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

The path of the reflected light ray is determined by the principle that the angle of incidence equals the angle of reflection, resulting in a reflected line with a slope of $\sqrt{3}$ after it hits the mirror $x+y=1$. The incident ray, originating at the origin with a $30^{\circ}$ angle, follows the line $y = \frac{x}{\sqrt{3}}$ and meets the mirror at the point $(\frac{\sqrt{3}}{\sqrt{3}+1}, \frac{1}{\sqrt{3}+1})$. Applying the point-slope form to the reflected ray, we get the equation $y - \frac{1}{\sqrt{3}+1} = \sqrt{3}(x - \frac{\sqrt{3}}{\sqrt{3}+1})$. Setting $y=0$ to find the $x$-intercept $Q$ where the ray hits the $x$-axis yields the final abscissa $x = \frac{2}{3+\sqrt{3}}$.

Finding the centroid of a triangle defined by three linear equations requires determining the intersection points of the lines, which act as the vertices of the triangle. By solving each pair of simultaneous linear equations, we pinpoint the three coordinates (6, 8), (1, 2), and (5, -7). Since the centroid ($\alpha, \beta$) is defined as the average of the vertices' coordinates, we calculate $\alpha = \frac{6+1+5}{3} = 4$ and $\beta = \frac{8+2-7}{3} = 1$.

With the centroid found, we identify the roots of the required quadratic equation by evaluating the expressions $\alpha + 2\beta$ and $2\alpha - \beta$. Substituting our values results in the roots 6 and 7. Applying the standard quadratic construction formula $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$, we sum these to get 13 and multiply them to get 42, resulting in the final equation $x^2 - 13x + 42 = 0$

An angular bisector serves as a line of symmetry between two intersecting lines, meaning any point on one line, when reflected across the bisector, will lie on the other line. Because the angular bisector $l_1: 3y - 2x = 3$ passes through the intersection of the two lines $l_2: x - y + 1 = 0$ and $l_3: \alpha x + \beta y + 17 = 0$, all three lines must intersect at the same coordinate.

Solving the equations $3y - 2x = 3$ and $x - y + 1 = 0$ simultaneously reveals that their point of intersection is $(0, 1)$. Since this point must also lie on $l_3$, substituting $x = 0$ and $y = 1$ into $\alpha x + \beta y + 17 = 0$ results in $0 + \beta + 17 = 0$, which gives $\beta = -17$.

To determine $\alpha$, select an arbitrary point $Q(-1, 0)$ on $l_2$. Its reflection $Q'$ across the mirror line $l_1$, which can be rewritten as $2x - 3y + 3 = 0$, must lie on $l_3$. Applying the reflection formula $\frac{x - (-1)}{2} = \frac{y - 0}{-3} = -2\frac{2(-1) - 3(0) + 3}{2^2 + (-3)^2}$ yields the coordinate $Q'\left(-\frac{17}{13}, \frac{6}{13}\right)$. Substituting these values into the line equation $\alpha x - 17y + 17 = 0$ leads to $\alpha\left(-\frac{17}{13}\right) - 17\left(\frac{6}{13}\right) + 17 = 0$, which simplifies to $17\alpha = 119$, confirming $\alpha = 7$.

Evaluating the expression $\alpha^2 + \beta^2 - \alpha - \beta$ using $\alpha = 7$ and $\beta = -17$ gives $7^2 + (-17)^2 - 7 - (-17)$, which equals $49 + 289 - 7 + 17$, resulting in 348.

When a ray of light reflects off a flat surface like the x-axis, the law of reflection dictates that the angle of incidence equals the angle of reflection. This implies that the normal to the mirror at the point of reflection $A$ serves as the angle bisector of the reflected path $\angle PAQ$. To find the coordinates of $A(x, 0)$, we equate the magnitude of the slopes of segments $PA$ and $AQ$ since they must be symmetric with respect to the normal. This gives the equation $\frac{3-0}{x-2} = \frac{4-0}{5-x}$, which simplifies to $3(5-x) = 4(x-2)$, resulting in $15 - 3x = 4x - 8$ and $x = 23/7$. Thus, the point of reflection is $A(23/7, 0)$, and the angle bisector is the vertical line $x = 23/7$.

