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

To determine the equation of the angle bisector for angle $ABC$, we first define the sides $AB$ and $BC$ as linear equations in the Cartesian plane. The line passing through $A(1,0)$ and $B(2,-1)$ has a slope of $-1$, leading to the equation $x+y-1=0$. Similarly, the line passing through $B(2,-1)$ and $C \left(\frac{7}{3}, \frac{4}{3}\right)$ has a slope of $7$, resulting in the equation $7x-y-15=0$. We relate these using the angle bisector formula $\frac{x+y-1}{\sqrt{1^2+1^2}} = \pm \frac{7x-y-15}{\sqrt{7^2+(-1)^2}}$.

To decide between the positive or negative sign, we evaluate the expressions $x+y-1$ and $7x-y-15$ at a test point like $(2,0)$, which is known to be located within the angle $ABC$. Because $x+y-1$ evaluates to $1$ and $7x-y-15$ evaluates to $-1$ at this point, we select the negative sign to ensure the calculation provides the internal angle bisector. The equation becomes:
$\frac{x+y-1}{\sqrt{2}} = -\frac{7x-y-15}{5\sqrt{2}}$
Multiplying both sides by $5\sqrt{2}$ yields $5(x+y-1) = -(7x-y-15)$, which simplifies to $12x+4y=20$. Dividing by $4$ results in the standard form $3x+y=5$, identifying $\alpha=3$ and $\beta=1$. Calculating the sum $\alpha^2+\beta^2$ gives $3^2+1^2$, which equals $10$.

The circumcenter of a triangle is the point of concurrency of its perpendicular bisectors. Given the vertex $A = (1, 2)$ and the midpoint of side $AB$ at $(5, -1)$, we determine the coordinates of vertex $B = (9, -4)$ using the midpoint formula. With the centroid $G = (3, 4)$ and the coordinates of vertices $A$ and $B$ known, we find the third vertex $C = (-1, 14)$ by applying the centroid definition $G = \frac{A+B+C}{3}$.

The perpendicular bisector of side $AB$ passes through the midpoint $(5, -1)$ with a slope of $\frac{4}{3}$, leading to the equation $4x - 3y = 23$. For side $BC$, which connects $(9, -4)$ and $(-1, 14)$, the midpoint is $(4, 5)$ and the perpendicular slope is $\frac{5}{9}$, resulting in the equation $5x - 9y = -25$. Solving these simultaneous linear equations provides the circumcenter $(\alpha, \beta) = \left( \frac{94}{7}, \frac{215}{21} \right)$. Substituting these values into $21(\alpha + \beta)$ gives $21 \left( \frac{282 + 215}{21} \right)$, which simplifies to 497.

A line that bisects the area of a rectangle must pass through its center. For the rectangle bounded by the lines $x=0$, $y=0$, $x=3$, and $y=4$, the center is the midpoint of the diagonal, located at $(\frac{3}{2}, 2)$. Because the line $L$ is perpendicular to $3x+y+6=0$, it must possess a slope of $\frac{1}{3}$, which is the negative reciprocal of the original line's slope of $-3$.

Using the point-slope form with the center $(\frac{3}{2}, 2)$ and slope $\frac{1}{3}$, the equation becomes $y - 2 = \frac{1}{3}(x - \frac{3}{2})$. Multiplying by $6$ to clear the fractions leads to $6y - 12 = 2x - 3$, which simplifies to $2x - 6y + 9 = 0$. The perpendicular distance from the point $(\frac{1}{2}, -5)$ to this line is found using the formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. Substituting the corresponding values gives $\frac{|2(\frac{1}{2}) - 6(-5) + 9|}{\sqrt{2^2 + (-6)^2}}$, which results in $\frac{|1 + 30 + 9|}{\sqrt{4 + 36}}$, or $\frac{40}{\sqrt{40}}$. This simplifies to $\sqrt{40}$, which is $2\sqrt{10}$.

