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

The area of a triangle formed by a set of coordinates is defined as half the product of the base length and the perpendicular height. Considering the points A(1, -1) and B(0, 2), the base length $AB$ is $\sqrt{(1-0)^2 + (-1-2)^2} = \sqrt{10}$. The equation of the line passing through these points is determined to be $3x + y - 2 = 0$. Given that the area of triangle PAB is 5, we can solve for the required height $h$ using the relationship $\frac{1}{2} \sqrt{10} h = 5$, which yields $h = \sqrt{10}$.

Point P resides on the line $3x + y - 4\lambda = 0$, meaning its perpendicular distance to the line $AB$ must also be $\sqrt{10}$. Utilizing the distance formula for parallel lines, we have $\frac{|-4\lambda - (-2)|}{\sqrt{3^2 + 1^2}} = \sqrt{10}$. This simplifies to $|2 - 4\lambda| = 10$, resulting in the linear equations $2 - 4\lambda = 10$ or $2 - 4\lambda = -10$. Solving for $\lambda$ provides the values $\lambda = -2$ and $\lambda = 3$.

The distance between two parallel lines defined as $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is calculated using the formula $\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. Because parallel lines maintain proportional coefficients, we must first normalize the given equations to match the coefficients of the reference line $2x - y + 3 = 0$.

Dividing $4x - 2y + \alpha = 0$ by $2$ transforms it into $2x - y + \frac{\alpha}{2} = 0$. Applying the distance formula $\frac{|\frac{\alpha}{2} - 3|}{\sqrt{2^2 + (-1)^2}} = \frac{1}{\sqrt{5}}$ simplifies to $\frac{|\alpha - 6|}{2\sqrt{5}} = \frac{1}{\sqrt{5}}$, which leads to the equation $|\alpha - 6| = 2$. This gives two solutions, $\alpha = 8$ and $\alpha = 4$, which sum to 12. Similarly, we normalize $6x - 3y + \beta = 0$ by dividing by $3$ to obtain $2x - y + \frac{\beta}{3} = 0$. Using the distance requirement $\frac{|\frac{\beta}{3} - 3|}{\sqrt{2^2 + (-1)^2}} = \frac{2}{\sqrt{5}}$ simplifies to $\frac{|\beta - 9|}{3\sqrt{5}} = \frac{2}{\sqrt{5}}$, resulting in $|\beta - 9| = 6$. This produces $\beta = 15$ and $\beta = 3$, with a sum of 18. Adding these two results together, the sum of all possible values for $\alpha$ and $\beta$ is 30.

A perpendicular bisector passes through the midpoint of a line segment and possesses a slope that is the negative reciprocal of the original segment's slope. The midpoint of $P(1,4)$ and $Q(k,3)$ is $(\frac{k+1}{2}, \frac{7}{2})$, and the slope of the line segment $PQ$ is $\frac{3-4}{k-1} = -\frac{1}{k-1}$, so the slope of the perpendicular bisector must be $k-1$. Substituting these values into the point-slope equation $y - \frac{7}{2} = (k-1)(x - \frac{k+1}{2})$ and setting the $y$-intercept to $-4$ by substituting $x=0$ and $y=-4$ gives $-4 - \frac{7}{2} = (k-1)(-\frac{k+1}{2})$. This simplifies to $-\frac{15}{2} = -\frac{k^2-1}{2}$, which leads to $15 = k^2 - 1$, resulting in $k^2 = 16$, or $k$ being $4$ or $-4$.

The reflection of light follows the principle that the angle of incidence equals the angle of reflection relative to the normal of the reflecting surface, which in this instance is a horizontal line perpendicular to the vertical wall $x=1$. Given the incident ray originates at $(2, 2\sqrt{3})$ and strikes the boundary at a $30^{\circ}$ angle, the point of contact $A$ is determined to be $(1, \sqrt{3})$ based on the tangent of the ray's inclination. Since the surface is vertical, the reflected ray follows a path with a slope of $-\sqrt{3}$, calculated from $\tan 120^{\circ}$. The equation for the reflected line $AB$ is therefore $y - \sqrt{3} = -\sqrt{3}(x - 1)$, which simplifies to the linear form $\sqrt{3}x + y = 2\sqrt{3}$. Testing the given coordinates confirms that the point $(3, -\sqrt{3})$ satisfies this equation exactly, placing it on the reflected ray.

The height of the intersection point P above the ground is determined by the properties of similar triangles formed by the two vertical poles and their connecting lines. By dropping a perpendicular from P to the ground AC, the total distance between the poles is divided into two segments that create proportional relationships with the heights of the poles.

