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

The reflection of a geometric figure across a line preserves its centroid, meaning the centroid of triangle $PQR$ is simply the reflected image of the centroid of the original triangle. Calculating the centroid $G$ of the given vertices $(1,3)$, $(3,1)$, and $(2,4)$ yields $G = \left(\frac{1+3+2}{3}, \frac{3+1+4}{3}\right) = \left(2, \frac{8}{3}\right)$.

The image of $G(2, 8/3)$ across the line $x+2y-2=0$ is found using the reflection formula $\frac{\alpha-x_1}{a} = \frac{\beta-y_1}{b} = -2\frac{ax_1+by_1+c}{a^2+b^2}$. Substituting the coordinates and line parameters gives $\frac{\alpha-2}{1} = \frac{\beta-8/3}{2} = -2\frac{2+16/3-2}{1+4}$, which simplifies to $-\frac{32}{15}$.

Equating the parts, $\alpha - 2 = -\frac{32}{15}$ results in $\alpha = -\frac{2}{15}$, and $\frac{\beta-8/3}{2} = -\frac{32}{15}$ leads to $\beta - \frac{8}{3} = -\frac{64}{15}$, or $\beta = \frac{40-64}{15} = -\frac{24}{15}$. The final expression $15(\alpha-\beta)$ is therefore $15\left(-\frac{2}{15} - \left(-\frac{24}{15}\right)\right) = 15\left(\frac{22}{15}\right) = 22$.

In an isosceles triangle, the third side must be perpendicular to the angle bisector of the two equal sides. We begin by determining the slopes of these angle bisectors between the lines $-x + 2y = 4$ and $x + y = 4$ using the distance-from-line formula, $\frac{-x+2y-4}{\sqrt{5}} = \pm \frac{x+y-4}{\sqrt{2}}$. This equation characterizes two distinct angle bisectors, and their slopes, $k$, satisfy the quadratic relation $k^2 - 6k - 1 = 0$.

Since the third side of the triangle is perpendicular to these bisectors, its slope $m$ must be the negative reciprocal of the bisector slope, $m = -1/k$. Substituting $k = -1/m$ into the quadratic relation $k^2 - 6k - 1 = 0$ yields $(-1/m)^2 - 6(-1/m) - 1 = 0$, which simplifies to $m^2 - 6m - 1 = 0$. The sum of all possible distinct values for $m$ corresponds to the sum of the roots of this quadratic equation, which equals 6.

Defining the coordinates of the endpoints $A$ and $B$ relative to their respective lines allows us to utilize the fixed distance condition to establish the locus of point $P$. Since $A$ lies on the line $x-y+2=0$, we can express it parametrically as $(a, a+2)$. Similarly, because $B$ lies on $y+2=0$, we denote it as $(b, -2)$. Given the rod length is eight units, the condition $AB^2=64$ implies $(a-b)^2 + (a+4)^2 = 64$.

Point $P(x, y)$ divides the segment $AB$ in a $2:1$ ratio. Applying the section formula for internal division, we have $x = \frac{2b+a}{3}$ and $y = \frac{2(-2)+1(a+2)}{3} = \frac{a-2}{3}$. From the expression for $y$, we find $a = 3y+2$, and by substituting this into the equation for $x$, we determine $b = \frac{3x-3y-2}{2}$.

Substituting these values of $a$ and $b$ into the distance constraint $(a-b)^2 + (a+4)^2 = 64$ leads to the equation $(3y+2 - \frac{3x-3y-2}{2})^2 + (3y+6)^2 = 64$. Multiplying this equation by four removes the fraction, yielding $9(x-3y-2)^2 + 36(y+2)^2 = 256$. Upon expanding and simplifying this result, we arrive at $9(x^2 - 6xy + 13y^2 - 4x + 28y) - 76 = 0$. By comparing this with the provided form $9(x^2 + \alpha y^2 + \beta xy + \gamma x + 28y) - 76 = 0$, we identify the constants as $\alpha = 13$, $\beta = -6$, and $\gamma = -4$. Calculating the required value, we find $\alpha - \beta - \gamma = 13 - (-6) - (-4) = 23$.