In an equilateral triangle, the orthocenter coincides with the centroid, placing the origin at this central point. The distance from the origin to the line $BC$ defined by $x + 2\sqrt{2} y = 4$ is the length of the apothem, found using the standard perpendicular distance formula as $\frac{|-4|}{\sqrt{1^2 + (2\sqrt{2})^2}}$, which simplifies to $\frac{4}{3}$. Because the centroid divides the triangle's altitude in a $2:1$ ratio, the distance from the centroid to vertex $A$ is exactly twice this apothem, resulting in a length of $\frac{8}{3}$.

Since the altitude passing through $A$ is perpendicular to line $BC$, its slope must be the negative reciprocal of the line's slope, which is $2\sqrt{2}$. Given that this altitude passes through the origin, any point $(\alpha, \beta)$ on it satisfies the relationship $\beta = 2\sqrt{2}\alpha$. We can represent the coordinates of $A$ as $(k, 2\sqrt{2}k)$ and solve for $k$ using the known distance $\frac{8}{3}$ from the origin: $\sqrt{k^2 + (2\sqrt{2}k)^2} = \sqrt{9k^2} = 3|k| = \frac{8}{3}$, leading to $|k| = \frac{8}{9}$. Evaluating the expression $|\alpha + \sqrt{2} \beta|$ by substituting the coordinates gives $|k + \sqrt{2}(2\sqrt{2}k)|$, which simplifies to $|k + 4k| = 5|k|$. Calculating this yields $5 \times \frac{8}{9} = \frac{40}{9}$, or approximately $4.44$, so the greatest integer less than or equal to this value is $4$.

In any triangle, the altitude from a vertex is perpendicular to the opposite side and must pass through the orthocentre. For vertex $A(5, -1)$ and orthocentre $O(0, 0)$, the altitude $AO$ has a slope of $-1/5$, implying that the opposite side $BC$ must have a perpendicular slope of $5$. By letting the third vertex be $C(h, k)$, the slope of $BC$ is $\frac{k-3}{h+2}$, which yields the equation $5h-k=-13$. Similarly, the altitude from vertex $B(-2, 3)$ to $O(0, 0)$ has a slope of $-3/2$, so the side $AC$ must have a perpendicular slope of $2/3$. Using the coordinates of $A(5, -1)$ and $C(h, k)$, the slope $\frac{k-(-1)}{h-5} = 2/3$ simplifies to $2h-3k=13$. Solving the system of equations $5h-k=-13$ and $2h-3k=13$ simultaneously confirms that the coordinates of the third vertex are $(-4, -7)$.

The condition that $2a, b, c$ are three consecutive terms of an arithmetic progression is given by the relation $2b = 2a + c$, which can be rearranged as $2a - 2b + c = 0$. For the family of lines defined by $ax + by + c = 0$ to be concurrent at a fixed point $(x, y)$, this point must satisfy the equation for any $a, b, c$ that adhere to the arithmetic progression constraint. Substituting $x=2$ and $y=-2$ into the line equation $ax + by + c = 0$ results in $2a - 2b + c = 0$. Since this matches the established arithmetic progression condition, all such lines are necessarily concurrent at the point $(2, -2)$.

In an equilateral triangle, the orthocentre and the centroid are identical, and this point divides the median drawn from any vertex to the opposite side in a ratio of 2:1. The first task is to determine the foot of the perpendicular dropped from vertex $P(3, 5)$ onto the line $x + y = 4$, which serves as the base for our altitude. Since the line segment $QR$ lies on $x + y = 4$ with a slope of $-1$, the altitude passing through $P$ must have a slope of $1$. The equation of this line is $y - 5 = 1(x - 3)$, simplifying to $y = x + 2$.

Finding the intersection of $x + y = 4$ and $y = x + 2$ provides the coordinates of the midpoint of $QR$. Substituting $y = x + 2$ into the first equation yields $x + x + 2 = 4$, which simplifies to $2x = 2$, meaning $x = 1$. Consequently, $y = 3$, placing the midpoint at $(1, 3)$. The orthocentre $(\alpha, \beta)$ lies on the line segment connecting $P(3, 5)$ and the midpoint $(1, 3)$, specifically two-thirds of the distance from the vertex $P$ toward the midpoint. Applying the section formula to divide the segment in a 2:1 ratio, $\alpha = \frac{2(1) + 1(3)}{3} = \frac{5}{3}$ and $\beta = \frac{2(3) + 1(5)}{3} = \frac{11}{3}$. Calculating the final requested value, $9(\alpha + \beta) = 9(\frac{5}{3} + \frac{11}{3}) = 48$.