The height $h$ and the total distance between the poles $x$ relate through the segments $x_1$ and $x_2$ created by the perpendicular from P. Using the geometric properties of similar triangles, the ratios follow these equations:

$\frac{h}{15} = \frac{x_2}{x} \quad \text{and} \quad \frac{h}{10} = \frac{x_1}{x}$

Summing these expressions results in $\frac{h}{15} + \frac{h}{10} = \frac{x_1 + x_2}{x}$. Because $x_1 + x_2$ equals the total distance $x$, the equation simplifies to $\frac{h}{15} + \frac{h}{10} = 1$. Combining the fractions yields $h \left( \frac{2+3}{30} \right) = 1$, which produces the height $h = 6$.

Finding the image of a point across a line relies on the principle of perpendicular symmetry: the line segment connecting the original point and its image must be perpendicular to the reflecting line, and the midpoint of this segment must lie on that line. To start, identify the equation of the line using the intercept form, $\frac{x}{3} + \frac{y}{1} = 1$, which simplifies to $x + 3y - 3 = 0$.

To calculate the coordinates $(x, y)$ of the reflected point from an original point $(x_0, y_0)$ in the line $ax + by + c = 0$, we utilize the relationship $\frac{x - x_0}{a} = \frac{y - y_0}{b} = -2 \frac{ax_0 + by_0 + c}{a^2 + b^2}$. Substituting the given point $(-1, -4)$ and the coefficients $a=1$, $b=3$, and $c=-3$, we determine the ratio $\frac{x+1}{1} = \frac{y+4}{3} = -2 \frac{1(-1) + 3(-4) - 3}{1^2 + 3^2}$. Simplifying the right side yields $-2 \frac{-16}{10}$, which equals $\frac{16}{5}$. Finally, setting $\frac{x+1}{1} = \frac{16}{5}$ gives $x = \frac{11}{5}$, and setting $\frac{y+4}{3} = \frac{16}{5}$ gives $y = \frac{48}{5} - 4$, which results in $y = \frac{28}{5}$, providing the final coordinates $\left(\frac{11}{5}, \frac{28}{5}\right)$.

The orthocenter of a triangle is the point where its altitudes intersect, and since an altitude is perpendicular to the base opposite its vertex, we find this intersection by constructing the equations for two such perpendicular lines. First, the slope of side $BC$ connecting $(-7,1)$ and $(5,-5)$ is calculated as $\frac{-5-1}{5-(-7)} = -\frac{1}{2}$, which means the altitude from vertex $A(-1,7)$ must have a perpendicular slope of $2$. Using the point-slope form, the equation for this altitude is $y-7 = 2(x+1)$, which simplifies to $2x-y+9=0$.

Next, determine the slope of side $AC$ connecting $(-1,7)$ and $(5,-5)$, which is $\frac{-5-7}{5-(-1)} = -2$, implying the altitude from vertex $B(-7,1)$ has a perpendicular slope of $\frac{1}{2}$. This produces the equation $y-1 = \frac{1}{2}(x+7)$, which rearranges to $x-2y+9=0$. Solving these two linear equations simultaneously yields $x=-3$ and $y=3$, identifying the orthocenter at $(-3,3)$.

Two points lie on the same side of a line $x + y - 1 = 0$ if their coordinates, when substituted into the linear expression, yield values with the same sign. Substituting the point (1, 2) into this expression yields $1 + 2 - 1 = 2$, which is positive, so the point $(\sin \theta, \cos \theta)$ must satisfy $\sin \theta + \cos \theta - 1 > 0$, or $\sin \theta + \cos \theta > 1$. Dividing both sides by $\sqrt{2}$ converts this inequality into $\sin(\theta + \frac{\pi}{4}) > \frac{1}{\sqrt{2}}$. Solving for the specified range of $\theta$, the angle $\theta + \frac{\pi}{4}$ must lie between $\frac{\pi}{4}$ and $\frac{3\pi}{4}$, which leads to the interval $0 < \theta < \frac{\pi}{2}$.

A point that divides a triangle into three triangles of equal area is necessarily the centroid, which is determined by calculating the arithmetic mean of the coordinates of the triangle's vertices. Using the vertices $A(1, 0)$, $B(6, 2)$, and $C\left(\frac{3}{2}, 6\right)$, the $x$-coordinate of the centroid $P$ is $\frac{1 + 6 + \frac{3}{2}}{3} = \frac{17}{6}$, while the $y$-coordinate is $\frac{0 + 2 + 6}{3} = \frac{8}{3}$, establishing $P$ at $\left(\frac{17}{6}, \frac{8}{3}\right)$. Computing the distance between this point and $Q\left(-\frac{7}{6}, -\frac{1}{3}\right)$ via the distance formula yields $\sqrt{\left(\frac{17}{6} - (-\frac{7}{6})\right)^2 + \left(\frac{8}{3} - (-\frac{1}{3})\right)^2}$, which simplifies to $\sqrt{\left(\frac{24}{6}\right)^2 + \left(\frac{9}{3}\right)^2}$. Squaring these differences leads to $\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$.

