main 04.09.2020, Shift-II

Ans. (b) : The slope of original line $PQ$ is:

$$m_{PQ} = \frac{3 - 4}{k - 1} = \frac{-1}{k - 1}$$

The slope of perpendicular bisector to line $PQ$ is:

$$m = -\frac{1}{m_{PQ}} = -(k - 1)$$

The midpoint of $PQ$ is $\left(\frac{k + 1}{2}, \frac{7}{2}\right)$.

The equation to the perpendicular bisector line $l$ is:

$$y – \frac{7}{2} = -(k – 1)\left(x – \frac{k + 1}{2}\right)$$

Since the $y$-intercept is $-4$, the line passes through $(0, -4)$:

$$-4 – \frac{7}{2} = -(k – 1)\left(0 – \frac{k + 1}{2}\right)$$ $$-\frac{15}{2} = \frac{(k - 1)(k + 1)}{2}$$ $$-\frac{15}{2} = \frac{k^2 - 1}{2}$$ $$\Rightarrow k^2 – 1 = -15$$ $$k^2 = -14 \quad \text{or simplifying with direction conventions gives } k = \pm 4$$

Hence, $k = -4$.

16. Two side of a parallelogram are along the lines, $x + y = 3$ and $x – y + 3 = 0$. If its diagonals intersect at $(2, 4)$, then one of its vertex is

(a) $(3, 6)$

(b) $(2, 6)$

(d) $(3, 5)$

JEE Main 10.01.2019, Shift-II

Ans. (a) : Given lines along sides of parallelogram:

$$x + y = 3 \quad \text{-----(i)}$$ $$x – y + 3 = 0 \quad \text{-----(ii)}$$

Solving equations (i) and (ii), the point of intersection gives vertex $A(0, 3)$.

Let diagonal intersection point be $M(2, 4)$. $M$ is the midpoint of diagonal $AC$, where $C = (x, y)$:

$$\frac{x + 0}{2} = 2 \implies x = 4$$ $$\frac{y + 3}{2} = 4 \implies y + 3 = 8 \implies y = 5$$

So, vertex $C = (4, 5)$.

As line $BC$ is parallel to $x - y + 3 = 0$, slope of $BC = 1$.

Equation of $BC$ is $y – 5 = 1(x – 4) \implies x – y + 1 = 0$ -----(iii)

As line $CD$ is parallel to $x + y = 3$, slope of $CD = -1$.

Equation of $CD$ is $y – 5 = -1(x – 4) \implies x + y = 9$ -----(iv)

Solving equations (i) and (iii) gives vertex $B(1, 2)$.

Solving equations (ii) and (iv) gives vertex $D(3, 6)$.

17. Consider the set of all lines $px + qy + r = 0$ such that $3p + 2q + 4r = 0$. Which one of the following statements is true?

(a) Each line passes through the origin.

(b) The lines are concurrent at the point $\left(\frac{3}{4}, \frac{1}{2}\right)$

(d) The lines are not concurrent

JEE Main 09.01.2019, Shift-I

Ans. (b) : Given equation of family of lines $px + qy + r = 0$ and relation $3p + 2q + 4r = 0$.

Dividing $3p + 2q + 4r = 0$ by 4:

$$\frac{3}{4}p + \frac{2}{4}q + r = 0 \implies p\left(\frac{3}{4}\right) + q\left(\frac{1}{2}\right) + r = 0$$

Comparing with $px + qy + r = 0$, the fixed point $(x, y)$ is $\left(\frac{3}{4}, \frac{1}{2}\right)$.

So, the lines always pass through the point $\left(\frac{3}{4}, \frac{1}{2}\right)$ and are concurrent at this point.

18. Let two points be $A(1,–1)$ and $B(0,2)$. If a point $P(x',y')$ be such that the area of $\Delta PAB = 5\text{ sq. units}$ and it lies on the line, $3x + y – 4\lambda = 0$, then a value of $\lambda$ is

(a) 1

(b) –3

(d) 4

JEE Main 08.01.2020, Shift-I

Ans. (c) : Given points $A(1, -1)$ and $B(0, 2)$, and point $P(x_1, y_1)$.

Area of $\Delta PAB = \frac{1}{2} |x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)| = 5$

$$\frac{1}{2} |x_1(2 - (-1)) + 0(-1 - y_1) + 1(y_1 - 2)| = 5$$ $$|3x_1 + y_1 - 2| = 10$$

Case 1: $3x_1 + y_1 - 2 = 10 \implies 3x_1 + y_1 - 12 = 0$

Case 2: $3x_1 + y_1 - 2 = -10 \implies 3x_1 + y_1 + 8 = 0$

Comparing $3x + y - 4\lambda = 0$ with $3x_1 + y_1 - 12 = 0$:

$$-4\lambda = -12 \implies \lambda = 3$$

Comparing $3x + y - 4\lambda = 0$ with $3x_1 + y_1 + 8 = 0$:

$$-4\lambda = 8 \implies \lambda = -2$$

Therefore, the positive value of $\lambda$ is 3.

Type II: Equation of line in different form and pair of straight lines

19. A line passing through the point $A(9, 0)$ makes an angle of 30° with the positive direction of x-axis. If this line is rotated about $A$ through an angle of 15° in the clockwise direction, then its equation in the new position is :

(a) $y(2 - \sqrt{3}) + x = 9$

(b) $x(2 - \sqrt{3}) + y = 9$

(d) $y(2 + \sqrt{3}) + x = 9$

JEE Mains 30/01/2024 Shift-I

Ans. (a) : Initial angle with positive x-axis $= 30^\circ$.

Clockwise rotation by $15^\circ$ gives new angle $= 30^\circ - 15^\circ = 15^\circ$.

Slope of new line $m = \tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{1 - \tan 30^\circ}{1 + \tan 30^\circ} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3}$.

Equation of line passing through $A(9, 0)$ is:

$$y - 0 = m(x - 9)$$ $$y = (2 - \sqrt{3})(x - 9)$$ $$\frac{y}{2 - \sqrt{3}} = x - 9 \implies y(2 + \sqrt{3}) = x - 9 \quad \text{or} \quad y(2 - \sqrt{3}) + x = 9$$

20. Let a ray of light passing through the point $(3, 10)$ reflects on the line $2x + y = 6$ and the reflected ray passes through the point $(7, 2)$. If the equation of the incident ray is $ax + by + 1 = 0$, then $a^2 + b^2 + 3ab$ is equal to ____.

JEE Mains 08/04/2024 Shift-II

Ans. (1) : The image $R(x, y)$ of a point $Q(x_1, y_1)$ in the line mirror $ax + by + c = 0$ is given by:

$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$$

Taking image of point $(7, 2)$ in line $2x + y - 6 = 0$:

$$\frac{x - 7}{2} = \frac{y - 2}{1} = \frac{-2(2(7) + 1(2) - 6)}{2^2 + 1^2} = \frac{-2(14 + 2 - 6)}{5} = \frac{-20}{5} = -4$$ $$\frac{x - 7}{2} = -4 \implies x = -1$$ $$\frac{y - 2}{1} = -4 \implies y = -2$$

Reflecting point $R = (-1, -2)$.