The locus of the midpoint $(h, k)$ of a line segment intersecting the axes can be efficiently determined by utilizing the line's intercept form, $\frac{x}{a} + \frac{y}{b} = 1$, where the midpoint coordinates relate to the intercepts as $a=2h$ and $b=2k$. Solving the system $4x + 3y = 1$ and $3x + 4y = 1$ identifies the fixed intersection point as $\left( \frac{1}{7}, \frac{1}{7} \right)$. Substituting this coordinate into the intercept equation gives $\frac{1/7}{2h} + \frac{1/7}{2k} = 1$. Simplifying this expression results in $\frac{1}{14h} + \frac{1}{14k} = 1$, and multiplying both sides by $14hk$ yields $k + h = 14hk$, which in terms of general coordinates $(x, y)$ is $x + y - 14xy = 0$.

In any rhombus, the diagonals bisect each other at a single common midpoint. This geometric property is useful here because the center of the rhombus is simply the midpoint of the given diagonal connecting $A(1,2)$ and $C(-3,-6)$, which is $M = \left( \frac{1-3}{2}, \frac{2-6}{2} \right) = (-1, -2)$. Because this point $M$ must also serve as the midpoint for the diagonal connecting $B(\alpha, \beta)$ and $D(\gamma, \delta)$, the midpoint coordinates lead to the relations $\frac{\alpha+\gamma}{2} = -1$ and $\frac{\beta+\delta}{2} = -2$. These equations simplify to $\alpha + \gamma = -2$ and $\beta + \delta = -4$. Summing these two values gives $\alpha + \beta + \gamma + \delta = -2 - 4 = -6$, and the absolute value is therefore $6$.

The determination of the two reflected rays relies on identifying the original rays passing through $(-1, -1)$ and then reflecting both the point of intersection and the lines themselves across the mirror $x+2y=1$. The incident rays make a $45^{\circ}$ angle with the line $x+y=0$, which has a slope of $-1$. Using the angle formula $\tan 45^{\circ} = \left| \frac{m - (-1)}{1 + m(-1)} \right| = 1$, we solve for the slopes by setting $\left| \frac{m+1}{1-m} \right| = 1$. This equation yields two solutions: $m=0$, corresponding to the horizontal line $y=-1$, and an undefined slope, corresponding to the vertical line $x=-1$.

To find the reflected equations, we first determine the image of the point $(-1, -1)$ across the mirror line $x+2y=1$ using the formula $\frac{x+1}{1} = \frac{y+1}{2} = -2\frac{-1+2(-1)-1}{1^{2}+2^{2}}$. This calculation simplifies to $\frac{x+1}{1} = \frac{y+1}{2} = \frac{8}{5}$, giving the image point $(\frac{3}{5}, \frac{11}{5})$, which must lie on both reflected rays. By calculating the images of these incident lines, the reflected lines are determined to be $4x+3y=9$ and $-3x+4y=7$.

Comparing these derived equations with the given forms $ax+by=9$ and $cx+dy=7$, we identify the integer coefficients as $a=4, b=3$ for the first line and $c=-3, d=4$ for the second. Substituting these values into the target expression $ad+bc$, we compute $(4)(4) + (3)(-3)$, which results in $16-9=7$.

The intersection of $L_1: x+3=0$ with $L_2: x-y=0$ and $L_3: 3x+y=0$ provides points $A(-3,3)$ and $B(-3,9)$, respectively. The obtuse angle bisector of these two lines is defined by the equation $\frac{x-y}{\sqrt{2}} = \frac{3x+y}{\sqrt{10}}$, as points on this bisector are equidistant from the given lines. Substituting $x=-3$ into this equation determines the intersection point $C$ to be $C\left(-3,\frac{9-3\sqrt{5}}{1+\sqrt{5}}\right)$. Evaluating the ratio of the squared distances from $C$ to $B$ and $A$ using the $y$-coordinate of $C$ yields a final result of $BC^2:AC^2 = 5:1$.