Point $R$ divides the line segment $AQ$ internally in the ratio $2:1$. Utilizing the section formula with $A(23/7, 0)$ and $Q(5, 4)$, the coordinates of $R$ are calculated as $\left( \frac{2(5) + 1(23/7)}{2+1}, \frac{2(4) + 1(0)}{2+1} \right)$, which simplifies to $\left( \frac{10 + 23/7}{3}, \frac{8}{3} \right)$ or $R = \left( \frac{31}{7}, \frac{8}{3} \right)$.

The foot of the perpendicular $M(\alpha, \beta)$ from $R$ to the vertical bisector line $x = 23/7$ is found by keeping the y-coordinate of $R$ constant and setting the x-coordinate to that of the bisector line. This yields $M = \left( \frac{23}{7}, \frac{8}{3} \right)$, where $\alpha = 23/7$ and $\beta = 8/3$. Substituting these values into the expression $7\alpha + 3\beta$ leads to $7(23/7) + 3(8/3) = 23 + 8$, which equals 31.

The orthocenter of a triangle lies at the intersection of its altitudes, which implies that the line segment connecting any vertex to the orthocenter is perpendicular to the opposite side. We first determine vertex $A$ by solving the system formed by sides $AB$ ($2x+y=0$) and $CA$ ($x-y=3$), resulting in $A(1, -2)$. The altitude $AH$ connects vertex $A$ to the orthocenter $H(2, a)$, so its slope is calculated as $(a - (-2))/(2 - 1) = a + 2$. Because this altitude is perpendicular to side $BC$ with equation $x+py=15a$ (and slope $-1/p$), the product of their slopes is $-1$, which establishes the relationship $p = a + 2$.

Next, we identify vertex $C$ by solving the intersection of $BC$ and $CA$, yielding coordinates $((18p-30)/(p+1), (15p-33)/(p+1))$ after substituting $a=p-2$. Since $HC$ is an altitude perpendicular to side $AB$, which has a slope of $-2$, the slope of $HC$ must be $1/2$. Setting the slope formula for the segment between $H(2, a)$ and $C$ equal to $1/2$ and simplifying the expression leads to the quadratic equation $p^2-8p+15=0$, giving two possible values, $p=3$ and $p=5$. Given the restriction that $-\frac{1}{2} < a < 2$ and noting that $a=p-2$, we find that $p=5$ results in $a=3$, which falls outside the permitted range, leaving $p=3$ as the valid solution.

Given vertices $A(\alpha, -2)$, $B(\alpha, 6)$, and $C(\frac{\alpha}{4}, -2)$, we identify that segment $AB$ is vertical and $AC$ is horizontal, which confirms that $\triangle ABC$ is a right-angled triangle with the right angle located at vertex $A$. For any right-angled triangle, the circumcenter is precisely the midpoint of the hypotenuse $BC$. Calculating the coordinates of this midpoint yields:
$\left( \frac{\alpha + \frac{\alpha}{4}}{2}, \frac{6 - 2}{2} \right) = \left( \frac{5\alpha}{8}, 2 \right)$
Comparing this to the given circumcenter $(5, \frac{\alpha}{4})$, we equate the vertical coordinates to find $2 = \frac{\alpha}{4}$, which gives $\alpha = 8$. Checking the horizontal coordinate, $\frac{5(8)}{8} = 5$ confirms that $\alpha = 8$ is the correct value.

Substituting $\alpha = 8$ into the vertex coordinates results in $A(8, -2)$, $B(8, 6)$, and $C(2, -2)$. The lengths of the perpendicular legs are $AB = 6 - (-2) = 8$ and $AC = 8 - 2 = 6$, making the hypotenuse $BC = \sqrt{8^2 + 6^2} = 10$. The area of this triangle is calculated as $\frac{1}{2} \times 8 \times 6 = 24$. The perimeter is the sum of these sides, $8 + 6 + 10 = 24$. The circumradius is half the hypotenuse, $\frac{10}{2} = 5$, and the inradius is the area divided by the semi-perimeter, which is $\frac{24}{12} = 2$. Since the perimeter is actually 24, the option stating it is 25 is the incorrect one.

