Straight Lines And Pair Of Straight Lines – JEE Main MathematicsPractice Questions & PYQs

The intercept form of a straight line is expressed as $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ and $b$ represent the x and y-intercepts respectively. Given that the sum of these intercepts is $-1$, we can relate the variables as $a + b = -1$, or $b = -1 - a$. Since the line passes through the point $(4, 3)$, these values must satisfy the equation $\frac{4}{a} + \frac{3}{b} = 1$. Substituting $b = -1 - a$ into this expression leads to $\frac{4}{a} + \frac{3}{-1 - a} = 1$. By multiplying the entire equation by $a(-1 - a)$, we obtain $4(-1 - a) + 3a = a(-1 - a)$, which further simplifies to $-4 - 4a + 3a = -a - a^2$. Rearranging the terms results in the quadratic equation $a^2 - 4 = 0$, which provides two possible solutions for the x-intercept: $a = 2$ and $a = -2$. When $a = 2$, the y-intercept is $b = -1 - 2 = -3$, yielding the line $\frac{x}{2} + \frac{y}{-3} = 1$, which is equivalently written as $\frac{x}{2} - \frac{y}{3} = 1$. When $a = -2$, the y-intercept is $b = -1 - (-2) = 1$, resulting in the line $\frac{x}{-2} + \frac{y}{1} = 1$.

Any homogeneous equation of the second degree in $x$ and $y$ of the form $Ax^2 + 2Hxy + By^2 = 0$ represents a pair of straight lines passing through the origin, where the slopes $m_1$ and $m_2$ satisfy $m_1+m_2 = -\frac{2H}{B}$ and $m_1m_2 = \frac{A}{B}$. Applying this to the equation $x^2 - 2cxy - 7y^2 = 0$, we identify $A=1$, $2H=-2c$, and $B=-7$, which yields the sum of slopes as $m_1+m_2 = -\frac{2c}{7}$ and the product as $m_1m_2 = -\frac{1}{7}$. Given that the sum of the slopes is four times their product, we equate these expressions as $-\frac{2c}{7} = 4\left(-\frac{1}{7}\right)$. Solving this linear equation by multiplying both sides by $-7$ leads to $2c = 4$, resulting in $c=2$.

A homogeneous quadratic equation representing a pair of lines must be satisfied by every coordinate pair $(x, y)$ that lies on either line, meaning that any point on the line $3x + 4y = 0$ must also satisfy the joint equation $6x^2 - xy + 4cy^2 = 0$. By expressing the line as $y = -\frac{3}{4}x$ and substituting this directly into the pair equation, we obtain:
$6x^2 - x\left(-\frac{3}{4}x\right) + 4c\left(-\frac{3}{4}x\right)^2 = 0$
Simplifying this expression and dividing by $x^2$ yields the following relation:
$6 + \frac{3}{4} + \frac{9c}{4} = 0$
Multiplying this entire equation by $4$ produces $24 + 3 + 9c = 0$, which simplifies to $9c = -27$ and ultimately gives $c = -3$.

When a pair of lines defined by $Ax^2 + 2Hxy + By^2 = 0$ is given, the equation representing their angle bisectors is expressed as $\frac{x^2 - y^2}{A - B} = \frac{xy}{H}$. Applying this principle to the second set of lines, $x^2 - 2qxy - y^2 = 0$, we identify $A=1$, $H=-q$, and $B=-1$, which yields $\frac{x^2 - y^2}{1 - (-1)} = \frac{xy}{-q}$. Simplifying this ratio leads to $q(x^2 - y^2) = -2xy$, which rearranges to $qx^2 + 2xy - qy^2 = 0$. Since this result must be identical to the first pair of lines $x^2 - 2pxy - y^2 = 0$, we compare the coefficients of corresponding terms to set $\frac{q}{1} = \frac{2}{-2p} = \frac{-q}{-1}$. Solving the proportionality $\frac{q}{1} = \frac{2}{-2p}$ simplifies directly to $pq = -1$.