The locus of the midpoints is determined by tracking the coordinates of a perpendicular projection from a moving point on the line $x = 2y$ onto the line $x = y$. Let a variable point on the first line be $(2t, t)$; the perpendicular line through this point to $x - y = 0$ has a slope of $-1$ and is defined by the equation $y - t = -1(x - 2t)$, which simplifies to $x + y = 3t$.

Determining the intersection of $x + y = 3t$ and $x - y = 0$ reveals the foot of the perpendicular at $(1.5t, 1.5t)$. The coordinates $(h, k)$ of the midpoint between the variable point $(2t, t)$ and the foot $(1.5t, 1.5t)$ are found by averaging their coordinates, resulting in $h = \frac{2t + 1.5t}{2} = 1.75t$ and $k = \frac{t + 1.5t}{2} = 1.25t$. Converting these to fractions yields $h = \frac{7t}{4}$ and $k = \frac{5t}{4}$, which implies a ratio of $\frac{h}{k} = \frac{7}{5}$. This simplifies to $5h = 7k$, leading to the final equation $5x - 7y = 0$.

The equation of a line passing through the intersection of two given lines is defined by the family of lines expression $L_1 + \lambda L_2 = 0$. First, calculate the centroid of the triangle by averaging the x and y coordinates of its vertices at (3, -1), (1, 3), and (2, 4), which results in the point $(2, 2)$. Because the line passes through $(2, 2)$, substitute these values into the expression $(x + 3y - 1) + \lambda(3x - y + 1) = 0$ to solve for $\lambda$. This substitution yields $7 + 5\lambda = 0$, so $\lambda = -7/5$. Replacing $\lambda$ with $-7/5$ in the expression and simplifying results in $5(x + 3y - 1) - 7(3x - y + 1) = 0$, which simplifies to $8x - 11y + 6 = 0$. Testing the coordinate $(-9, -6)$ in this final equation gives $8(-9) - 11(-6) + 6 = -72 + 66 + 6 = 0$, confirming that this point satisfies the line equation.

The intercept form of a line equation, $\frac{x}{a} + \frac{y}{b} = 1$, identifies where the line meets the axes at points $(a, 0)$ and $(0, b)$, meaning the midpoint of the segment between them is located at $(\frac{a}{2}, \frac{b}{2})$. Since the point $P(-3, 4)$ bisects this intercepted segment, we set $\frac{a}{2} = -3$ and $\frac{b}{2} = 4$ to determine the intercepts $a = -6$ and $b = 8$. Substituting these values into the intercept form produces $\frac{x}{-6} + \frac{y}{8} = 1$, and multiplying through by 24 yields the final equation $4x - 3y + 24 = 0$.

The equation of a straight line at a perpendicular distance $p$ from the origin, where the normal makes an angle $\alpha$ with the positive $x$-axis, is expressed by the normal form $x \cos \alpha + y \sin \alpha = p$. The line $x + y = 0$ corresponds to a slope of $-1$, which makes an angle of $135^\circ$ with the positive $x$-axis, so the normal, which is inclined at $60^\circ$ to this line, has an inclination of $\alpha = 135^\circ - 60^\circ = 75^\circ$. Substituting $p = 4$ and $\alpha = 75^\circ$ into the normal form gives $x \cos 75^\circ + y \sin 75^\circ = 4$. Using the exact values $\cos 75^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}$ and $\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$ transforms the equation into $x(\sqrt{6} - \sqrt{2}) + y(\sqrt{6} + \sqrt{2}) = 16$, which simplifies to $(\sqrt{3} - 1)x + (\sqrt{3} + 1)y = 8\sqrt{2}$ after dividing both sides by $\sqrt{2}$.

In any parallelogram, the diagonals bisect each other at a common midpoint. Using the coordinates of $B(3, 4)$ and $C(2, 5)$, the midpoint of diagonal $BC$ is calculated as $(\frac{3+2}{2}, \frac{4+5}{2}) = (\frac{5}{2}, \frac{9}{2})$. Because diagonal $AD$ shares this same midpoint, we set the midpoint of $AD$ as $(\frac{1+x}{2}, \frac{2+y}{2}) = (\frac{5}{2}, \frac{9}{2})$, which determines the coordinates of point $D$ to be $(4, 7)$. Calculating the line equation through $A(1, 2)$ and $D(4, 7)$ involves finding the slope $\frac{7-2}{4-1} = \frac{5}{3}$ and applying the point-slope form $y - 2 = \frac{5}{3}(x - 1)$, which simplifies to the final equation:

