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

The interior of the triangle with vertices $(0, 0)$, $(0, 41)$, and $(41, 0)$ is defined by the set of all points $(x, y)$ that satisfy the conditions $x > 0$, $y > 0$, and $x + y < 41$. Since the points must have integer coordinates, finding the total number of such points is equivalent to finding the number of positive integer pairs $(x, y)$ such that $x + y \le 40$.

This problem can be represented by the equation $x + y + k = 41$, where $x, y, k \ge 1$ are positive integers and $k$ acts as a slack variable to account for the space between the point and the hypotenuse. Using the combinatorial formula for the number of positive integer solutions to an equation of the form $x_1 + x_2 + \dots + x_r = n$, which is $\binom{n-1}{r-1}$, we set $n = 41$ and $r = 3$. This leads to the expression $\binom{41-1}{3-1}$, or $\binom{40}{2}$. Calculating this value results in 780.

A point positioned in the fourth quadrant that is equidistant from both coordinate axes necessarily has coordinates of the form $(h, -h)$ where $h$ is a positive value. Substituting these coordinates into the equation $4ax+2ay+c=0$ leads to:

$4ah+2a(-h)+c=0$

This simplifies to $2ah+c=0$, which implies $h = -\frac{c}{2a}$. Similarly, substituting the point $(h, -h)$ into the equation $5bx+2by+d=0$ gives:

$5bh+2b(-h)+d=0$

This expression simplifies to $3bh+d=0$, or $h = -\frac{d}{3b}$. Equating these two expressions for $h$ derived from the lines results in the relation:

$-\frac{c}{2a} = -\frac{d}{3b}$

Cross-multiplying these terms produces $-3bc = -2ad$, which rearranges to the final condition $3bc-2ad=0$.

The median $PS$ connects vertex $P(2, 2)$ to the midpoint $S$ of side $QR$, where $Q$ is $(6, -1)$ and $R$ is $(7, 3)$. Calculating the midpoint of $QR$ results in $S\left(\frac{13}{2}, 1\right)$. The slope of $PS$ is determined by the vertical change over the horizontal change, providing a slope of $\frac{1-2}{\frac{13}{2}-2} = -\frac{2}{9}$.

Because parallel lines share the same slope, the line passing through $(1, -1)$ with a slope of $-\frac{2}{9}$ can be written using the point-slope formula $\frac{y+1}{x-1} = -\frac{2}{9}$. Clearing the denominator by cross-multiplying leads to $9(y+1) = -2(x-1)$, which simplifies to $2x+9y+7=0$.

When a ray of light reflects off a flat boundary such as the x-axis, the angle of reflection matches the angle of incidence, effectively negating the sign of the slope. The incident ray equation $x+\sqrt{3}y=\sqrt{3}$ can be rewritten as $y=-\frac{1}{\sqrt{3}}x+1$, indicating a slope of $-\frac{1}{\sqrt{3}}$, which means the reflected ray must possess a slope of $\frac{1}{\sqrt{3}}$. Setting $y=0$ in the incident equation reveals the point of impact at $(\sqrt{3},0)$. Applying the point-slope formula with a slope of $\frac{1}{\sqrt{3}}$ through the point $(\sqrt{3},0)$ results in $y=\frac{1}{\sqrt{3}}(x-\sqrt{3})$, which simplifies to $\sqrt{3}y=x-\sqrt{3}$.

The x-coordinate of the incenter of a triangle is determined by the weighted average of the x-coordinates of its vertices, where the weights are the lengths of the sides opposite each vertex according to the formula $\frac{ax_1+bx_2+cx_3}{a+b+c}$. Given the midpoints (0, 1), (1, 1), and (1, 0), the vertices of the triangle are determined to be (0, 0), (2, 0), and (0, 2), which form a right-angled isosceles triangle with side lengths 2, 2, and $2\sqrt{2}$. Mapping each vertex $(x_i, y_i)$ to its opposite side length, we use $x_1=2$ with $a=2$, $x_2=0$ with $b=2\sqrt{2}$, and $x_3=0$ with $c=2$. Substituting these values yields the following expression:

$\frac{2(2) + 2\sqrt{2}(0) + 2(0)}{2+2+2\sqrt{2}} = \frac{4}{4+2\sqrt{2}}$

Simplifying this fraction results in $2-\sqrt{2}$.

A point that divides a line segment into a specific ratio is found using the section formula, which effectively calculates the weighted average of the endpoints. For the segment connecting $(1,1)$ and $(2,4)$ with a ratio of $3:2$, the $x$-coordinate is determined by $\frac{3(2) + 2(1)}{3+2} = \frac{8}{5}$ and the $y$-coordinate is determined by $\frac{3(4) + 2(1)}{3+2} = \frac{14}{5}$. Since the line $2x+y=k$ includes this specific point, substituting these coordinates into the equation yields $2\left(\frac{8}{5}\right) + \frac{14}{5} = k$. Simplifying the expression results in $\frac{16}{5} + \frac{14}{5} = \frac{30}{5}$, which confirms that $k=6$.