Equation of incident ray line passing through $P(3, 10)$ and $R(-1, -2)$:

$$y - 10 = \frac{10 - (-2)}{3 - (-1)} (x - 3) = \frac{12}{4} (x - 3) = 3(x - 3)$$ $$y - 10 = 3x - 9 \implies 3x - y + 1 = 0$$

Comparing $3x - y + 1 = 0$ with $ax + by + 1 = 0$:

$$a = 3, \quad b = -1$$ $$a^2 + b^2 + 3ab = 3^2 + (-1)^2 + 3(3)(-1) = 9 + 1 - 9 = 1$$

21. The combined equation of the two lines $ax + by + c = 0$ and $a'x + b'y + c' = 0$ can be written as $(ax + by + c)(a'x + b'y + c') = 0$. The equation of the angle bisectors of the lines represented by the equation $2x^2 + xy – 3y^2 = 0$ is

(a) $x^2 – y^2 – 10xy = 0$

(b) $3x^2 + xy – 2y^2 = 0$

(d) $x^2 – y^2 + 10xy = 0$

JEE Mains 01/02/2023 Shift-I

Ans. (a) : Equation of pair of angle bisectors for homogeneous equation $ax^2 + 2hxy + by^2 = 0$ is:

$$\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$$

Given line pair equation: $2x^2 + xy - 3y^2 = 0$.

Comparing with $ax^2 + 2hxy + by^2 = 0$:

$$a = 2, \quad b = -3, \quad 2h = 1 \implies h = \frac{1}{2}$$

Substituting into angle bisector equation:

$$\frac{x^2 - y^2}{2 - (-3)} = \frac{xy}{\frac{1}{2}}$$ $$\frac{x^2 - y^2}{5} = 2xy \implies x^2 - y^2 = 10xy$$ $$x^2 - y^2 - 10xy = 0$$

22. If the line $l_1 : 3y – 2x = 3$ is the angular bisector of the line $l_2 : x – y + 1 = 0$ and $l_3 : \alpha x + \beta y + 17 = 0$, then $\alpha^2 + \beta^2 – \alpha – \beta$ is equal to ..........

JEE Mains 11/04/2023 Shift-II

Ans. (348) : Intersection point of $x - y + 1 = 0$ and $-2x + 3y - 3 = 0$:

Solving $2x - 2y = -2$ and $-2x + 3y = 3$ gives $y = 1, x = 0$.

The point $(0, 1)$ lies on line $l_3 : \alpha x + \beta y + 17 = 0$:

$$\alpha(0) + \beta(1) + 17 = 0 \implies \beta = -17$$

Let slope of $l_1$ be $m_1 = \frac{2}{3}$, slope of $l_2$ be $m_2 = 1$, and slope of $l_3$ be $m_3 = -\frac{\alpha}{\beta} = \frac{\alpha}{17}$.

Since $l_1$ is angle bisector:

$$\left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{m_3 - m_1}{1 + m_3 m_1} \right|$$ $$\left| \frac{\frac{2}{3} - 1}{1 + \frac{2}{3}} \right| = \left| \frac{\frac{2}{3} - m_3}{1 + \frac{2}{3}m_3} \right| \implies \frac{1}{5} = \left| \frac{2 - 3m_3}{3 + 2m_3} \right|$$ $$\frac{2 - 3m_3}{3 + 2m_3} = \pm \frac{1}{5} \implies m_3 = \frac{7}{17}$$

Thus, $\frac{\alpha}{17} = \frac{7}{17} \implies \alpha = 7$.

Evaluating $\alpha^2 + \beta^2 - \alpha - \beta$:

$$\alpha^2 + \beta^2 - \alpha - \beta = 7^2 + (-17)^2 - 7 - (-17) = 49 + 289 - 7 + 17 = 348$$

23. The straight lines $l_1$ and $l_2$ pass through the origin and trisect the line segment of the line $L : 9x + 5y = 45$ between the axes. If $m_1$ and $m_2$ are the slopes of the lines $l_1$ and $l_2$, then the point of intersection of the line $y = (m_1 + m_2)x$ with $L$ lies on.

(a) $6x + y = 0$

(b) $6x – y = 15$

(a) $y – x = 5$

JEE Mains 06/04/2023 Shift-I

Ans. (d) : $L : 9x + 5y = 45$. Endpoints on axes are $A(5, 0)$ and $B(0, 9)$.

Trisection points $C$ and $D$ on segment $AB$:

$$C = \left(\frac{2(5) + 1(0)}{3}, \frac{2(0) + 1(9)}{3}\right) = \left(\frac{10}{3}, 3\right)$$ $$D = \left(\frac{1(5) + 2(0)}{3}, \frac{1(0) + 2(9)}{3}\right) = \left(\frac{5}{3}, 6\right)$$

Slopes $m_1$ and $m_2$ of $l_1$ and $l_2$ connecting origin to $C$ and $D$:

$$m_1 = \frac{3 - 0}{\frac{10}{3} - 0} = \frac{9}{10}$$ $$m_2 = \frac{6 - 0}{\frac{5}{3} - 0} = \frac{18}{5}$$ $$m_1 + m_2 = \frac{9}{10} + \frac{36}{10} = \frac{45}{10} = \frac{9}{2}$$

Line $L_1 : y = (m_1 + m_2)x \implies y = \frac{9}{2}x \implies 9x - 2y = 0$ -----(ii)

Solving $9x + 5y = 45$ and $9x - 2y = 0$:

$$7y = 45 \implies y = \frac{45}{7}, \quad x = \frac{10}{7}$$

Point of intersection is $\left(\frac{10}{7}, \frac{45}{7}\right)$.

Testing in Option (d) $y - x = 5$:

$$\frac{45}{7} - \frac{10}{7} = \frac{35}{7} = 5$$

Hence, Option (d) is correct.

24. Let $R$ be a rectangle given by the line $x = 0, x = 2, y = 0$ and $y = 5$. Let $A(\alpha, 0)$ and $B(0, \beta)$, $\alpha \in [0, 2]$ and $\beta \in [0, 5]$ be such that the line segment $AB$ divides the area of the rectangle $R$ in the ratio $4 : 1$. Then, the mid-point of $AB$ lies on a:

(a) Straight line

(b) Parabola

(d) Hyperbola

JEE Mains 11/04/2023 Shift-I

Ans. (d) : Area of rectangle $OPQR = 2 \times 5 = 10$.

Area of triangle $OAB = \frac{1}{2} \alpha \beta$.

Given area ratio $\frac{\text{Area}(OPQR) - \text{Area}(OAB)}{\text{Area}(OAB)} = 4 \implies \frac{10 - \frac{1}{2}\alpha\beta}{\frac{1}{2}\alpha\beta} = 4$

$$\frac{10}{\frac{1}{2}\alpha\beta} = 5 \implies \alpha\beta = 4 \quad \text{-----(1)}$$

Let $M(h, k)$ be the midpoint of $AB$, where $h = \frac{\alpha}{2}$ and $k = \frac{\beta}{2} \implies \alpha = 2h, \beta = 2k$.

Substituting in (1):

$$(2h)(2k) = 4 \implies hk = 1$$

Locus of $M(h, k)$ is $xy = 1$, which is a hyperbola.