Positioning an equilateral triangle with side length $a$ such that vertex $A$ lies between two parallel lines requires observing the geometric constraints provided by the distances of 6 and 3 units. The sum of these distances, 9 units, represents the total separation between the parallel lines. If side $AB$ creates an angle $\theta$ with the line at distance 3, the height relationship is defined as:
$a \sin \theta = 3 \implies \sin \theta = \frac{3}{a}$
Since the triangle is equilateral, side $AC$ is angled at $60^\circ + \theta$ relative to that same line, satisfying the condition that the total projection onto the perpendicular reaches the other line:
$a \sin(60^\circ + \theta) = 9$
Expanding the trigonometric expression using the sine addition identity $a (\sin 60^\circ \cos \theta + \cos 60^\circ \sin \theta) = 9$ and substituting $\sin \theta = \frac{3}{a}$ along with $\cos \theta = \sqrt{1 - \frac{9}{a^2}}$ results in the equation:
$\frac{\sqrt{3}}{2} \sqrt{1 - \frac{9}{a^2}} + \frac{1}{2} \left( \frac{3}{a} \right) = \frac{9}{a}$
Multiplying by 2 and isolating the radical term gives $\sqrt{3} \sqrt{1 - \frac{9}{a^2}} = \frac{15}{a}$. Squaring both sides of this equation leads to $3(1 - \frac{9}{a^2}) = \frac{225}{a^2}$, which simplifies to $3 - \frac{27}{a^2} = \frac{225}{a^2}$, or $3 = \frac{252}{a^2}$. Consequently, $a^2$ equals 84. Applying the formula for the area of an equilateral triangle provides the final result:
$\text{Area} = \frac{\sqrt{3}}{4} \times 84 = 21 \sqrt{3}$

The vertices of a triangle are determined by the midpoints of its sides, which form the medial triangle; by calculating the coordinates of the original vertices based on these midpoints, we identify the vertices as $(4, 1)$, $(4, 9)$, and $(1, 5)$. Using the distance formula between these vertices, the lengths of the sides opposite to them are found to be $a = 5$, $b = 5$, and $c = 8$. We calculate the incenter $(h, k)$ using the standard formula $(h, k) = \left(\frac{ax_A + bx_B + cx_C}{a+b+c}, \frac{ay_A + by_B + cy_C}{a+b+c}\right)$, substituting our values to obtain $h = \frac{5(4) + 5(4) + 8(1)}{18} = \frac{48}{18} = \frac{8}{3}$ and $k = \frac{5(1) + 5(9) + 8(5)}{18} = \frac{90}{18} = 5$. Substituting these coordinates into the requested expression yields $3h + k = 3\left(\frac{8}{3}\right) + 5 = 8 + 5 = 13$.

Any line passing through the point $P(2, 3)$ and making an angle $\theta$ with the positive $x$-axis can be defined by the parametric coordinates $x = 2 + r \cos \theta$ and $y = 3 + r \sin \theta$, where $r$ represents the distance from $P$. Substituting these coordinates into the given line equation $x + y = 6$ results in $2 + r \cos \theta + 3 + r \sin \theta = 6$, which simplifies to $r(\cos \theta + \sin \theta) = 1$. With the provided distance $r = \sqrt{\frac{2}{3}}$, this relationship becomes $\cos \theta + \sin \theta = \frac{1}{\sqrt{\frac{2}{3}}} = \sqrt{\frac{3}{2}}$.