The geometry of this problem relies on locating the vertices of the triangle $AMN$ within the coordinate plane defined by the axes and the line $x+y=1$. Since the line intersects the $x$ and $y$ axes at $A(1, 0)$ and $B(0, 1)$, we define the point $N$ dividing $AB$ in the ratio $\lambda:1$ using the section formula, resulting in coordinates $N = \left(\frac{1}{\lambda+1}, \frac{\lambda}{\lambda+1}\right)$. Because $\triangle AMN$ is right-angled and $M$ lies on $OB$, the segment $MN$ must be perpendicular to the line $AB$. Given the slope of $AB$ is $-1$, the slope of $MN$ must be $1$, allowing us to define the line $MN$ as $y - \frac{\lambda}{\lambda+1} = 1\left(x - \frac{1}{\lambda+1}\right)$, which simplifies to $y = x + \frac{\lambda-1}{\lambda+1}$.

Setting $x=0$ provides the coordinates of $M$ as $\left(0, \frac{\lambda-1}{\lambda+1}\right)$. For $M$ to lie on the segment $OB$, its $y$-coordinate must satisfy the condition $0 < \frac{\lambda-1}{\lambda+1} < 1$, which strictly requires $\lambda > 1$. The area of the right-angled triangle $AMN$ is $\frac{1}{2} \times AN \times MN$. Calculating the lengths of the legs using the distance formula gives $AN = \frac{\lambda}{\lambda+1}\sqrt{2}$ and $MN = \frac{\sqrt{2}}{\lambda+1}$, which yields an area of $\frac{\lambda}{(\lambda+1)^2}$. Equating this to $\frac{4}{9}$ of the area of $\triangle OAB$, which is $\frac{1}{2} \times 1 \times 1 = \frac{1}{2}$, we obtain:
$\frac{\lambda}{(\lambda+1)^2} = \frac{4}{9} \cdot \frac{1}{2} = \frac{2}{9}$
This simplifies to the quadratic equation $2\lambda^2 - 5\lambda + 2 = 0$, which factors as $(2\lambda-1)(\lambda-2) = 0$. Since our geometric constraint requires $\lambda > 1$, we must discard $\lambda = \frac{1}{2}$ and accept $\lambda = 2$ as the only valid value.

The centroid of a triangle is determined by the average coordinates of its vertices, so if one vertex moves along a given line, the locus of the centroid can be found by relating its coordinates to that line's equation. To determine the fixed points $A$ and $B$, solve the system formed by the circle $x^2 + y^2 - 8x = 0$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$. Substituting $y^2 = 8x - x^2$ into the hyperbola equation leads to $13x^2 - 72x - 36 = 0$, which results in $x = 6$. Evaluating the circle equation at this $x$-coordinate confirms the intersection points as $A(6, \sqrt{12})$ and $B(6, -\sqrt{12})$.

The point $P$ can be represented as $(h, k)$, constrained by the equation $2h - 3k + 4 = 0$. The centroid $(x, y)$ of $\triangle PAB$ is calculated by averaging the vertices, resulting in $x = \frac{h + 6 + 6}{3}$ and $y = \frac{k + \sqrt{12} - \sqrt{12}}{3}$. Rearranging these to express $h$ and $k$ in terms of the centroid's coordinates gives $h = 3x - 12$ and $k = 3y$. Substituting these expressions into the constraint equation yields $2(3x - 12) - 3(3y) + 4 = 0$. Expanding and simplifying this equation produces $6x - 24 - 9y + 4 = 0$, which reduces to $6x - 9y = 20$.