The locus of a point $P(x,y)$ whose sum of squared distances to two fixed points is constant represents a circle, and the geometry of this curve relative to the coordinate axes allows us to define the area of the resulting quadrilateral. By expressing the distance condition algebraically, we can determine the specific equation of the curve and subsequently extract the intercepts, which serve as the vertices for the quadrilateral.

Given the points $(1,2)$ and $(-2,1)$, the condition $(x-1)^2 + (y-2)^2 + (x+2)^2 + (y-1)^2 = 14$ simplifies to the circle equation $x^2 + y^2 + x - 3y - 2 = 0$. Setting $y=0$ identifies the x-intercepts as the roots of $x^2 + x - 2 = 0$, which are $1$ and $-2$, creating a segment of length $3$. Setting $x=0$ identifies the y-intercepts as the roots of $y^2 - 3y - 2 = 0$, where the difference between roots-representing the length of the segment on the y-axis-is determined by the discriminant formula $\sqrt{b^2 - 4ac}$ as $\sqrt{(-3)^2 - 4(1)(-2)}$, which equals $\sqrt{17}$. Because these segments lie on the perpendicular coordinate axes, they function as the diagonals of the quadrilateral, and the area is calculated as $\frac{1}{2} \times 3 \times \sqrt{17}$, resulting in $\frac{3\sqrt{17}}{2}$.

The centroid of a triangle is the arithmetic mean of its vertex coordinates, and since the orthocenter represents the intersection of the triangle's altitudes, determining the vertices requires finding the intersection of these altitude lines with the respective sides. Vertex $A$ is determined by solving the system of equations for the two provided sides, $x-2y+1=0$ and $2x-y-1=0$, which yields the point $(1,1)$.

With vertex $A$ established, we define the altitude lines that pass through the orthocenter $H\left(\frac{7}{3},\frac{7}{3}\right)$ and are perpendicular to the sides. The side $x-2y+1=0$ has a slope of $\frac{1}{2}$, meaning the altitude from vertex $B$ must have a slope of $-2$. Using the point-slope form, the equation for this altitude $BH$ is $2x+y-7=0$, and solving this simultaneously with the side $2x-y-1=0$ locates vertex $B$ at $(2,3)$. Similarly, the side $2x-y-1=0$ has a slope of $2$, so the altitude from vertex $C$ perpendicular to this side must have a slope of $-\frac{1}{2}$, resulting in the equation $x+2y-7=0$. Intersecting this altitude line with the side $x-2y+1=0$ identifies vertex $C$ at $(3,2)$. Finally, the centroid $G$ is calculated as the average of the vertices $(1,1)$, $(2,3)$, and $(3,2)$, which results in $G(2,2)$. The distance from the origin to this centroid is $\sqrt{2^2+2^2}$, yielding $2\sqrt{2}$.

In an isosceles triangle, the vertex $A$ is equidistant from the endpoints of the base $BC$, and the centroid is the arithmetic mean of the three vertices' coordinates. To find the location of vertex $B$, we solve for the intersection of the base line $2x + y = 4$ and the line $x + 3y = 7$. Solving these simultaneous linear equations, we find $x = 1$ and $y = 2$, establishing point $B$ as $(1, 2)$.

Any point $C$ on the line $2x + y = 4$ can be parameterized as $(k, 4 - 2k)$. Since the triangle is isosceles with vertex $A(6, 1)$, the square of the distances $AB$ and $AC$ must be equal. Calculating $AB^2 = (6 - 1)^2 + (1 - 2)^2 = 25 + 1 = 26$, we equate this to $AC^2 = (6 - k)^2 + (1 - (4 - 2k))^2 = (6 - k)^2 + (2k - 3)^2$. Expanding the right side gives $36 - 12k + k^2 + 4k^2 - 12k + 9 = 5k^2 - 24k + 45$. Setting this equal to $26$ results in the quadratic equation $5k^2 - 24k + 19 = 0$. Factoring this leads to $(5k - 19)(k - 1) = 0$, where $k = 1$ corresponds to vertex $B$ and $k = \frac{19}{5}$ identifies the coordinates of vertex $C$ as $(\frac{19}{5}, -\frac{18}{5})$.