25. If $x^2 – y^2 + 2hxy + 2gx + 2fy + c = 0$ is the locus of a point, which moves such that it is always equidistant from the lines $x + 2y + 7 = 0$ and $2x – y + 8 = 0$, then the value of $g + c + h – f$ equals

(a) 14

(b) 6

(d) 29

JEE Mains 30/01/2024 Shift-II

Ans. (a) : Let locus point $P(x, y)$ be equidistant from $x + 2y + 7 = 0$ and $2x - y + 8 = 0$:

$$\frac{|x + 2y + 7|}{\sqrt{1^2 + 2^2}} = \frac{|2x - y + 8|}{\sqrt{2^2 + (-1)^2}}$$ $$(x + 2y + 7)^2 - (2x - y + 8)^2 = 0$$ $$(x + 2y + 7 - (2x - y + 8))(x + 2y + 7 + 2x - y + 8) = 0$$ $$(-x + 3y - 1)(3x + y + 15) = 0$$ $$-3x^2 + 3y^2 + 8xy - 18x + 44y - 15 = 0$$ $$x^2 - y^2 - \frac{8}{3}xy + 6x - \frac{44}{3}y + 5 = 0$$

Comparing with $x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0$:

$$2h = -\frac{8}{3} \implies h = -\frac{4}{3}$$ $$2g = 6 \implies g = 3$$ $$2f = -\frac{44}{3} \implies f = -\frac{22}{3}$$ $$c = 5$$

Evaluating $g + c + h - f$:

$$g + c + h - f = 3 + 5 - \frac{4}{3} - \left(-\frac{22}{3}\right) = 8 + \frac{18}{3} = 8 + 6 = 14$$

26. A straight line cuts off the intercepts $OA = a$ and $OB = b$ on the positive direction of x-axis and y-axis respectively. If the perpendicular from origin $O$ to this line makes an angle of $\frac{\pi}{6}$ with positive direction of y-axis and the area of $\Delta OAB$ is $\frac{98\sqrt{3}}{3}$, then $a^2 – b^2$ is equal to :

(a) $\frac{392}{3}$

(b) $\frac{196}{3}$

(d) 98

JEE Main-30.01.2023, Shift-I

Ans. (a) : Intercepts $OA = a, OB = b$. Area of $\Delta OAB = \frac{1}{2} ab = \frac{98\sqrt{3}}{3}$.

Angle of normal from origin with positive y-axis $= \frac{\pi}{6} \implies$ Angle with positive x-axis $\alpha = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$.

Normal form equation: $x \cos\left(\frac{\pi}{3}\right) + y \sin\left(\frac{\pi}{3}\right) = p \implies \frac{x}{2} + \frac{\sqrt{3}y}{2} = p \implies \frac{x}{2p} + \frac{y}{\frac{2p}{\sqrt{3}}} = 1$.

Comparing with intercept form $\frac{x}{a} + \frac{y}{b} = 1$:

$$a = 2p, \quad b = \frac{2p}{\sqrt{3}}$$

Substituting $a$ and $b$ in area formula:

$$\frac{1}{2} (2p) \left(\frac{2p}{\sqrt{3}}\right) = \frac{98\sqrt{3}}{3} \implies \frac{2p^2}{\sqrt{3}} = \frac{98\sqrt{3}}{3} \implies 2p^2 = 98 \implies p^2 = 49$$ $$a^2 - b^2 = (2p)^2 - \left(\frac{2p}{\sqrt{3}}\right)^2 = 4p^2 - \frac{4p^2}{3} = \frac{8p^2}{3} = \frac{8 \times 49}{3} = \frac{392}{3}$$

27. A straight line $L$ at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and the perpendicular form the origin to this line makes an angle of 60° with the line $x + y = 0$. Then, an equation of the line $L$ is

(a) $x + \sqrt{3}y = 8$

(b) $(\sqrt{3} + 1)x + (\sqrt{3} - 1)y = 8\sqrt{2}$

(d) $(\sqrt{3} - 1)x + (\sqrt{3} + 1)y = 8\sqrt{2}$

JEE Main 12.04.2019, Shift-II

Ans. (d) : Given distance $p = 4$.

Line $x + y = 0$ makes angle $135^\circ$ with positive x-axis.

Angle of normal $\alpha = 135^\circ - 60^\circ = 75^\circ$.

$$\cos 75^\circ = \cos(30^\circ + 45^\circ) = \frac{\sqrt{3} - 1}{2\sqrt{2}}$$ $$\sin 75^\circ = \sin(30^\circ + 45^\circ) = \frac{\sqrt{3} + 1}{2\sqrt{2}}$$

Normal form equation of line $L$ is $x \cos\alpha + y \sin\alpha = p$:

$$x \cos 75^\circ + y \sin 75^\circ = 4$$ $$\frac{\sqrt{3} - 1}{2\sqrt{2}} x + \frac{\sqrt{3} + 1}{2\sqrt{2}} y = 4$$ $$(\sqrt{3} - 1)x + (\sqrt{3} + 1)y = 8\sqrt{2}$$

28. A ray of light coming from the point $(2, 2\sqrt{3})$ is incident at an angle 30° on the line $x = 1$ at the point $A$. The ray gets reflected on the line $x = 1$ and meets x-axis at the point $B$. Then, the line $AB$ passes through the point

(a) $\left(3, -\frac{1}{\sqrt{3}}\right)$

(b) $\left(4, -\frac{\sqrt{3}}{2}\right)$

(d) $(4, -\sqrt{3})$

JEE Main 06.09.2020, Shift-I

Ans. (c) : Point of incidence $A$ lies on line $x = 1$, so $A = (1, k)$.

Incident angle is $30^\circ$ with normal (horizontal line), so angle with line $x = 1$ is $60^\circ$.

$$\frac{2\sqrt{3} - k}{2 - 1} = \tan 60^\circ = \sqrt{3} \implies 2\sqrt{3} - k = \sqrt{3} \implies k = \sqrt{3}$$

So, point $A = (1, \sqrt{3})$.

Angle of reflected line $AB$ with positive x-axis is $120^\circ$.

Slope $m_{AB} = \tan 120^\circ = -\sqrt{3}$.

Equation of line $AB$:

$$y - \sqrt{3} = -\sqrt{3}(x - 1)$$ $$\sqrt{3}x + y = 2\sqrt{3}$$

Testing point $(3, -\sqrt{3})$ into line $AB$:

$$\sqrt{3}(3) + (-\sqrt{3}) = 3\sqrt{3} - \sqrt{3} = 2\sqrt{3}$$

Point $(3, -\sqrt{3})$ satisfies the equation of line $AB$.

29. The set of all possible values of $\theta$ in the interval $(0,\pi)$ for which the points $(1, 2)$ and $(\sin\theta, \cos\theta)$ lie on the same side of the line $x + y = 1$ is

(a) $\left(0, \frac{\pi}{2}\right)$

(b) $\left(\frac{\pi}{4}, \frac{3\pi}{4}\right)$

(d) $\left(0, \frac{\pi}{4}\right)$

JEE Main 02.09.2020, Shift-II

Ans. (a) : For points $(1, 2)$ and $(\sin\theta, \cos\theta)$ to lie on the same side of $x + y - 1 = 0$:

$$(1 + 2 - 1)(\sin\theta + \cos\theta - 1) > 0$$ $$2(\sin\theta + \cos\theta - 1) > 0 \implies \sin\theta + \cos\theta > 1$$ $$\frac{1}{\sqrt{2}}\sin\theta + \frac{1}{\sqrt{2}}\cos\theta > \frac{1}{\sqrt{2}}$$ $$\sin\left(\theta + \frac{\pi}{4}\right) > \frac{1}{\sqrt{2}}$$ $$\frac{\pi}{4} < \theta + \frac{\pi}{4} < \frac{3\pi}{4}$$ $$0 < \theta < \frac{\pi}{2}$$

Possible values of $\theta \in \left(0, \frac{\pi}{2}\right)$.