The ratio $AM/BM$ in the right-angled triangle formed when a transversal intersects two parallel lines is equal to the tangent of the angle between the transversal and the lines. The line passing through the origin that makes equal angles with the positive coordinate axes has the equation $y = x$, which corresponds to a slope of $m_1 = 1$. The given lines $L_1: 2x+y+6=0$ and $L_2: 4x+2y-p=0$ share the same slope $m_2 = -2$, confirming that $L_1$ and $L_2$ are parallel. Since the transversal $y=x$ intersects these parallel lines at $A$ and $B$, and $M$ is the foot of the perpendicular from $A$ to $L_2$, the segment $AM$ is the side opposite to the angle $\theta$ at vertex $B$ in the right-angled triangle $\triangle AMB$, while $BM$ is the adjacent side. The ratio $AM/BM$ is therefore equal to $\tan \theta$, where $\theta$ is the angle between the transversal and the parallel lines. Calculating this using the slope formula, $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{1 - (-2)}{1 + (1)(-2)} \right| = 3$. Consequently, the ratio $AM/BM$ is 3.

The orthocenter of a triangle lies at the intersection of its altitudes, meaning the line segment connecting any vertex to the orthocenter must be perpendicular to the opposite side. Given the sides $y = x + 1$ and $y = 4x - 8$ with the orthocenter at $(3, -1)$, the altitude from vertex $C$ to the side $y = x + 1$ must have a slope of $-1$, leading to the equation $y = -x + 2$. Solving for the intersection of this altitude with the side $y = 4x - 8$ yields vertex $C$ at $(2, 0)$. The altitude from vertex $A$ to the side $y = 4x - 8$ has a slope of $-\frac{1}{4}$, which results in the equation $x + 4y + 1 = 0$. Intersecting this altitude with the side $y = x + 1$ provides vertex $A$ at $(-1, 0)$. Since both vertices $A$ and $C$ lie on the x-axis, the third side of the triangle is $y = 0$, which implies $m = 0$ and $c = 0$, thus making $m - c = 0$.

The linear equation involving the parameter $\lambda$ describes a family of lines that all pass through a common intersection point $P$. By rearranging the expression as $(x+2y-5) + \lambda(3x+7y-17) = 0$, we identify $P$ by solving the system $x+2y-5=0$ and $3x+7y-17=0$, which yields the coordinates $(1,2)$. Geometrically, among all lines passing through a specific point $P$ not at the origin, the line farthest from the origin is the unique line perpendicular to the vector connecting the origin to $P$. Since the slope of the line segment from the origin to $(1,2)$ is $2$, the slope of the required line $L$ must be the negative reciprocal, $-\frac{1}{2}$. Using the point-slope form $y-2 = -\frac{1}{2}(x-1)$, the equation of line $L$ simplifies to $x+2y-5=0$. The distance $d$ from the point $(3,6)$ to this line is calculated using the formula $d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$, resulting in $d = \frac{|1(3)+2(6)-5|}{\sqrt{1^2+2^2}} = \frac{10}{\sqrt{5}}$. Squaring this distance gives $d^2 = \frac{100}{5} = 20$.

The orthocentre of a right-angled triangle is the vertex where the right angle is formed. For the first triangle, the lines $4x-7y+10=0$ and $7x+4y=15$ have slopes $\frac{4}{7}$ and $-\frac{7}{4}$, respectively, indicating they are perpendicular. Solving these equations simultaneously, $4x-7y=-10$ and $7x+4y=15$, yields the intersection point $(1, 2)$, which is the orthocentre of the first triangle. For the second triangle formed by the lines $x=0$, $y=0$, and $x+y=1$, the lines $x=0$ and $y=0$ are perpendicular, meeting at the origin $(0, 0)$, which serves as the orthocentre. The distance between the two orthocentres $(1, 2)$ and $(0, 0)$ is calculated as:
$d = \sqrt{(1-0)^2 + (2-0)^2} = \sqrt{1^2 + 2^2} = \sqrt{5}$

Rotating a line that makes an angle $\alpha$ with the $x$-axis about a point on it by an angle $\alpha/2$ clockwise reduces its inclination to $\alpha/2$. Given the slope in this new orientation is $2-\sqrt{3}$, we establish $\tan(\alpha/2) = 2-\sqrt{3}$. Applying the double-angle identity for tangent, $\tan \alpha = \frac{2 \tan(\alpha/2)}{1-\tan^2(\alpha/2)}$, we find that $\tan \alpha = 1/\sqrt{3}$.