The centroid $(\alpha, \beta)$ is the average of the coordinates of vertices $A(6, 1)$, $B(1, 2)$, and $C(\frac{19}{5}, -\frac{18}{5})$. Summing the $x$-coordinates yields $\alpha = \frac{6 + 1 + \frac{19}{5}}{3} = \frac{18}{5}$, and summing the $y$-coordinates yields $\beta = \frac{1 + 2 - \frac{18}{5}}{3} = -\frac{1}{5}$. Evaluating the final expression $15(\alpha + \beta)$ gives $15(\frac{18}{5} - \frac{1}{5})$, which simplifies to $15(\frac{17}{5}) = 51$.

To find the positions of the points $A$ and $A'$ on the line $y=2$, we rely on the geometric property of the angle subtended by segments at the origin. With the origin $O(0,0)$ and point $B(2,3)$, the slope of the line segment $OB$ is $m_2 = \frac{3}{2}$. Any point $A$ on the line $y=2$ has coordinates $(x, 2)$, meaning the slope of the segment $OA$ is $m_1 = \frac{2}{x}$. The problem states that the angle between the segments $OA$ and $OB$ is $\frac{\pi}{4}$, and we use the standard formula for the tangent of the angle between two lines to find the value of $x$:

$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$

Substituting the known values into this equation gives $\left| \frac{\frac{3}{2} - \frac{2}{x}}{1 + \left(\frac{3}{2}\right)\left(\frac{2}{x}\right)} \right| = 1$. Simplifying the fraction inside the absolute value leads to $\left| \frac{3x - 4}{2x + 6} \right| = 1$. This equation generates two distinct scenarios: $\frac{3x - 4}{2x + 6} = 1$ and $\frac{3x - 4}{2x + 6} = -1$. Solving the first equation yields $x = 10$, while the second results in $x = -\frac{2}{5}$. Since both points $A$ and $A'$ lie on the line $y=2$, the distance between them is simply the difference between their $x$-coordinates, which is $10 - \left(-\frac{2}{5}\right) = \frac{52}{5}$.

The altitude of an equilateral triangle formed by a vertex and a line is equal to the perpendicular distance from that vertex to the line. Using the point-to-line distance formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$, the height $h$ from the point $(3, 7)$ to the line $x+y-5=0$ is determined as
$h = \frac{|3+7-5|}{\sqrt{1^2+1^2}} = \frac{5}{\sqrt{2}}$

For any equilateral triangle with side length $a$, the altitude and side are related by $h = a \sin 60^{\circ} = \frac{a\sqrt{3}}{2}$. Solving for $a$ gives $a = \frac{2h}{\sqrt{3}}$. Substituting the value for $h$, the side length is
$a = \frac{2}{\sqrt{3}} \times \frac{5}{\sqrt{2}} = \frac{10}{\sqrt{6}}$

The area of the triangle is calculated using the formula $\frac{\sqrt{3}}{4} a^2$. Substituting the value for $a$ yields
$\text{Area} = \frac{\sqrt{3}}{4} \left( \frac{10}{\sqrt{6}} \right)^2 = \frac{\sqrt{3}}{4} \times \frac{100}{6} = \frac{25\sqrt{3}}{6}$
Simplifying this expression by dividing the numerator and denominator by $\sqrt{3}$ results in an area of $\frac{25}{2\sqrt{3}}$.