In a square where one vertex is at the origin and an adjacent side $OA$ of length $a$ makes an angle $\alpha$ with the positive $x$-axis, the coordinates of vertex $A$ are $(a\cos\alpha, a\sin\alpha)$. Because the side adjacent to $OA$ must be perpendicular to it, the other vertex adjacent to the origin, which we can call $C$, is located at $(-a\sin\alpha, a\cos\alpha)$. The diagonal that does not pass through the origin is the line segment connecting these two points, $A$ and $C$. To find its equation, we first determine the slope $m$ by calculating the change in $y$ over the change in $x$: $m = \frac{a\cos\alpha - a\sin\alpha}{-a\sin\alpha - a\cos\alpha} = -\frac{\cos\alpha - \sin\alpha}{\cos\alpha + \sin\alpha}$.

Applying the point-slope formula $y - y_1 = m(x - x_1)$ with the point $A(a\cos\alpha, a\sin\alpha)$ and our calculated slope $m$, we have:
$y - a\sin\alpha = -\frac{\cos\alpha - \sin\alpha}{\cos\alpha + \sin\alpha} (x - a\cos\alpha)$
Multiplying both sides by $-(\cos\alpha + \sin\alpha)$ allows us to clear the denominator and organize the terms:
$-(y - a\sin\alpha)(\cos\alpha + \sin\alpha) = (x - a\cos\alpha)(\cos\alpha - \sin\alpha)$
Expanding both sides of the equation yields:
$-y(\cos\alpha + \sin\alpha) + a\sin\alpha\cos\alpha + a\sin^2\alpha = x(\cos\alpha - \sin\alpha) - a\cos^2\alpha + a\cos\alpha\sin\alpha$
By moving the $x$ and $y$ terms to one side and grouping the constants, the $a\sin\alpha\cos\alpha$ terms cancel out, leaving:
$y(\cos\alpha + \sin\alpha) + x(\cos\alpha - \sin\alpha) = a\sin^2\alpha + a\cos^2\alpha$
Since $\sin^2\alpha + \cos^2\alpha = 1$, the final equation simplifies to $y(\cos\alpha + \sin\alpha) + x(\cos\alpha - \sin\alpha) = a$.

The locus of points equidistant from two fixed points $(a_1, b_1)$ and $(a_2, b_2)$ satisfies the condition that the squared distances to each point are equal, given by the equation $(x-a_1)^2+(y-b_1)^2=(x-a_2)^2+(y-b_2)^2$. Expanding both sides of this equation results in $x^2-2a_1x+a_1^2+y^2-2b_1y+b_1^2 = x^2-2a_2x+a_2^2+y^2-2b_2y+b_2^2$. Canceling the squared terms and rearranging the variables to one side produces $2(a_2-a_1)x + 2(b_2-b_1)y + (a_1^2+b_1^2-a_2^2-b_2^2) = 0$. Dividing the entire equation by 2 and adjusting the sign of the variables to match the standard linear form yields $(a_1-a_2)x + (b_1-b_2)y + \frac{1}{2}(a_2^2+b_2^2-a_1^2-b_1^2) = 0$. By comparing this expression to the general form of the equation, the constant term is identified as $c = \frac{1}{2}(a_2^2+b_2^2-a_1^2-b_1^2)$.

Collinearity for three points is determined by verifying that the slope between any two pairs is identical, as this indicates a constant rate of change across the sequence. Representing the points as $(x_1, y_1)$, $(x_1 r, y_1 r)$, and $(x_1 r^2, y_1 r^2)$ based on the common geometric progression ratio $r$, the slope of the line joining the first two points is calculated as $m_1 = \frac{y_1 r - y_1}{x_1 r - x_1} = \frac{y_1(r-1)}{x_1(r-1)} = \frac{y_1}{x_1}$. In the same manner, the slope of the line connecting the second and third points yields $m_2 = \frac{y_1 r^2 - y_1 r}{x_1 r^2 - x_1 r} = \frac{y_1 r(r-1)}{x_1 r(r-1)} = \frac{y_1}{x_1}$. Because the slopes $m_1$ and $m_2$ are equivalent, the points necessarily lie on a straight line.