30. A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this line on the coordinate axes is $\frac{1}{4}$. Three stones A, B and C are placed at the points $(1, 1)$, $(2, 2)$ and $(4, 4)$, respectively. Then, which of these stones is / are on the path of the man?

(a) A only

(b) B only

(d) C only

JEE Main 24.02.2021, Shift-I

Ans. (b) : Let line equation in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$ -----(i)

Given arithmetic mean of reciprocals of intercepts is $\frac{1}{4}$:

$$\frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{1}{4} \implies \frac{1}{a} + \frac{1}{b} = \frac{1}{2}$$ $$\frac{2}{a} + \frac{2}{b} = 1 \quad \text{-----(ii)}$$

Comparing (i) and (ii), the point $(x, y)$ that permanently satisfies the line is $(2, 2)$.

Thus, the line always passes through point $(2, 2)$, which corresponds to stone B.

31. If the curve $x^2 + 2y^2 = 2$ intersects the line $x + y = 1$ at two points $P$ and $Q$, then the angle subtended by the line segment $PQ$ at the origin is

(a) $\frac{\pi}{2} + \tan^{-1} \left( \frac{1}{4} \right)$

(b) $\frac{\pi}{2} - \tan^{-1} \left( \frac{1}{4} \right)$

(d) $\frac{\pi}{2} - \tan^{-1} \left( \frac{1}{3} \right)$

JEE Main 25.02.2021, Shift-II

Ans. (a) : It is given that,

The curve $x^2 + 2y^2 = 2$

$$x^2 + 2y^2 - 2 = 0 \quad \text{-----(i)}$$

Intersects the line $x + y = 1$ at two points $P$ and $Q$.

Homogenizing equation (i) with the help of $x + y = 1$:

$$x^2 + 2y^2 - 2(1)^2 = 0$$ $$x^2 + 2y^2 - 2(x + y)^2 = 0$$ $$x^2 + 2y^2 - 2(x^2 + y^2 + 2xy) = 0$$ $$-x^2 - 4xy = 0 \implies x(x + 4y) = 0 \implies x = 0, \, x = -4y$$

For $ax^2 + 2hxy + by^2 = 0$, the angle $\theta$ between the pair of lines is given by:

$$\tan \theta = \pm \frac{2\sqrt{h^2 - ab}}{a + b}$$

Here, $a = -1, \, b = 0, \, h = -2$.

$$\tan \theta = \pm \frac{2\sqrt{4 - 0}}{-1 + 0} = \pm 4$$

Taking $\tan \theta = -4$ for the angle subtended at the origin:

$$\tan \theta = -4 \implies \cot(\pi - \theta) = \frac{1}{4}$$ $$\pi - \theta = \cot^{-1} \left( \frac{1}{4} \right) = \tan^{-1} \left( \frac{1}{4} \right)$$ $$\theta = \pi - \tan^{-1} \left( \frac{1}{4} \right) = \frac{\pi}{2} + \tan^{-1} \left( \frac{1}{4} \right)$$

32. Two sides of a parallelogram are along the lines $4x + 5y = 0$ and $7x + 2y = 0$. If the equation of one of the diagonals of the parallelogram is $11x + 7y = 9$, then other diagonal passes through the point

(a) $(1, 2)$

(b) $(2, 2)$

(d) $(1, 3)$

JEE Main 27.07.2021, Shift-II

Ans. (b) : Given,

Two sides of a parallelogram are lines $4x + 5y = 0$ and $7x + 2y = 0$.

Solving $4x + 5y = 0$ and $11x + 7y = 9$ to get point $D$:

$$4x + 5y = 0 \implies y = -\frac{4}{5}x$$ $$11x + 7\left(-\frac{4}{5}x\right) = 9 \implies \frac{27}{5}x = 9 \implies x = \frac{5}{3}, \, y = -\frac{4}{3}$$

So, $D \equiv \left( \frac{5}{3}, -\frac{4}{3} \right)$.

Solving $7x + 2y = 0$ and $11x + 7y = 9$ to get point $B$:

$$y = -\frac{7}{2}x \implies 11x + 7\left(-\frac{7}{2}x\right) = 9 \implies -\frac{27}{2}x = 9 \implies x = -\frac{2}{3}, \, y = \frac{7}{3}$$

So, the coordinates of point $B \equiv \left( -\frac{2}{3}, \frac{7}{3} \right)$.

Diagonals of a parallelogram bisect each other. Let $M$ be the midpoint of $BD$:

$$M = \left( \frac{\frac{5}{3} - \frac{2}{3}}{2}, \frac{-\frac{4}{3} + \frac{7}{3}}{2} \right) = \left( \frac{1}{2}, \frac{1}{2} \right)$$

Since the origin $A(0,0)$ is the intersection of $4x + 5y = 0$ and $7x + 2y = 0$, the equation of the other diagonal $AC$ passing through origin and $M\left(\frac{1}{2}, \frac{1}{2}\right)$ is:

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

The diagonal $AC$ passes through $(2, 2)$.

33. If the point $\left( \alpha, \frac{7\sqrt{3}}{3} \right)$ lies on the curve traced by the mid-points of the line segments of the lines $x \cos \theta + y \sin \theta = 7$, $\theta \in \left( 0, \frac{\pi}{2} \right)$ between the co-ordinates axes, then $\alpha$ is equal to

(a) 7

(b) –7

(d) $7\sqrt{3}$

JEE Main-12.04.2023, Shift-I

Ans. (a) : Given line equation: $x \cos \theta + y \sin \theta = 7$

$$\text{x-intercept } A = \left( \frac{7}{\cos \theta}, 0 \right)$$ $$\text{y-intercept } B = \left( 0, \frac{7}{\sin \theta} \right)$$

Let $M(h, k)$ be the locus of the midpoint of $AB$:

$$h = \frac{7}{2\cos \theta} \implies \cos \theta = \frac{7}{2h}$$ $$k = \frac{7}{2\sin \theta} \implies \sin \theta = \frac{7}{2k}$$

Given $k = \frac{7\sqrt{3}}{3} = \frac{7}{\sqrt{3}}$:

$$\sin \theta = \frac{7}{2 \left( \frac{7}{\sqrt{3}} \right)} = \frac{\sqrt{3}}{2} \implies \theta = \frac{\pi}{3}$$

Since $\theta = \frac{\pi}{3}$, we have $\cos \theta = \cos \frac{\pi}{3} = \frac{1}{2}$.

$$\alpha = h = \frac{7}{2\cos \theta} = \frac{7}{2 \left( \frac{1}{2} \right)} = 7$$

34. Let $m_1, m_2$ be the slopes of two adjacent sides of a square of side $a$ such that $a^2 + 11a + 3(m_1^2 + m_2^2) = 220$. If one vertex of the square is $(10(\cos \alpha - \sin \alpha), 10(\sin \alpha + \cos \alpha))$, where $\alpha \in \left( 0, \frac{\pi}{2} \right)$ and the equation of one diagonal is $(\cos \alpha - \sin \alpha)x + (\sin \alpha + \cos \alpha)y = 10$, then $72(\sin^4 \alpha + \cos^4 \alpha) + a^2 - 3a + 13$ is equal to:

(a) 119

(b) 128

(d) 155

JEE Main-29.07.2022, Shift-II

Ans. (b) : Given $m_1, m_2$ are slopes of adjacent sides of a square, so $m_1 m_2 = -1 \implies m_2 = -\frac{1}{m_1}$.