The circumcentre of a triangle is equidistant from all its vertices and must lie on the perpendicular bisector of each side. Since the line segment $AC$ connects the vertices $A(a, 3)$ and $C(a, b)$, it is a vertical line segment, meaning its perpendicular bisector is the horizontal line $y = \frac{3+b}{2}$. Because the circumcentre $P(1, 1)$ must lie on this line, we immediately obtain the equation $1 = \frac{3+b}{2}$, which yields $b = -1$. With $b$ determined, the vertex $C$ becomes $(a, -1)$, and because the distance from $P$ to any vertex must be equal, we can equate $PA^2 = PB^2$. Using the coordinates $A(a, 3)$, $B(-1, 5)$, and $P(1, 1)$, this equality is $(a-1)^2 + (3-1)^2 = (-1-1)^2 + (5-1)^2$, which simplifies to $(a-1)^2 + 4 = 20$. This leads to $(a-1)^2 = 16$, and given the condition $ab > 0$ where $b = -1$, we must have $a < 0$, confirming that $a = -3$.

With the vertices fixed as $A(-3, 3)$, $B(-1, 5)$, and $C(-3, -1)$, we can now determine the equations of the required lines. The line $BC$ passes through $(-1, 5)$ and $(-3, -1)$, resulting in a slope of $\frac{-1-5}{-3-(-1)} = 3$, giving the equation $y - 5 = 3(x + 1)$, or $y = 3x + 8$. The line $AP$ passes through $A(-3, 3)$ and $P(1, 1)$, having a slope of $\frac{1-3}{1-(-3)} = -\frac{1}{2}$, which results in the equation $y - 1 = -\frac{1}{2}(x - 1)$, or $y = \frac{3-x}{2}$. To find the intersection point $Q(k_1, k_2)$, we equate the two expressions for $y$ to get $3x + 8 = \frac{3-x}{2}$, which simplifies to $6x + 16 = 3 - x$, yielding $x = -\frac{13}{7}$. Substituting this back into the equation for the line $BC$ gives $y = 3(-\frac{13}{7}) + 8 = \frac{17}{7}$. The value of $k_1 + k_2$ is therefore $-\frac{13}{7} + \frac{17}{7} = \frac{4}{7}$.

The properties of the circumcentre $P(2,3)$ and the intersection of side equations allow us to locate all three vertices of triangle $ABC$. Calculating the intersection of $2x+y=0$ and $x-y=3$ provides vertex $A(1,-2)$. Given the circumcentre $P(2,3)$, we determine vertices $B(-13/5, 26/5)$ and $C(7,4)$ using the perpendicular bisectors $x-2y=-4$ and $x+y=5$. Substituting the coordinates of $C(7,4)$ into the side equation $x+py=39$ results in $7+4p=39$, confirming $p=8$.

Calculating $(AC)^2$ gives $(7-1)^2+(4-(-2))^2 = 72$. This verifies that $(AC)^2=9(8)=72$ and $(AC)^2+p^2 = 72+64=136$ are both true. Finally, the area of the triangle is $32.4$, which makes the inequality $32 < 32.4 < 36$ true, while the assertion that the area lies between 34 and 38 is false.

The set of points equidistant from two intersecting lines corresponds to the angle bisectors of those lines, found by setting the perpendicular distance from a coordinate pair $(x, y)$ to each line equal: $\frac{|3x-4y+12|}{5} = \frac{|8x+6y+11|}{10}$. This equation simplifies to the condition $2|3x-4y+12| = |8x+6y+11|$. Given the positional constraints for point $P(\alpha, \beta)$, where it lies below the first line and above the second, the appropriate angle bisector equation is $2(3x-4y+12) = 8x+6y+11$, which simplifies to $2x+14y-13=0$.

Simultaneously, the unit distance condition from the line $L_1$ requires that $\frac{3\alpha-4\beta+12}{5} = 1$, which simplifies to $3\alpha-4\beta+7=0$. Solving this system of two linear equations yields $\alpha = -\frac{23}{25}$ and $\beta = \frac{53}{50}$. Calculating the final expression $100(\alpha+\beta)$ with these coordinates provides $100\left(-\frac{23}{25} + \frac{53}{50}\right) = 100\left(-\frac{46}{50} + \frac{53}{50}\right) = 100\left(\frac{7}{50}\right)$, which equals $14$.