Finding the locus of a geometric point defined by a parameter typically involves eliminating that parameter using fundamental trigonometric identities, specifically the property that $\sin^2 t + \cos^2 t = 1$. The centroid $(x, y)$ of a triangle is determined by calculating the average of the coordinates of its vertices, which are $(a \cos t, a \sin t)$, $(b \sin t, -b \cos t)$, and $(1, 0)$. This yields the expressions $x = \frac{a \cos t + b \sin t + 1}{3}$ and $y = \frac{a \sin t - b \cos t}{3}$.

To remove the variable $t$, we isolate the trigonometric components by writing $a \cos t + b \sin t = 3x - 1$ and $a \sin t - b \cos t = 3y$. Squaring both of these equations and summing them results in $a^2(\cos^2 t + \sin^2 t) + b^2(\sin^2 t + \cos^2 t)$ on the left side, as the mixed terms $2ab \sin t \cos t$ cancel each other out perfectly. This simplifies to $a^2 + b^2$, which establishes that the coordinates of the centroid must satisfy the equation $(3x - 1)^2 + (3y)^2 = a^2 + b^2$.

Determining the characteristics of a triangle defined by coordinates requires calculating the lengths of all three sides using the distance formula. We identify an isosceles triangle by checking if at least two sides are equal in length, while applying the Pythagorean theorem confirms if the square of the longest side equals the sum of the squares of the shorter sides, satisfying the condition for a right-angled triangle.

For the vertices at $(4,0)$, $(-1,-1)$, and $(3,5)$, calculating the distance between the first two points yields $\sqrt{(4 - (-1))^2 + (0 - (-1))^2} = \sqrt{5^2 + 1^2} = \sqrt{26}$. The distance between the vertices $(4,0)$ and $(3,5)$ is calculated as $\sqrt{(4-3)^2 + (0-5)^2} = \sqrt{1^2 + (-5)^2} = \sqrt{26}$. Since these two sides are equal, the triangle is isosceles. Finally, the segment connecting $(-1,-1)$ and $(3,5)$ has a length of $\sqrt{(3 - (-1))^2 + (5 - (-1))^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52}$. Because the square of this side, $52$, is equal to the sum of the squares of the two equal sides, $(\sqrt{26})^2 + (\sqrt{26})^2 = 26 + 26 = 52$, the triangle also satisfies the Pythagorean theorem, confirming it is right-angled.

To find the locus of the midpoint, we first rewrite the line equation $x \cos\alpha + y \sin\alpha = p$ in the intercept form $\frac{x}{p/\cos\alpha} + \frac{y}{p/\sin\alpha} = 1$, which reveals that the line intersects the axes at points $\left(\frac{p}{\cos\alpha}, 0\right)$ and $\left(0, \frac{p}{\sin\alpha}\right)$. Since the midpoint $(x, y)$ of this line segment is calculated by averaging these intercepts, we have $x = \frac{p}{2\cos\alpha}$ and $y = \frac{p}{2\sin\alpha}$. Rearranging these to isolate the trigonometric functions gives $\cos\alpha = \frac{p}{2x}$ and $\sin\alpha = \frac{p}{2y}$. By substituting these into the fundamental identity $\cos^2\alpha + \sin^2\alpha = 1$, we obtain $\left(\frac{p}{2x}\right)^2 + \left(\frac{p}{2y}\right)^2 = 1$, which simplifies to $\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}$.