The equation of the line in its new position passing through $P(a,0)$ is $(y-0)=(2-\sqrt{3})(x-a)$, which simplifies to $(2-\sqrt{3})x - y - a(2-\sqrt{3}) = 0$. The distance from the origin to this line is given by $\frac{|-a(2-\sqrt{3})|}{\sqrt{(2-\sqrt{3})^2 + (-1)^2}}$, which equals $1/\sqrt{2}$. Squaring both sides yields $\frac{a^2(2-\sqrt{3})^2}{8-4\sqrt{3}} = 1/2$. Observing that the denominator $8-4\sqrt{3}$ simplifies to $4(2-\sqrt{3})$, we find $\frac{a^2(2-\sqrt{3})}{4} = 1/2$, leading to $a^2 = \frac{2}{2-\sqrt{3}}$. Multiplying by the conjugate $2+\sqrt{3}$ gives $a^2 = 4+2\sqrt{3}$. Substituting these results into the final expression $3 a^2 \tan^2 \alpha - 2 \sqrt{3}$ results in $3(4+2\sqrt{3})(1/3) - 2\sqrt{3}$, which equals 4.

The area of a triangle with vertices $A(-1, 1)$, $B(2, 3)$, and $P(x, y)$ is defined by the formula $\frac{1}{2} |x(1 - 3) - 1(3 - y) + 2(y - 1)| = 10$, which simplifies to the condition $|-2x + 3y - 5| = 20$. Since the variable point $P$ must lie above the line segment $AB$, the equation simplifies to $-2x + 3y = 25$. To align this with the required locus form $ax + by = 15$, we multiply the equation by $\frac{3}{5}$ to obtain $-\frac{6}{5}x + \frac{9}{5}y = 15$. This identifies the constants as $a = -\frac{6}{5}$ and $b = \frac{9}{5}$, allowing for the final calculation $5a + 2b = 5(-\frac{6}{5}) + 2(\frac{9}{5}) = -6 + \frac{18}{5} = -\frac{12}{5}$.

The line passing through $(4, -9)$ with slope $m$ satisfies the equation $y+9=m(x-4)$. Identifying the intercepts by setting $x=0$ and $y=0$ provides the coordinate $A = (\frac{9+4m}{m}, 0)$ and $B = (0, -9-4m)$. Because $m>0$, the lengths $OA$ and $OB$ are $\frac{9}{m} + 4$ and $9+4m$ respectively, making their sum $13 + 4m + \frac{9}{m}$. Utilizing the arithmetic mean-geometric mean inequality on the terms $4m$ and $\frac{9}{m}$, the expression $4m + \frac{9}{m}$ reaches a minimum value of $2\sqrt{4m \cdot \frac{9}{m}}$, which calculates to $12$. Adding the constant $13$ results in an overall minimum sum of $25$.

Rotating a line about a fixed point changes its angle of inclination, and moving the line 15 degrees clockwise from an original inclination of 30 degrees results in a new slope of $m = \tan(15^{\circ})$, which evaluates to $2-\sqrt{3}$. Using the point-slope form with the fixed point $(9,0)$, the line is defined by the equation $y = (2-\sqrt{3})(x-9)$. Rearranging this relationship to isolate the constant 9 gives $x - 9 = \frac{y}{2-\sqrt{3}}$, and since $2-\sqrt{3}$ is equivalent to $-(\sqrt{3}-2)$, the expression becomes $x - 9 = -\frac{y}{\sqrt{3}-2}$, which simplifies to $x + \frac{y}{\sqrt{3}-2} = 9$.