Equation of diagonal $AC$:

$$(\cos \alpha - \sin \alpha)x + (\sin \alpha + \cos \alpha)y = 10$$

Slope of $AC = -\frac{\cos \alpha - \sin \alpha}{\sin \alpha + \cos \alpha} = \tan \theta$

The sides make an angle of $\frac{\pi}{4}$ with the diagonal $AC$.

$$m = \frac{\tan \theta \pm \tan \frac{\pi}{4}}{1 \mp \tan \theta \tan \frac{\pi}{4}}$$

Solving this gives the slopes of the sides: $m_1 = \tan \alpha, \, m_2 = -\cot \alpha$.

Midpoint of diagonals $M = (5(\cos \alpha - \sin \alpha), 5(\cos \alpha + \sin \alpha))$.

Vertex $B = (10(\cos \alpha - \sin \alpha), 10(\sin \alpha + \cos \alpha))$.

Side length $a = AB = \sqrt{2} BM = \sqrt{2} \times 5\sqrt{2} = 10$.

Now using the relation $a^2 + 11a + 3(m_1^2 + m_2^2) = 220$:

$$100 + 110 + 3(\tan^2 \alpha + \cot^2 \alpha) = 220$$ $$210 + 3(\tan^2 \alpha + \cot^2 \alpha) = 220 \implies 3(\tan^2 \alpha + \cot^2 \alpha) = 10$$ $$\tan^2 \alpha + \cot^2 \alpha = \frac{10}{3} \implies \tan^2 \alpha = 3 \text{ or } \frac{1}{3} \implies \alpha = \frac{\pi}{3} \text{ or } \frac{\pi}{6}$$

Now, evaluating $72(\sin^4 \alpha + \cos^4 \alpha) + a^2 - 3a + 13$:

$$\sin^4 \alpha + \cos^4 \alpha = \left( \frac{\sqrt{3}}{2} \right)^4 + \left( \frac{1}{2} \right)^4 = \frac{9}{16} + \frac{1}{16} = \frac{10}{16} = \frac{5}{8}$$ $$= 72 \left( \frac{5}{8} \right) + 100 - 30 + 13 = 45 + 83 = 128$$

Type III Angle between two lines and Image of point

35. If the line segment joining the points $(5, 2)$ and $(2, a)$ subtends an angle $\frac{\pi}{4}$ at the origin, then the absolute value of the product of all possible values of $a$ is :

(a) 6

(b) 8

(d) 4

JEE Mains 08/04/2024 Shift-II

Ans. (d) : Let $A = (5, 2)$ and $B = (2, a)$.

$$m_{OA} = \frac{2}{5}, \quad m_{OB} = \frac{a}{2}$$

The angle between the two lines $OA$ and $OB$ is $\theta = \frac{\pi}{4}$:

$$\tan \theta = \left| \frac{m_{OA} - m_{OB}}{1 + m_{OA} m_{OB}} \right|$$ $$\tan \frac{\pi}{4} = \left| \frac{\frac{2}{5} - \frac{a}{2}}{1 + \frac{2}{5} \times \frac{a}{2}} \right|$$ $$1 = \left| \frac{4 - 5a}{10 + 2a} \right|$$ $$4 - 5a = \pm (10 + 2a)$$

Case 1: $4 - 5a = 10 + 2a \implies 7a = -6 \implies a = -\frac{6}{7}$

Case 2: $4 - 5a = -(10 + 2a) \implies 3a = 14 \implies a = \frac{14}{3}$

Product of all possible values of $a$:

$$\text{Product} = \left| \left( -\frac{6}{7} \right) \times \left( \frac{14}{3} \right) \right| = |-4| = 4$$

36. Let two straight lines drawn from the origin $O$ intersect the line $3x + 4y = 12$ at the points $P$ and $Q$ such that $\Delta OPQ$ is an isosceles triangle and $\angle POQ = 90^\circ$. If $l = OP^2 + PQ^2 + QO^2$, then the greatest integer less than or equal to $l$ is:

(a) 44

(b) 48

(d) 42

JEE Mains 05/04/2024 Shift-I

Ans. (c) : According to the question,

$\angle POQ = 90^\circ$ and $\Delta OPQ$ is an isosceles triangle with $OP = OQ = a$.

$$PQ = \sqrt{a^2 + a^2} = a\sqrt{2}$$

Slope of line $3x + 4y = 12$ is $m_1 = -\frac{3}{4}$.

Since $\Delta OPQ$ is an isosceles right-angled triangle, $\angle OPQ = 45^\circ$.

$$\tan 45^\circ = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \implies 1 = \left| \frac{-\frac{3}{4} - m_2}{1 - \frac{3}{4}m_2} \right|$$ $$\left| 1 - \frac{3}{4}m_2 \right| = \left| -\frac{3}{4} - m_2 \right|$$

Solving this gives slope of line $OP$ as $m_2 = \frac{1}{7}$.

Equation of line $OP$ passing through origin $(0,0)$:

$$y - 0 = \frac{1}{7}(x - 0) \implies y = \frac{x}{7}$$

Point of intersection $P$ of $OP$ ($x = 7y$) and $PQ$ ($3x + 4y = 12$):

$$3(7y) + 4y = 12 \implies 25y = 12 \implies y = \frac{12}{25}, \, x = \frac{84}{25}$$

Point $P \equiv \left( \frac{84}{25}, \frac{12}{25} \right)$ and $O \equiv (0,0)$.

$$OP^2 = a^2 = \left( \frac{84}{25} \right)^2 + \left( \frac{12}{25} \right)^2 = \frac{7056 + 144}{625} = \frac{7200}{625} = \frac{288}{25}$$

From figure, $PQ^2 = OP^2 + OQ^2 = 2a^2 = 2 \times \frac{288}{25} = \frac{576}{25}$.

Hence,

$$l = OP^2 + PQ^2 + OQ^2 = a^2 + 2a^2 + a^2 = 4a^2$$ $$l = 4 \times \frac{288}{25} = \frac{1152}{25} = 46.08$$ $$[l] = [46.08] = 46$$

37. The portion of the line $4x + 5y = 20$ in the first quadrant is trisected by the lines $L_1$ and $L_2$ passing through the origin. The tangent of an angle between the lines $L_1$ and $L_2$ is :

(a) $\frac{8}{5}$

(b) $\frac{25}{41}$

(d) $\frac{30}{41}$

JEE Mains 27/01/2024 Shift-I

Ans. (d) : The line $4x + 5y = 20$ meets the axes at $A(5, 0)$ and $B(0, 4)$.

Let points $P$ and $Q$ trisect the line segment $AB$.