A pair of lines represented by the homogeneous equation $Ax^2 + 2Hxy + By^2 = 0$ are perpendicular to each other if and only if the sum of the coefficients of $x^2$ and $y^2$ is zero, which requires $A + B = 0$. Applying this principle to the expression $3 a x^2 + 5 x y + (a^2 - 2) y^2 = 0$, we set the sum of the coefficients equal to zero, resulting in $3a + a^2 - 2 = 0$. Rearranging this into the standard quadratic form $a^2 + 3a - 2 = 0$ and solving via the quadratic formula gives $a = \frac{-3 \pm \sqrt{17}}{2}$. Because this calculation yields two distinct real roots, there are exactly two values of $a$ for which the lines are perpendicular.

When the pair of lines $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ intersect on the $y$-axis, their point of intersection must satisfy the condition $x=0$. Substituting this value into the equation yields $b y^2+2 f y+c=0$, and because this quadratic must yield a unique intersection point, its discriminant must be zero, which leads to the relation $f^2=b c$. Incorporating this fact into the general condition for the equation to represent a pair of lines, $a b c+2 f g h-a f^2-b g^2-c h^2=0$, allows us to replace $a f^2$ with $a b c$, causing the $a b c$ and $a f^2$ terms to cancel out. This simplification leaves behind the equation $2 f g h-b g^2-c h^2=0$, which directly rearranges to $2 f g h=b g^2+c h^2$.

The triangle $\text{ABC}$ is isosceles with $\angle A = \frac{2\pi}{3}$, which implies the base angles $\angle B$ and $\angle C$ are each $\frac{\pi}{6}$. Applying the law of sines, $\frac{AB}{\sin(\pi/6)} = \frac{BC}{\sin(2\pi/3)}$ leads to $\frac{AB}{1/2} = \frac{4\sqrt{3}}{\sqrt{3}/2}$, resulting in $AB = 4$. Given $A$ is located at $(-1, 0)$ and $B$ lies on the positive x-axis, the coordinates of $B$ must be $(3, 0)$.

The line $\text{BC}$ passes through $B(3, 0)$ and, because $\angle B = \pi/6$ and the side $AB$ lies on the x-axis, it makes an angle of $5\pi/6$ with the positive x-axis, giving it a slope of $\tan(5\pi/6) = -\frac{1}{\sqrt{3}}$. The equation for line $\text{BC}$ is $y - 0 = -\frac{1}{\sqrt{3}}(x - 3)$, which rearranges to $x + \sqrt{3}y = 3$. Solving this simultaneously with $y = x + 3$ by substituting $x = y - 3$ gives $(y - 3) + \sqrt{3}y = 3$, or $y(1 + \sqrt{3}) = 6$, yielding $\beta = \frac{6}{\sqrt{3} + 1}$. The corresponding coordinate $\alpha$ is $\beta - 3 = \frac{6 - 3(\sqrt{3} + 1)}{\sqrt{3} + 1} = \frac{3 - 3\sqrt{3}}{\sqrt{3} + 1}$.

Calculating the ratio $\frac{\beta^4}{\alpha^2}$ involves simplifying $\frac{(\frac{6}{\sqrt{3} + 1})^4}{(\frac{3(1 - \sqrt{3})}{\sqrt{3} + 1})^2}$. This expression becomes $\frac{6^4}{(\sqrt{3} + 1)^4} \cdot \frac{(\sqrt{3} + 1)^2}{9(1 - \sqrt{3})^2}$, which reduces to $\frac{1296}{9 \cdot [(\sqrt{3} + 1)(1 - \sqrt{3})]^2} = \frac{144}{(1 - 3)^2} = \frac{144}{4}$, resulting in 36.

Translating a point along a line by a specific distance involves moving along the line's direction vector. For the line $L$ given by $x - y = 4$, the slope is $1$, which corresponds to an inclination angle where $\tan\theta = 1$. The unit vector along this line is either $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ or its negative counterpart $(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$.