The Angle Bisector Theorem provides the most direct approach to this problem, as it states that an angle bisector of a triangle divides the opposite side into segments proportional to the lengths of the other two sides. By identifying the intersection of lines $L_1: y-x=0$ and $L_2: 2x+y=0$ at the origin $O(0,0)$ and finding their intersection points with $L_3: y+2=0$ as $P(-2,-2)$ and $Q(1,-2)$, we determine the lengths of the adjacent sides originating from the origin as $OP = \sqrt{(-2)^2+(-2)^2} = 2\sqrt{2}$ and $OQ = \sqrt{1^2+(-2)^2} = \sqrt{5}$. Applying the theorem, the ratio $PR:RQ$ must equal the ratio of the side lengths $OP:OQ$, which is $2\sqrt{2}:\sqrt{5}$, confirming that the first statement is true. The second statement is false, however, as bisecting an angle does not inherently partition a triangle into two similar triangles; this is not a general property of geometry.

Parallel lines are characterized by identical slopes, and the perpendicular distance between two lines in the standard form $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is calculated using the expression $\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.

For line $L$, given by $\frac{x}{5} + \frac{y}{b} = 1$, substituting the point $(13, 32)$ yields $\frac{13}{5} + \frac{32}{b} = 1$. Solving this for $b$ gives $\frac{32}{b} = -\frac{8}{5}$, resulting in $b = -20$, which corresponds to the standard form $4x - y - 20 = 0$. Since line $K$ is parallel to $L$, it must share the same slope of $4$. Given $K$ is expressed as $\frac{x}{c} + \frac{y}{3} = 1$, its slope is determined by $-\frac{3}{c}$, so setting $-\frac{3}{c} = 4$ reveals $c = -\frac{3}{4}$, making the equation of $K$ equivalent to $4x - y + 3 = 0$.

Applying the distance formula with $A = 4$, $B = -1$, $C_1 = -20$, and $C_2 = 3$, we calculate the perpendicular distance as $\frac{|-20 - 3|}{\sqrt{4^2 + (-1)^2}}$, which simplifies to $\frac{23}{\sqrt{17}}$.

The shortest distance between a line and a curve is realized at the point on the curve where the tangent is parallel to the given line, which can be found by minimizing the perpendicular distance from an arbitrary point on the curve to the line. Representing any point on the parabola $x=y^2$ as $(a^2, a)$, we apply the perpendicular distance formula for the line $x-y+1=0$ to obtain $D = \frac{|a^2-a+1|}{\sqrt{1^2+(-1)^2}}$, which simplifies to $D = \frac{a^2-a+1}{\sqrt{2}}$ since the expression $a^2-a+1$ is always positive. To minimize this, we locate the vertex of the quadratic expression $a^2-a+1$ by setting its derivative to zero, which occurs at $a = 1/2$. Substituting this value into our distance formula, we calculate $D = \frac{1/4-1/2+1}{\sqrt{2}}$, simplifying to $\frac{3/4}{\sqrt{2}}$. Rationalizing the denominator yields the shortest distance of $\frac{3\sqrt{2}}{8}$.

When two lines are perpendicular to the same third line, they must be parallel to each other. For the lines $p(p^2+1) x - y + q = 0$ and $(p^2+1)^2 x + (p^2+1) y + 2 q = 0$ to satisfy this condition, their slopes must be identical. Equating the slope of the first line, $p(p^2+1)$, to the slope of the second line, $-(p^2+1)$, yields the equation:
$p(p^2+1) = -(p^2+1)$
This simplifies to $(p^2+1)(p+1) = 0$, and since $p^2+1$ is never zero, the only valid solution is $p = -1$. Therefore, there is exactly one value of $p$ for which the lines are perpendicular to a common line.

A perpendicular bisector is defined by two geometric requirements: it must pass through the midpoint of the given segment and possess a slope equal to the negative reciprocal of that segment's slope. The line segment connecting $P(1,4)$ and $Q(k,3)$ has a slope of $\frac{3-4}{k-1} = -\frac{1}{k-1}$, so the perpendicular bisector's slope must be $k-1$. Its midpoint is $\left(\frac{k+1}{2}, \frac{7}{2}\right)$, which leads to the line equation:
$y - \frac{7}{2} = (k-1)\left(x - \frac{k+1}{2}\right)$
To determine the $y$-intercept, set $x=0$, which yields $y = -\frac{(k-1)(k+1)}{2} + \frac{7}{2}$, simplifying to $y = \frac{8-k^2}{2}$. Equating this to the given $y$-intercept of $-4$ results in $\frac{8-k^2}{2} = -4$, or $8-k^2 = -8$, confirming $k^2 = 16$ and thus $k = -4$ as a possible solution.