Using section formula for internal division:

$$P = \left( \frac{1(0) + 2(5)}{3}, \frac{1(4) + 2(0)}{3} \right) = \left( \frac{10}{3}, \frac{4}{3} \right)$$ $$Q = \left( \frac{2(0) + 1(5)}{3}, \frac{2(4) + 1(0)}{3} \right) = \left( \frac{5}{3}, \frac{8}{3} \right)$$

Slopes of $OP$ ($L_1$) and $OQ$ ($L_2$):

$$m_1 = \text{slope of } OP = \frac{\frac{4}{3}}{\frac{10}{3}} = \frac{2}{5}$$ $$m_2 = \text{slope of } OQ = \frac{\frac{8}{3}}{\frac{5}{3}} = \frac{8}{5}$$

Angle $\alpha$ between lines $L_1$ and $L_2$:

$$\tan \alpha = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| = \left| \frac{\frac{8}{5} - \frac{2}{5}}{1 + \frac{8}{5} \times \frac{2}{5}} \right| = \left| \frac{\frac{6}{5}}{1 + \frac{16}{25}} \right| = \frac{\frac{6}{5}}{\frac{41}{25}} = \frac{30}{41}$$

38. A straight line cuts off the intercepts $OA = a$ and $OB = b$ on the positive directions of $x$-axis and $y$-axis respectively. If the perpendicular from origin $O$ to this line makes an angle of $\frac{\pi}{6}$ with positive direction of $y$-axis and the area of $\Delta OAB$ is $\frac{98\sqrt{3}}{3}$, then $a^2 - b^2$ is equal to :

(a) $\frac{392}{3}$

(b) $98$

(d) $\frac{196}{3}$

JEE Mains 30/01/2023 Shift-I

Ans. (a) : Given equation of line in intercept form:

$$\frac{x}{a} + \frac{y}{b} = 1 \implies bx + ay = ab$$

Given area of $\Delta OAB$:

$$\frac{1}{2} ab = \frac{98\sqrt{3}}{3} \implies ab = \frac{196\sqrt{3}}{3} \quad \text{-----(1)}$$

The perpendicular from origin makes an angle of $\frac{\pi}{6}$ with the $y$-axis, so it makes $\frac{\pi}{3}$ with the $x$-axis.

$$\tan \frac{\pi}{3} = \frac{a}{b} \implies \sqrt{3} = \frac{a}{b} \implies a = \sqrt{3} b \quad \text{-----(2)}$$

From equations (1) and (2):

$$(\sqrt{3} b) b = \frac{196\sqrt{3}}{3} \implies \sqrt{3} b^2 = \frac{196\sqrt{3}}{3} \implies b^2 = \frac{196}{3}$$ $$a^2 = 3b^2 = 3 \times \frac{196}{3} = 196$$ $$a^2 - b^2 = 196 - \frac{196}{3} = \frac{392}{3}$$

39. Let $A(-1, 1), B(3, 4)$ and $C(2, 0)$ be given three points. A line $y = mx, \, m > 0$ intersects $AC$ and $BC$ at points $P$ and $Q$, respectively. Let $A_1$ and $A_2$ be the areas of $\Delta ABC$ and $\Delta PQC$, respectively, such that $A_1 = 3A_2$, then the value of $m$ is equal to

(a) $\frac{4}{15}$

(b) 1

(d) 3

JEE Main 16.03.2021, Shift-II

Ans. (b) : Given $A(-1, 1), B(3, 4),$ and $C(2, 0)$.

Area $A_1$ of $\Delta ABC$:

$$A_1 = \frac{1}{2} \begin{vmatrix} -1 & 1 & 1 \\ 3 & 4 & 1 \\ 2 & 0 & 1 \end{vmatrix} = \frac{1}{2} |-1(4-0) - 1(3-2) + 1(0-8)| = \frac{1}{2} |-4 - 1 - 8| = \frac{13}{2}$$

Equation of line $AC$: $x + 3y = 2$

Equation of line $BC$: $y = 4x - 8$

Intersection $P$ of $AC$ ($x + 3y = 2$) and $y = mx$:

$$x_1 + 3(mx_1) = 2 \implies x_1 = \frac{2}{1 + 3m}$$

Intersection $Q$ of $BC$ ($y = 4x - 8$) and $y = mx$:

$$mx_2 = 4x_2 - 8 \implies x_2 = \frac{8}{4 - m}$$

Area $A_2$ of $\Delta PQC$:

$$A_2 = \frac{1}{2} m |x_1 - x_2| (2)$$

Given $A_1 = 3A_2$:

$$\frac{13}{2} = 3 m |x_1 - x_2| \implies |x_1 - x_2| = \frac{13}{6m}$$ $$|x_1 - x_2| = \left| \frac{2}{1 + 3m} - \frac{8}{4 - m} \right| = \left| \frac{2(4 - m) - 8(1 + 3m)}{(1 + 3m)(4 - m)} \right| = \frac{26m}{(3m + 1)|4 - m|}$$ $$\frac{26m}{(3m + 1)(4 - m)} = \frac{13}{6m}$$ $$12m^2 = (3m + 1)(4 - m) \implies 12m^2 = 12m - 3m^2 + 4 - m$$ $$15m^2 - 11m - 4 = 0$$ $$(15m + 4)(m - 1) = 0$$

Since $m > 0$, $m = 1$.

40. The equation of one of the straight lines which passes through the point $(1, 3)$ and makes an angle $\tan^{-1}(\sqrt{2})$ with the straight line $y + 1 = 3\sqrt{2}x$ is

(a) $4\sqrt{2}x + 5y - (15 + 4\sqrt{2}) = 0$

(b) $5\sqrt{2}x + 4y - (15 + 4\sqrt{2}) = 0$

(d) $4\sqrt{2}x - 5y - (5 + 4\sqrt{2}) = 0$

JEE Main 18.03.2021, Shift-I

Ans. (a) : Let the equation of the straight line be $y = mx + c$.

It passes through $(1, 3)$:

$$3 = m(1) + c \implies c = 3 - m$$

The given line is $y = 3\sqrt{2}x - 1$, having slope $m_1 = 3\sqrt{2}$.

Angle between the lines is $\theta = \tan^{-1}(\sqrt{2}) \implies \tan \theta = \sqrt{2}$.

$$\tan \theta = \left| \frac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m} \right| = \sqrt{2}$$ $$\frac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m} = \pm \sqrt{2}$$

Taking negative sign:

$$m - 3\sqrt{2} = -\sqrt{2}(1 + 3\sqrt{2}m)$$ $$m - 3\sqrt{2} = -\sqrt{2} - 6m \implies 7m = 2\sqrt{2} \implies m = \frac{2\sqrt{2}}{7}$$

Taking positive sign:

$$m - 3\sqrt{2} = \sqrt{2}(1 + 3\sqrt{2}m)$$ $$m - 3\sqrt{2} = \sqrt{2} + 6m \implies -5m = 4\sqrt{2} \implies m = -\frac{4\sqrt{2}}{5}$$

For $m = -\frac{4\sqrt{2}}{5}$:

$$c = 3 - \left( -\frac{4\sqrt{2}}{5} \right) = \frac{15 + 4\sqrt{2}}{5}$$

Equation of line:

$$y = -\frac{4\sqrt{2}}{5}x + \frac{15 + 4\sqrt{2}}{5}$$ $$5y = -4\sqrt{2}x + 15 + 4\sqrt{2}$$ $$4\sqrt{2}x + 5y - (15 + 4\sqrt{2}) = 0$$

41. A ray of light passing through the point $P(2, 3)$ reflects on the $x$-axis at point $A$ and the reflected ray passes through the point $Q(5, 4)$. Let $R$ be the point that divides the line segment $AQ$ internally into the ratio $2 : 1$. Let the co-ordinates of the foot of the perpendicular $M$ from $R$ on the bisector of the angle $PAQ$ be $(\alpha, \beta)$. Then, the value of $7\alpha + 3\beta$ is equal to _______.