A point lies in the interior region between two lines containing the origin if the point and the origin yield the same algebraic sign when substituted into the equations of both lines. For the line $3x-y+1=0$, substituting the origin results in a positive value of $1$, so the coordinates $(a^2, a+1)$ must satisfy the inequality:
$3a^2 - (a+1) + 1 > 0$
This expression simplifies to $3a^2 - a > 0$, which factors as $a(3a-1) > 0$, indicating that $a$ must be less than $0$ or greater than $1/3$.

For the second line $x+2y-5=0$, substituting the origin yields a negative value of $-5$, which means that substituting the point $(a^2, a+1)$ into this equation must also produce a negative result:
$a^2 + 2(a+1) - 5 < 0$
Simplifying this inequality leads to $a^2 + 2a - 3 < 0$, which factors into $(a+3)(a-1) < 0$ and requires $a$ to be strictly between $-3$ and $1$. Combining the conditions from both lines by finding their intersection, we determine that the valid values for $a$ are those in the interval $(-3, 0) \cup (1/3, 1)$.

The centroid of a triangle lies on the line segment joining its orthocentre and circumcentre, dividing the segment in a $2:1$ ratio. Using the given orthocentre $(-6,-8)$ and circumcentre $(3,4)$, the centroid is calculated as $\left( \frac{-6 + 2(3)}{3}, \frac{-8 + 2(4)}{3} \right)$, which simplifies to the origin $(0,0)$. This centroid serves as the orthocentre for the triangle defined by the three given lines.

By solving the system formed by $2x+3y-1=0$ and $x+2y-1=0$, we find the vertex of the triangle to be at $(-1,1)$. The altitude from this vertex to the line $ax+by-1=0$ must pass through the orthocentre $(0,0)$, so the slope of this altitude is $\frac{1-0}{-1-0} = -1$. Because this altitude is perpendicular to the line $ax+by-1=0$, whose slope is $-\frac{a}{b}$, the product of their slopes must be $-1$, which leads to $(-1)\left(-\frac{a}{b}\right) = -1$, or $a = -b$. Substituting this relationship into the equation of the lines and utilizing the condition that the altitudes intersect at $(0,0)$, we solve the resulting equations to determine $a = -8$ and $b = 8$. Finally, the absolute difference $|a-b|$ is calculated as $|-8 - 8| = 16$.

The number of intersection points for a set of lines with restricted geometry is determined by considering the total potential intersections in general position and then adjusting for specific constraints. With 20 distinct lines, there are ${}^{20}C_2 = 190$ potential intersection points. Since the 10 lines with odd indices are parallel, they never intersect, requiring us to subtract the ${}^{10}C_2 = 45$ points that would have formed among them in a general arrangement. Similarly, the 10 lines with even indices are concurrent at point $P$, meaning they generate only 1 intersection point rather than the ${}^{10}C_2 = 45$ points they would otherwise produce. Accounting for these adjustments leads to the final count of ${}^{20}C_2 - {}^{10}C_2 - {}^{10}C_2 + 1 = 101$ points.

The locus of a point equidistant from two intersecting lines corresponds to the angle bisectors of those lines. For any point $(x, y)$, the perpendicular distance to a line $ax+by+c=0$ is given by $\frac{|ax+by+c|}{\sqrt{a^2+b^2}}$. Equating the distances from the point $(x, y)$ to the lines $x+2y+7=0$ and $2x-y+8=0$ results in the equation $\frac{|x+2y+7|}{\sqrt{5}} = \frac{|2x-y+8|}{\sqrt{5}}$. Since the denominators are identical, this simplifies to $|x+2y+7| = |2x-y+8|$. Squaring both sides and setting them equal gives $(x+2y+7)^2 - (2x-y+8)^2 = 0$.