Squaring both sides of this equation leads to $(\cos \theta + \sin \theta)^2 = \frac{3}{2}$, which expands to $1 + 2 \sin \theta \cos \theta = 1.5$. Using the double-angle identity, this simplifies to $1 + \sin 2\theta = 1.5$, or $\sin 2\theta = \frac{1}{2}$. The equation $\sin 2\theta = \frac{1}{2}$ provides two solutions for $2\theta$, namely $30^\circ$ and $150^\circ$, meaning the possible angles are $\theta = 15^\circ$ and $\theta = 75^\circ$. Expressed in radians, these angles are $\theta_1 = \frac{\pi}{12}$ and $\theta_2 = \frac{5\pi}{12}$. Therefore, their sum is $\theta_1 + \theta_2 = \frac{\pi}{12} + \frac{5\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2}$.

The side length $s$ of the equilateral triangle $PQR$ is constrained by the perpendicular distances from point $P$ to the two parallel lines. Letting $\theta$ represent the angle $PQ$ makes with the normal to the first line, we establish the equations $s \sin \theta = 1$ and $s \cos(30^\circ + \theta) = 4$, derived from the fact that the triangle is equilateral and the total gap between the lines is 5. Dividing these relations gives the ratio:
$\frac{\cos(30^\circ + \theta)}{\sin \theta} = 4$
Expanding this expression with the cosine addition formula yields the equation:
$\frac{\sqrt{3}}{2} \cot \theta - \frac{1}{2} = 4$
This simplifies to $\cot \theta = 3\sqrt{3}$, and because $(QR)^2$ is equivalent to $s^2 = \csc^2 \theta = 1 + \cot^2 \theta$, calculating $1 + (3\sqrt{3})^2$ results in 28.

Determining the reflection of a triangle's centroid involves finding the intersection points of the bounding lines to establish the vertices, calculating the centroid as the average of these coordinates, and then applying the reflection formula across the specified line. Solving the system of equations $7x-6y+3=0$, $x+2y-31=0$, and $9x-2y-19=0$ yields the vertices $(9,11)$, $(3,4)$, and $(5,13)$, which leads to a centroid coordinate of $(\frac{17}{3}, \frac{28}{3})$.

The reflection $(h, k)$ of a point $(x_0, y_0)$ in a line $ax+by+c=0$ is calculated using the relation:
$\frac{h-x_0}{a} = \frac{k-y_0}{b} = -2\frac{ax_0+by_0+c}{a^2+b^2}$
Substituting the centroid $(\frac{17}{3}, \frac{28}{3})$ and the line $3x+6y-53=0$ into this expression, we evaluate the term $-2\frac{3(\frac{17}{3})+6(\frac{28}{3})-53}{3^2+6^2}$. This simplifies to $-2\frac{17+56-53}{9+36}$, which equals $-2\frac{20}{45}$ or $-\frac{8}{9}$. Setting $\frac{h-\frac{17}{3}}{3} = -\frac{8}{9}$ gives $h=3$, and setting $\frac{k-\frac{28}{3}}{6} = -\frac{8}{9}$ gives $k=4$. The final value is computed by evaluating $h^2+k^2+hk$, which equals 37.

A point lies within a triangular region if its coordinates are bounded by the lines that form the sides of the triangle. For a fixed x-coordinate of $\frac{11}{2}$, the range of possible $\alpha$ values is determined by the intersection of the vertical line $x = \frac{11}{2}$ with the boundary lines that enclose the triangle at that position.

Substituting $x = \frac{11}{2}$ into the first boundary equation $x+y=11$ results in $\frac{11}{2} + y = 11$, which simplifies to $\alpha = \frac{11}{2}$. Applying the same x-coordinate to the boundary equation $2x+3y=29$ leads to $2(\frac{11}{2}) + 3y = 29$, which simplifies to $11 + 3y = 29$ and yields $3y = 18$, giving $\alpha = 6$. These points establish the interval $[\frac{11}{2}, 6]$ for $\alpha$. Calculating the product of the smallest and largest values within this range, $\frac{11}{2} \times 6$, results in $33$.

To identify the vertices of the triangle, we first determine where the lines $AB$ and $AC$ intersect the x-axis, as points $B$ and $C$ are constrained to this axis. Setting $y = 0$ in the equation for line $A B$, $3y - x = 2$, yields $x = -2$, confirming $B$ is at $(-2, 0)$. Similarly, applying $y = 0$ to the equation for line $A C$, $x + y = 2$, gives $x = 2$, confirming $C$ is at $(2, 0)$.