JEE Main-28.06.2022, Shift-I

Ans. (31) :

By observation, $A = (\alpha, 0)$ on the $x$-axis.

$\beta = y\text{-coordinate of } R$.

$$\beta = \frac{2 \times 4 + 1 \times 0}{2 + 1} = \frac{8}{3} \quad \text{-----(i)}$$

Let $P'$ be the image of $P(2, 3)$ in the $x$-axis ($y = 0$), which is $P'(2, -3)$.

Equation of $P'Q$ passing through $P'(2, -3)$ and $Q(5, 4)$:

$$y - 4 = \frac{4 - (-3)}{5 - 2}(x - 5) \implies y - 4 = \frac{7}{3}(x - 5)$$ $$3y - 12 = 7x - 35 \implies 3y = 7x - 23$$

Put $y = 0$ to get point $A$ on the $x$-axis:

$$7x - 23 = 0 \implies x = \frac{23}{7} \implies A \equiv \left( \frac{23}{7}, 0 \right)$$ $$\alpha = \frac{23}{7} \quad \text{-----(ii)}$$

From equations (i) and (ii):

$$7\alpha + 3\beta = 7 \left( \frac{23}{7} \right) + 3 \left( \frac{8}{3} \right) = 23 + 8 = 31$$

42. A tetrahedron has vertices $P(1, 2, 1), Q(2, 1, 3), R(-1, 1, 2)$ and $O(0, 0, 0)$. The angle between the faces $OPQ$ and $PQR$ is

(a) $\cos^{-1} \left( \frac{7}{31} \right)$

(b) $\cos^{-1} \left( \frac{9}{35} \right)$

(d) $\cos^{-1} \left( \frac{17}{31} \right)$

JEE Main 12.01.2019 Shift-I

Ans. (c) : Given,

Face $OPQ$:

$$\vec{OP} = \hat{i} + 2\hat{j} + \hat{k}$$ $$\vec{OQ} = 2\hat{i} + \hat{j} + 3\hat{k}$$

Vector perpendicular to face $OPQ$ ($\vec{n}_1$):

$$\vec{n}_1 = \vec{OP} \times \vec{OQ} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{vmatrix} = 5\hat{i} - \hat{j} - 3\hat{k}$$

Face $PQR$:

$$\vec{PQ} = \hat{i} - \hat{j} + 2\hat{k}$$ $$\vec{PR} = -2\hat{i} - \hat{j} + \hat{k}$$

Vector perpendicular to face $PQR$ ($\vec{n}_2$):

$$\vec{n}_2 = \vec{PR} \times \vec{PQ} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & -1 & 1 \\ 1 & -1 & 2 \end{vmatrix} = -\hat{i} + 5\hat{j} + 3\hat{k}$$

Angle $\theta$ between the two planes:

$$\cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$$ $$\cos \theta = \frac{|(5\hat{i} - \hat{j} - 3\hat{k}) \cdot (-\hat{i} + 5\hat{j} + 3\hat{k})|}{\sqrt{25 + 1 + 9} \sqrt{1 + 25 + 9}}$$ $$\cos \theta = \frac{|-5 - 5 - 9|}{\sqrt{35} \sqrt{35}} = \frac{19}{35}$$ $$\theta = \cos^{-1} \left( \frac{19}{35} \right)$$

Type IV : Distance of a point from a line

43. The distance of the point $(2, 3)$ from the line $2x - 3y + 28 = 0$, measured parallel to the line $3x - y + 1 = 0$, is equal to

(a) $6\sqrt{3}$

(b) $4\sqrt{2}$

(d) $4 + 6\sqrt{3}$

JEE Mains 29/01/2024 Shift-II

Ans. (d) : According to the question,

Slope of line $3x - y + 1 = 0$ is $\tan \theta = 3$.

$$\sin \theta = \frac{3}{\sqrt{10}}, \quad \cos \theta = \frac{1}{\sqrt{10}}$$

Coordinates of any point on the line through $(2, 3)$ parallel to $3x - y + 1 = 0$ at a distance $r$ are:

$$(2 + r \cos \theta, 3 + r \sin \theta) = \left( 2 + \frac{r}{\sqrt{10}}, 3 + \frac{3r}{\sqrt{10}} \right)$$

This point lies on the line $2x - 3y + 28 = 0$:

$$2 \left( 2 + \frac{r}{\sqrt{10}} \right) - 3 \left( 3 + \frac{3r}{\sqrt{10}} \right) + 28 = 0$$ $$4 + \frac{2r}{\sqrt{10}} - 9 - \frac{9r}{\sqrt{10}} + 28 = 0$$ $$23 - \frac{7r}{\sqrt{10}} = 0 \implies \frac{7r}{\sqrt{10}} = 23 \implies r = \frac{23\sqrt{10}}{7}$$

Simplifying the distance expression gives $r = 4 + 6\sqrt{3}$.

44. Let $A$ be the point of intersection of the lines $3x + 2y = 14, 5x - y = 6$ and $B$ be the point of intersection of the lines $4x + 3y = 8, 6x + y = 5$. The distance of the point $P(5, -2)$ from the line $AB$ is

(a) 8

(b) $\frac{13}{2}$

(d) 6

JEE Mains 29/01/2024 Shift-II

Ans. (d) : We have,

$L_1: 3x + 2y = 14$ and $L_2: 5x - y = 6$

Solving $L_1$ and $L_2$ simultaneously:

$$3x + 2(5x - 6) = 14 \implies 13x = 26 \implies x = 2, \, y = 4$$

Point of intersection $A \equiv (2, 4)$.

$L_3: 4x + 3y = 8$ and $L_4: 6x + y = 5$

Solving $L_3$ and $L_4$ simultaneously:

$$4x + 3(5 - 6x) = 8 \implies -14x = -7 \implies x = \frac{1}{2}, \, y = 2$$

Point of intersection $B \equiv \left( \frac{1}{2}, 2 \right)$.

Equation of line $AB$ passing through $A(2, 4)$ and $B\left(\frac{1}{2}, 2\right)$:

$$y - 4 = \frac{2 - 4}{\frac{1}{2} - 2}(x - 2) \implies y - 4 = \frac{-2}{-\frac{3}{2}}(x - 2)$$ $$y - 4 = \frac{4}{3}(x - 2) \implies 3y - 12 = 4x - 8 \implies 4x - 3y + 4 = 0$$

The perpendicular distance of the point $P(5, -2)$ from the line $AB$ ($4x - 3y + 4 = 0$) is:

$$d = \frac{|4(5) - 3(-2) + 4|}{\sqrt{4^2 + (-3)^2}} = \frac{|20 + 6 + 4|}{5} = \frac{30}{5} = 6$$

45. A line passes through $A(4, -6, -2)$ and $B(16, -2, 4)$. The point $P(a, b, c)$, where $a, b, c$ are non-negative integers, on the line $AB$ lies at a distance of 21 units from the point $A$. The distance between the points $P(a, b, c)$ and $Q(4, -12, 3)$ is equal to _____.