Expanding this expression yields $(x+2y+7-2x+y-8)(x+2y+7+2x-y+8) = 0$, which simplifies to $(-x+3y-1)(3x+y+15) = 0$. This can be rewritten as $-(x-3y+1)(3x+y+15) = 0$, or simply $(x-3y+1)(3x+y+15) = 0$. Further expanding this product leads to $3x^2-3y^2-8xy+18x-44y+15 = 0$. Dividing the entire equation by 3 to match the form $x^2-y^2+2hxy+2gx+2fy+c = 0$, we obtain $x^2-y^2-\frac{8}{3}xy+6x-\frac{44}{3}y+5 = 0$. Comparing coefficients, we identify $h = -\frac{4}{3}$, $g = 3$, $f = -\frac{22}{3}$, and $c = 5$. Substituting these values into $g+c+h-f$ gives $3+5+(-\frac{4}{3}) - (-\frac{22}{3})$, which results in 14.

The defining characteristic of a parallelogram is that its diagonals bisect each other, which implies that the midpoint of diagonal $AC$ coincides with the midpoint of diagonal $BD$. Given the vertices $A(-2,-1)$, $B(1,0)$, $C(\alpha, \beta)$, and $D(\gamma, \delta)$, the midpoint of $AC$ is $(\frac{\alpha-2}{2}, \frac{\beta-1}{2})$ and the midpoint of $BD$ is $(\frac{\gamma+1}{2}, \frac{\delta}{2})$. Equating these coordinates yields the relations $\alpha - \gamma = 3$ and $\beta - \delta = 1$.

Since points $C$ and $D$ lie on specific lines, they must satisfy $2\alpha - \beta = 5$ and $3\gamma - 2\delta = 6$. By substituting $\alpha = \gamma + 3$ and $\beta = \delta + 1$ into the first equation, we get $2(\gamma + 3) - (\delta + 1) = 5$, which simplifies to $2\gamma - \delta = 0$, meaning $\delta = 2\gamma$. Replacing $\delta$ with $2\gamma$ in the second line equation results in $3\gamma - 2(2\gamma) = 6$, giving $\gamma = -6$. Substituting this back allows us to find $\delta = -12$, $\alpha = -3$, and $\beta = -11$. Finally, the sum of these four values is $\alpha+\beta+\gamma+\delta = -3 - 11 - 6 - 12 = -32$, and the magnitude of this sum is 32.

When determining the angle subtended by a line segment at the origin, we apply the formula relating the slopes of the two lines connecting the origin to the endpoints, which is $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$. For the points $(5,2)$ and $(2,a)$, the slopes of the lines originating from the origin are $m_1 = \frac{2}{5}$ and $m_2 = \frac{a}{2}$. Substituting these values into the formula with $\tan \frac{\pi}{4} = 1$, we get $1 = \left| \frac{\frac{2}{5} - \frac{a}{2}}{1 + (\frac{2}{5})(\frac{a}{2})} \right|$, which simplifies to $1 = \left| \frac{4-5a}{10+2a} \right|$.

This absolute value equation produces two distinct scenarios to solve for $a$. In the first scenario, $4-5a = 10+2a$, which simplifies to $7a = -6$ and yields $a = -\frac{6}{7}$. In the second scenario, $4-5a = -(10+2a)$, which simplifies to $4-5a = -10-2a$, leading to $3a = 14$ or $a = \frac{14}{3}$. The product of these two possible values for $a$ is $-\frac{6}{7} \times \frac{14}{3} = -4$, and the absolute value of this result is 4.

Finding the required distance begins by identifying the coordinates for points $A$ and $B$. By solving the system $3 x+2 y=14$ and $5 x-y=6$, we find that point $A$ is $(2, 4)$. Similarly, solving the system $4 x+3 y=8$ and $6 x+y=5$ determines that point $B$ is located at $(\frac{1}{2}, 2)$.

Once these two points are established, the equation of the line passing through them can be determined. The slope is calculated as $m = \frac{2-4}{\frac{1}{2}-2} = \frac{-2}{-\frac{3}{2}} = \frac{4}{3}$, leading to the linear equation $4 x-3 y+4=0$.