The slopes $m_1$ and $m_2$ of adjacent sides in a square are perpendicular, satisfying the relationship $m_1 m_2 = -1$. By analyzing the provided diagonal equation and vertex coordinate, the side length of the square is identified as $a=10$. Substituting $a=10$ into the given condition $a^2+11 a+3(m_1^2+m_2^2)=220$ simplifies the expression to $100 + 110 + 3(m_1^2 + \frac{1}{m_1^2}) = 220$, which leads to $3(m_1^2 + \frac{1}{m_1^2}) = 10$. Given that the slopes are related to the angle $\alpha$ as $\tan \alpha$ and $\cot \alpha$, the equation becomes $3(\tan^2 \alpha + \cot^2 \alpha) = 10$, resulting in $\tan^2 \alpha = 3$ or $\tan^2 \alpha = \frac{1}{3}$.

With these values, we can determine the expression $72(\sin^4 \alpha + \cos^4 \alpha) + a^2 - 3a + 13$. Utilizing the identity $\sin^4 \alpha + \cos^4 \alpha = 1 - 2\sin^2 \alpha \cos^2 \alpha$, we substitute $\tan^2 \alpha = 3$ to find $\sin^2 \alpha = \frac{3}{4}$ and $\cos^2 \alpha = \frac{1}{4}$. This yields $\sin^4 \alpha + \cos^4 \alpha = (\frac{3}{4})^2 + (\frac{1}{4})^2 = \frac{9}{16} + \frac{1}{16} = \frac{5}{8}$. The first part of the expression calculates to $72(\frac{5}{8}) = 45$, and the second part evaluates to $10^2 - 3(10) + 13 = 100 - 30 + 13 = 83$. Adding these values together, $45 + 83$ equals $128$.

The vertices of the triangle are identified by finding the intersection of lines $L_1: 2x+5y=10$ and $L_2: -4x+3y=12$. Solving this system yields the vertex $C \equiv \left( -\frac{15}{13}, \frac{32}{13} \right)$. Points $A$ and $B$ lie on $L_2$ and $L_1$ respectively, and point $P(2,3)$ divides the segment $AB$ in the ratio $1:3$. Applying the section formula with $A = \left( x_1, \frac{12+4x_1}{3} \right)$ and $B = \left( x_2, \frac{10-2x_2}{5} \right)$ leads to the linear relations $3x_1+x_2=8$ and $10x_1-x_2=-5$.

These equations determine the coordinates $x_1 = \frac{3}{13}$ and $x_2 = \frac{95}{13}$, establishing the vertex positions $A = \left( \frac{3}{13}, \frac{56}{13} \right)$ and $B = \left( \frac{95}{13}, -\frac{12}{13} \right)$. With all three vertices known, the area of the triangle is calculated using the coordinate geometry formula $\frac{1}{2} |x_C(y_A - y_B) + x_A(y_B - y_C) + x_B(y_C - y_A)|$. Substituting these coordinates results in an area of $\frac{132}{13}$ sq. units.

When two triangles share the same vertex and their bases lie on the same straight line, the ratio of their areas is identical to the ratio of their bases. Since $P$ lies on $BC$ and the ratio $\Delta_1 / \Delta_2 = 4/7$, the segment $BP$ must represent $4/7$ of the total length $BC$, meaning $P$ divides the segment $BC$ internally in a ratio of $4:3$. By applying the section formula to the coordinates of $B(-4,3)$ and $C(-2,-5)$, the point $P$ is located at $(-\frac{20}{7}, -\frac{11}{7})$.

With the position of $P$ established, we identify the equations for lines $AC$ and $AP$. The line passing through $A(1,1)$ and $C(-2,-5)$ has a slope of $2$, resulting in the equation $y - 1 = 2(x - 1)$, which intersects the $x$-axis at $(\frac{1}{2}, 0)$. Similarly, the line passing through $A(1,1)$ and $P(-\frac{20}{7}, -\frac{11}{7})$ has a slope of $\frac{2}{3}$, leading to the equation $y - 1 = \frac{2}{3}(x - 1)$, which intersects the $x$-axis at $(-\frac{1}{2}, 0)$.