Moving the point $(2,1)$ by a distance of $2\sqrt{3}$ in the direction of the negative unit vector gives $Q = (2 - 2\sqrt{3} \cdot \frac{1}{\sqrt{2}}, 1 - 2\sqrt{3} \cdot \frac{1}{\sqrt{2}})$, which simplifies to $(2 - \sqrt{6}, 1 - \sqrt{6})$. Because $\sqrt{6}$ is approximately $2.45$, both coordinates are negative, confirming that $Q$ lies in the third quadrant as required.

The line perpendicular to $L$ must have a slope of $-1$, as it is the negative reciprocal of the slope of $L$. Using the point-slope form $y - y_1 = m(x - x_1)$ with the point $Q$ and slope $m = -1$, we get $y - (1 - \sqrt{6}) = -1(x - (2 - \sqrt{6}))$. Rearranging these terms leads to the equation $x + y = 3 - 2\sqrt{6}$.

The coordinates of a point situated at a specific distance from a given point on a line are best described using the parametric form of the line equation. Considering point $B$ lies at a distance $r = \frac{\sqrt{29}}{3}$ from $A(4, 3)$ along a line with inclination $\theta$, we express the coordinates of $B$ as $(4 + r \cos \theta, 3 + r \sin \theta)$. Substituting these expressions into the linear equation $x - y - 2 = 0$ leads to $(4 + r \cos \theta) - (3 + r \sin \theta) - 2 = 0$, which simplifies to $r(\cos \theta - \sin \theta) = 1$. By substituting the distance value $r = \frac{\sqrt{29}}{3}$, we isolate the trigonometric components to find that $\cos \theta - \sin \theta = \frac{3}{\sqrt{29}}$.

Squaring both sides of this relation results in $1 - 2 \sin \theta \cos \theta = \frac{9}{29}$, which simplifies further to show that $\sin 2\theta = \frac{20}{29}$. Utilizing the identity $\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$ yields the quadratic equation $10 \tan^2 \theta - 29 \tan \theta + 10 = 0$. Factoring this quadratic gives $(2 \tan \theta - 5)(5 \tan \theta - 2) = 0$, resulting in potential slope values of $\frac{5}{2}$ and $\frac{2}{5}$. Given the constraint that the slope must be greater than one, we select $\tan \theta = \frac{5}{2}$.

With $\tan \theta = \frac{5}{2}$, the trigonometric ratios are $\sin \theta = \frac{5}{\sqrt{29}}$ and $\cos \theta = \frac{2}{\sqrt{29}}$. Substituting these values into our parametric expressions with $r = -\frac{\sqrt{29}}{3}$ (derived from the orientation of point $B$ on the line) yields the coordinates $(10/3, 4/3)$ for point $B$. Inserting these coordinates into the given linear options reveals that the point satisfies the equation $x + 2y = 6$.

To find the coordinates of the midpoint of a line segment intercepted between the coordinate axes, we first calculate the intercepts by setting $x$ and $y$ to zero in the equation $x \cos \theta + y \sin \theta = 7$. The x-intercept is $\frac{7}{\cos \theta}$ and the y-intercept is $\frac{7}{\sin \theta}$, which means the midpoint $(h, k)$ has coordinates defined by $h = \frac{7}{2 \cos \theta}$ and $k = \frac{7}{2 \sin \theta}$.

We are given that the midpoint has a y-coordinate of $\frac{7\sqrt{3}}{3}$. Setting $k = \frac{7}{2 \sin \theta} = \frac{7\sqrt{3}}{3}$ allows us to solve for $\sin \theta$, yielding $\sin \theta = \frac{\sqrt{3}}{2}$, which implies $\theta = \frac{\pi}{3}$. Substituting this value of $\theta$ into the expression for $h$, we calculate $\alpha = \frac{7}{2 \cos \left( \frac{\pi}{3} \right)}$. Since $\cos \left( \frac{\pi}{3} \right) = \frac{1}{2}$, this results in $\alpha = \frac{7}{2(1/2)}$, which simplifies to $7$.