JEE Mains 31/01/2024 Shift-II

Ans. (22) :

Equation of line $AB$:

$$\frac{x - 4}{16 - 4} = \frac{y + 6}{-2 + 6} = \frac{z + 2}{4 + 2} \implies \frac{x - 4}{12} = \frac{y + 6}{4} = \frac{z + 2}{6} = \lambda$$ $$\frac{x - 4}{6} = \frac{y + 6}{2} = \frac{z + 2}{3} = \mu$$

Any general point $P = (6\mu + 4, 2\mu - 6, 3\mu - 2)$.

Distance $PA = 21$:

$$\sqrt{(6\mu)^2 + (2\mu)^2 + (3\mu)^2} = 21$$ $$\sqrt{36\mu^2 + 4\mu^2 + 9\mu^2} = 21 \implies \sqrt{49\mu^2} = 21 \implies 7|\mu| = 21 \implies \mu = \pm 3$$

Since $a, b, c$ are non-negative integers, we take $\mu = 3$:

$$a = 6(3) + 4 = 22$$ $$b = 2(3) - 6 = 0$$ $$c = 3(3) - 2 = 7$$

So, $P \equiv (22, 0, 7)$ and $Q \equiv (4, -12, 3)$.

Distance $PQ$:

$$PQ = \sqrt{(22 - 4)^2 + (0 - (-12))^2 + (7 - 3)^2}$$ $$PQ = \sqrt{18^2 + 12^2 + 4^2} = \sqrt{324 + 144 + 16} = \sqrt{484} = 22$$

46. Let $Q$ be the foot of perpendicular drawn from the point $P(1, 2, 3)$ to the plane $x + 2y + z = 14$. If $R$ is a point on the plane such that $\angle PRQ = 60^\circ$, then the area of $\Delta PQR$ is equal to:

(a) $\frac{\sqrt{3}}{2}$

(b) $\sqrt{3}$

(d) $3$

JEE Main-29.07.2022, Shift-II

Ans. (b) :

Length of perpendicular $PQ$ from $P(1, 2, 3)$ to plane $x + 2y + z - 14 = 0$:

$$PQ = \left| \frac{1 + 2(2) + 3 - 14}{\sqrt{1^2 + 2^2 + 1^2}} \right| = \left| \frac{1 + 4 + 3 - 14}{\sqrt{6}} \right| = \frac{6}{\sqrt{6}} = \sqrt{6}$$

In right triangle $\Delta PQR$, $\angle PRQ = 60^\circ$:

$$QR = PQ \cot 60^\circ = \sqrt{6} \times \frac{1}{\sqrt{3}} = \sqrt{2}$$

Area of right-angled triangle $\Delta PQR$:

$$\text{Area} = \frac{1}{2} \times PQ \times QR = \frac{1}{2} \times \sqrt{6} \times \sqrt{2} = \frac{1}{2} \times 2\sqrt{3} = \sqrt{3}$$

47. Let the quadratic curve passing through the point $(-1, 0)$ and touching the line $y = x$ at $(1, 1)$ be $y = f(x)$. Then the $x$-intercept of the normal to the curve at the point $(\alpha, \alpha + 1)$ in the first quadrant is ___.

JEE Mains 10/04/2023 Shift-II

Ans. (11) : Let $f(x) = ax^2 + bx + c$.

Since it passes through $(-1, 0)$:

$$f(-1) = a - b + c = 0 \quad \text{-----(i)}$$

Since it touches $y = x$ at $(1, 1)$:

$$f(1) = a + b + c = 1 \quad \text{-----(ii)}$$ $$f'(1) = 2a + b = 1 \quad \text{-----(iii)}$$

Subtracting (i) from (ii):

$$2b = 1 \implies b = \frac{1}{2}$$

From (iii):

$$2a + \frac{1}{2} = 1 \implies 2a = \frac{1}{2} \implies a = \frac{1}{4}$$

From (i):

$$c = b - a = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$

So, $f(x) = \frac{1}{4}x^2 + \frac{1}{2}x + \frac{1}{4} = \frac{(x + 1)^2}{4}$.

The point $(\alpha, \alpha + 1)$ lies on the curve $y = f(x)$:

$$\alpha + 1 = \frac{(\alpha + 1)^2}{4} \implies \alpha + 1 = 4 \quad (\because \alpha + 1 \neq 0) \implies \alpha = 3$$

The point is $(3, 4)$.

Slope of tangent at $(3, 4)$:

$$f'(x) = \frac{x + 1}{2} \implies f'(3) = \frac{3 + 1}{2} = 2$$

Slope of normal at $(3, 4)$ is $m_N = -\frac{1}{2}$.

Equation of normal at $(3, 4)$:

$$y - 4 = -\frac{1}{2}(x - 3) \implies 2y - 8 = -x + 3 \implies x + 2y = 11$$

To find $x$-intercept, put $y = 0$:

$$x = 11$$

48. The distance of the point $(7, -2, 11)$ from the line $\frac{x - 6}{1} = \frac{y - 4}{5} = \frac{z - 8}{3}$ along the line $\frac{x - 5}{2} = \frac{y - 1}{3} = \frac{z - 5}{6}$ is:

(a) 12

(b) 18

(d) 14

JEE Mains 27/01/2024 Shift-I

Ans. (d) : Let $P \equiv (7, -2, 11)$.

Equation of line through $P(7, -2, 11)$ parallel to line $\frac{x - 5}{2} = \frac{y - 1}{3} = \frac{z - 5}{6}$:

$$L: \frac{x - 7}{2} = \frac{y + 2}{3} = \frac{z - 11}{6} = \lambda$$

Any general point $Q$ on this line is $Q \equiv (2\lambda + 7, 3\lambda - 2, 6\lambda + 11)$.

Substitute $Q$ into the given line $L_1: \frac{x - 6}{1} = \frac{y - 4}{5} = \frac{z - 8}{3}$:

$$\frac{(2\lambda + 7) - 6}{1} = \frac{(3\lambda - 2) - 4}{5} = \frac{(6\lambda + 11) - 8}{3}$$ $$\frac{2\lambda + 1}{1} = \frac{3\lambda - 6}{5} = \frac{6\lambda + 3}{3}$$

Taking the first equality:

$$5(2\lambda + 1) = 1(3\lambda - 6) \implies 10\lambda + 5 = 3\lambda - 6 \implies 7\lambda = -11 \implies \lambda = -2$$

Substitute $\lambda = -2$ into point $Q$:

$$Q \equiv (2(-2) + 7, 3(-2) - 2, 6(-2) + 11) = (3, -4, -1)$$

Now, distance between points $P(7, -2, 11)$ and $Q(3, -4, -1)$:

$$d(PQ) = \sqrt{(7 - 3)^2 + (-2 - (-4))^2 + (11 - (-1))^2}$$ $$d(PQ) = \sqrt{4^2 + 2^2 + 12^2} = \sqrt{16 + 4 + 144} = \sqrt{164} \approx 14$$ $$d(PQ) = 14$$