The orthocentre $P$ is found at the intersection of the triangle's altitudes. First, solving the equations $3y - x = 2$ and $x + y = 2$ simultaneously gives the coordinates of vertex $A$ as $(1, 1)$. Because the base $BC$ lies on the x-axis, the altitude from $A$ is the vertical line $x = 1$. The altitude from vertex $C$ to line $AB$ must have a slope perpendicular to the slope of $AB$, which is $1/3$; thus, this altitude has a slope of $-3$ and passes through $(2, 0)$, defined by the equation $y = -3x + 6$. Substituting $x = 1$ into this equation confirms the orthocentre $P$ is located at $(1, 3)$.

The area of $\triangle P B C$ can be determined using the base $BC$ and the height provided by the orthocentre. With $B$ at $(-2, 0)$ and $C$ at $(2, 0)$, the base length is $4$. The height of the triangle is the y-coordinate of the orthocentre $P(1, 3)$, which is $3$. Calculating the area with $\frac{1}{2} \times \text{base} \times \text{height}$ gives $\frac{1}{2} \times 4 \times 3$, resulting in an area of $6$.

The area of a triangle formed by a line intercepting the coordinate axes is determined by half the product of the absolute values of its x-intercept and y-intercept. For the line $x+by+c=0$, setting $y=0$ yields an x-intercept of $-c$, while setting $x=0$ yields a y-intercept of $-\frac{c}{b}$, resulting in an area calculation of $\frac{1}{2} |(-c) \cdot (-\frac{c}{b})| = 48$, which simplifies to $c^2 = 96|b|$.

Because the perpendicular line segment from the origin to $L$ makes an angle of $45^{\circ}$ with the positive x-axis, its slope is $\tan(45^{\circ}) = 1$. Given that the slope of $L$ is $-\frac{1}{b}$, the slope of the line perpendicular to $L$ is $b$, which directly implies that $b=1$. Substituting this value into the area relationship gives $c^2 = 96(1)$, leading to the final calculation $b^2 + c^2 = 1^2 + 96 = 97$.

Three lines are concurrent if they share a common point of intersection, a condition satisfied when the determinant of their coefficients is equal to 0. Additionally, the reflection of a point across a line relies on the perpendicular relationship between the segment connecting the point and its image and the line itself.

The value of $\lambda$ is identified by the image formula for a point $(x_0, y_0)$ reflected across the line $ax+by+c=0$, which is $\frac{x-x_0}{a} = \frac{y-y_0}{b} = -2\frac{ax_0+by_0+c}{a^2+b^2}$. Substituting the point $(1, 2)$ and the line $2x-3y+\lambda=0$ into this expression provides an x-coordinate of $\frac{29-4\lambda}{13}$. Setting this value equal to $\frac{57}{13}$ leads to the equation $29-4\lambda = 57$, which results in $\lambda = -7$.

With $\lambda$ determined, the concurrency condition requires the determinant

. Evaluating this determinant produces the relationship $2\alpha - \lambda = 33$. Substituting $\lambda = -7$ gives $2\alpha - (-7) = 33$, which simplifies to $2\alpha = 26$, yielding $\alpha = 13$. The final magnitude $|\alpha \lambda|$ is $|13 \times -7|$, which equals 91.

The diagonals of a square bisect each other at the center, providing a fixed reference point equidistant from all vertices. By solving the system of linear equations representing these diagonals, we identify the intersection point $P$ as $(3-\sqrt{3}, 3+\sqrt{3})$. Because the origin $O$ is one vertex, the squared distance $OP^2$ from the origin to the center is $(3-\sqrt{3})^2 + (3+\sqrt{3})^2$, which simplifies to $12+12=24$. Since this distance is exactly half of the diagonal $OB$, the square of the diagonal $OB^2$ equals $4 \cdot OP^2 = 96$, and with $OB^2 = 2a^2$, it follows that $2a^2 = 96$ and $a^2 = 48$.