Reflecting the point $A\left(\frac{3}{\sqrt{a}}, \sqrt{a}\right)$ across the y-axis results in point $B\left(-\frac{3}{\sqrt{a}}, \sqrt{a}\right)$, and subsequently reflecting $B$ across the x-axis gives point $C\left(-\frac{3}{\sqrt{a}}, -\sqrt{a}\right)$. With the third vertex $D$ defined as $(3 \cos \theta, a \sin \theta)$, the area of $\triangle ACD$ is determined by the coordinates of these three points using the standard determinant formula.

The area $\Delta$ can be expressed as:
$\Delta = \frac{1}{2} \left| \begin{matrix} \frac{3}{\sqrt{a}} & \sqrt{a} & 1 \\ -\frac{3}{\sqrt{a}} & -\sqrt{a} & 1 \\ 3\cos\theta & a\sin\theta & 1 \end{matrix} \right|$
Evaluating the determinant simplifies the expression to $\Delta = 3\sqrt{a} |\sin \theta + \cos \theta|$. The trigonometric component $|\sin \theta + \cos \theta|$ reaches a maximum value of $\sqrt{2}$. Since the problem states the maximum area is 12, we set $3\sqrt{a} \cdot \sqrt{2} = 12$. Dividing by $3\sqrt{2}$ yields $\sqrt{a} = \frac{4}{\sqrt{2}}$, which simplifies to $\sqrt{a} = 2\sqrt{2}$. Squaring both sides of this equation results in $a = (2\sqrt{2})^2 = 8$.

The lines defined by the equation $x - \sqrt{3} |y| = \alpha$ represent two rays originating from the point $P(\alpha, 0)$. Given that the angle between these lines is $60^{\circ}$ and the points $A$ and $B$ lie on these rays at a distance of $\alpha$ from $P$, the triangle $\triangle PAB$ is equilateral with a side length of $\alpha$. The point $Q$ lies on the angle bisector of the two lines, which corresponds to the altitude of this equilateral triangle dropped from vertex $P$ to the base $AB$.

The altitude of an equilateral triangle with side length $\alpha$ is expressed as $PQ = \frac{\sqrt{3}}{2} \alpha$. By equating this to the given distance $\frac{9}{2}$, we find $\frac{\sqrt{3}}{2} \alpha = \frac{9}{2}$, which simplifies to $\alpha = \frac{9}{\sqrt{3}} = 3\sqrt{3}$. The circumradius $R$ of an equilateral triangle is related to the side length $\alpha$ by the formula $R = \frac{\alpha}{\sqrt{3}}$. Substituting our value for $\alpha$, we get $R = \frac{3\sqrt{3}}{\sqrt{3}} = 3$. Finally, evaluating the required expression $\frac{\alpha^2}{R}$ yields $\frac{(3\sqrt{3})^2}{3} = \frac{27}{3} = 9$.

The fundamental property of any parallelogram is that its diagonals bisect each other, meaning the midpoint of diagonal $AC$ must coincide with the midpoint of diagonal $BD$. Given the vertices $B(1, 0)$ and $D(1, 2)$, the midpoint of diagonal $BD$ is calculated by averaging the coordinates, resulting in $(\frac{1+1}{2}, \frac{0+2}{2})$, which simplifies to $(1, 1)$.

Since $A(\alpha, \beta)$ and $C(\gamma, \delta)$ are the endpoints of diagonal $AC$, their midpoint is $(\frac{\alpha+\gamma}{2}, \frac{\beta+\delta}{2})$. Equating this midpoint to $(1, 1)$ establishes the relations $\frac{\alpha+\gamma}{2} = 1$ and $\frac{\beta+\delta}{2} = 1$, which simplify to $\alpha+\gamma = 2$ and $\beta+\delta = 2$. Substituting these values into the expression $2(\alpha+\gamma+\beta+\delta)$ yields $2(2+2)$, which results in 8.