$5x - 3y + 1 = 0$

When a collection of lines is governed by a linear dependency between their coefficients, they invariably pass through a common intersection point. Normalizing the equation $3p + 2q + 4r = 0$ involves dividing the entire expression by $4$ to align it with the standard linear form $px + qy + r = 0$. This division yields the equivalent expression $p(\frac{3}{4}) + q(\frac{2}{4}) + r = 0$, which simplifies to $p(\frac{3}{4}) + q(\frac{1}{2}) + r = 0$. Comparing this result directly to the general line equation shows that $x = \frac{3}{4}$ and $y = \frac{1}{2}$ satisfy the condition for all possible values of $p, q,$ and $r$, confirming that the lines are concurrent at the point $\left(\frac{3}{4}, \frac{1}{2}\right)$.

Systems of linear equations often allow for variable elimination by identifying patterns where terms cancel out through simple arithmetic operations. Since the first and third equations contain $kz$ and $-kz$ respectively, adding the equations $x - 2y + kz = 1$ and $3x - y - kz = 3$ eliminates the $z$ term entirely, yielding the equation $4x - 3y = 4$. Rearranging this relationship into standard linear form results in $4x - 3y - 4 = 0$, which defines the straight line where the solution coordinates must reside.

Lines are perpendicular when the product of their slopes equals $-1$. Rewriting the equations $x+(a-1)y=1$ and $2x+a^2y=1$ into slope-intercept form gives slopes of $-\frac{1}{a-1}$ and $-\frac{2}{a^2}$. Multiplying these slopes and setting the result to $-1$ produces the equation $\frac{2}{a^2(a-1)} = -1$, which simplifies to $a^3 - a^2 + 2 = 0$. This cubic expression is satisfied when $a = -1$.

Substituting $a = -1$ into the original line equations results in $x - 2y = 1$ and $2x + y = 1$. Solving this system for $x$ and $y$ yields the intersection coordinates $(\frac{3}{5}, -\frac{1}{5})$. The distance from the origin is calculated as $\sqrt{x^2+y^2}$, resulting in $\sqrt{(\frac{3}{5})^2 + (-\frac{1}{5})^2} = \sqrt{\frac{10}{25}}$, which simplifies to $\sqrt{\frac{2}{5}}$.

Since the triangle OAB is formed by the coordinate axes and the line $3x + 4y - 24 = 0$, it is a right-angled triangle with the right angle at the origin, meaning the incentre must have coordinates $(r, r)$, where $r$ is the inradius. This radius represents the perpendicular distance from the incentre to the line $3x + 4y - 24 = 0$, allowing us to set up the equation $\frac{|3r + 4r - 24|}{\sqrt{3^2 + 4^2}} = r$, which simplifies to $\frac{|7r - 24|}{5} = r$. Solving the absolute value equation yields two possibilities, $7r - 24 = 5r$ or $7r - 24 = -5r$, and since $r = 12$ places the incentre outside the triangle, the valid solution is $12r = 24$, giving $r = 2$ and thus the incentre $(2, 2)$.

The inequalities $|x-y| \leq 2$ and $|x+y| \leq 2$ define a region bounded by the four lines $x-y = 2$, $x-y = -2$, $x+y = 2$, and $x+y = -2$. These boundary lines intersect at the vertices (2, 0), (0, 2), (-2, 0), and (0, -2), which form a square centered at the origin. Calculating the distance between any two adjacent vertices, such as (2, 0) and (0, 2), relies on the distance formula shown below:
$d = \sqrt{(2-0)^2 + (0-2)^2} = 2\sqrt{2}$
This configuration results in a square of side length $s = 2\sqrt{2}$ units.

The centroid of a triangle is defined by the average of the coordinates of its vertices, providing a direct link between the moving point P and the centroid's position. Let the coordinates of point P be $(\alpha, \beta)$ and the centroid be $(x, y)$. Using the given vertices Q(1, 4) and R(3, -2), the centroid is determined by the relations $x = \frac{\alpha + 1 + 3}{3}$ and $y = \frac{\beta + 4 - 2}{3}$. Rearranging these to solve for $\alpha$ and $\beta$ yields $\alpha = 3x - 4$ and $\beta = 3y - 2$.

Because point P must lie on the line $2x - 3y + 4 = 0$, substituting these expressions for $\alpha$ and $\beta$ into the line equation defines the path of the centroid: $2(3x - 4) - 3(3y - 2) + 4 = 0$. Expanding this expression results in $6x - 8 - 9y + 6 + 4 = 0$, which simplifies to the linear equation $6x - 9y + 2 = 0$. Rearranging this equation into slope-intercept form reveals the slope to be $-\frac{6}{-9}$, which simplifies to $\frac{2}{3}$.