The points $P=(-1,0)$ and $Q=(0,0)$ indicate that the line segment $QP$ lies exactly on the negative x-axis, creating an angle of $180^{\circ}$ with the positive x-axis. For the segment $QR$ connecting $(0,0)$ to $(3,3\sqrt{3})$, the slope is $\frac{3\sqrt{3}-0}{3-0} = \sqrt{3}$, which means the line makes a $60^{\circ}$ angle with the positive x-axis. Because the interior angle $\angle PQR$ is the supplementary angle to this $60^{\circ}$ inclination, it measures $180^{\circ} - 60^{\circ} = 120^{\circ}$.

The angle bisector divides this $120^{\circ}$ angle into two $60^{\circ}$ segments. Starting from the negative x-axis and rotating $60^{\circ}$ towards the line $QR$, the angle the bisector makes with the positive x-axis is $180^{\circ} - 60^{\circ} = 120^{\circ}$. Consequently, the slope of this line is $\tan(120^{\circ}) = -\sqrt{3}$. Since the line passes through the origin, the equation is $y = -\sqrt{3}x$, which rearranges to $\sqrt{3}x + y = 0$.

The lines defined by $xy=0$ are the coordinate axes, and their angle bisectors are given by $y=x$ and $y=-x$. Since one of the lines represented by $my^2 + (1-m^2)xy - mx^2 = 0$ is a bisector, substituting $y=x$ must satisfy the equation, leading to:
$mx^2 + (1-m^2)x^2 - mx^2 = 0$
This simplifies to $(1-m^2)x^2 = 0$, which implies $1-m^2 = 0$, resulting in $m^2 = 1$ and therefore $m = \pm 1$. Consequently, $1$ is the value that satisfies the condition.

To identify the possible values for the coordinate $k$, we utilize the geometric property that a right-angled triangle with hypotenuse $AC$ must have its right angle at vertex $B$. Because the vertices $B(1, 1)$ and $C(2, 1)$ lie on a horizontal line, the length of the base $BC$ is 1. To satisfy the right-angle requirement at $B$, the leg $AB$ must be vertical, which means $A$ must have an $x$-coordinate of 1, establishing $A$ as $(1, k)$.

The area of a right-angled triangle is defined as $\frac{1}{2} \times \text{base} \times \text{height}$, or $\frac{1}{2} \times BC \times AB$. Substituting the known area of 1 and base of 1, we obtain the equation $\frac{1}{2} \times 1 \times |k-1| = 1$. Solving for the absolute value $|k-1|$ yields 2, which implies two possible scenarios: $k-1 = 2$ or $k-1 = -2$. This results in $k = 3$ and $k = -1$ as the set of viable values.

When a point lies inside the angular region formed by two rays extending from the origin, its vertical position relative to those lines determines the bounds on its coordinates. For the point $(a, a^2)$ to be caught between the steeper line $y=3x$ and the shallower line $y=\frac{x}{2}$, the $y$-value $a^2$ must be lower than the output of the steeper line but higher than the output of the shallower one. This requires the simultaneous satisfaction of $a^2 < 3a$ and $a^2 > \frac{a}{2}$.

Analyzing the first inequality $a^2 - 3a < 0$ leads to $a(a-3) < 0$, which holds when $0 < a < 3$. For the second condition, $a^2 - \frac{a}{2} > 0$ factors into $a(a - \frac{1}{2}) > 0$, which is satisfied when $a < 0$ or $a > \frac{1}{2}$. Since the problem restricts the domain to $x > 0$, we already have the condition $a > 0$, narrowing our focus to the intersection of these valid regions. Combining these constraints isolates the valid range for $a$ to the interval $(\frac{1}{2}, 3)$.

A line intersecting the coordinate axes at points $(a, 0)$ and $(0, b)$ is described by the intercept form equation $\frac{x}{a} + \frac{y}{b} = 1$. Given that the point $(3, 4)$ bisects the segment between these axes, it must satisfy the midpoint relations $\frac{a+0}{2} = 3$ and $\frac{0+b}{2} = 4$, which leads to $a = 6$ and $b = 8$. Substituting these values into the intercept form gives $\frac{x}{6} + \frac{y}{8} = 1$, and multiplying the entire equation by 24 to clear the denominators results in the final equation $4x + 3y = 24$.

The centroid of a triangle is defined as the average of the coordinates of its three vertices. Since we are given one vertex $A(1, 1)$ and the midpoints of the two sides adjacent to this vertex, we first determine the coordinates of the other two vertices, $B$ and $C$, using the midpoint formula. For the first side, setting the midpoint equal to the average of $A$ and $B$ gives the equations $\frac{1+x_B}{2} = -1$ and $\frac{1+y_B}{2} = 2$, which reveal that $B$ is $(-3, 3)$. Similarly, for the second side with the given midpoint $(3, 2)$, solving $\frac{1+x_C}{2} = 3$ and $\frac{1+y_C}{2} = 2$ identifies vertex $C$ as $(5, 3)$. With all three vertices known, the centroid is calculated by averaging their $x$ and $y$ values: $x = \frac{1-3+5}{3} = 1$ and $y = \frac{1+3+3}{3} = \frac{7}{3}$, resulting in a final coordinate of $(1, \frac{7}{3})$.

Any line passing through the intersection of the two provided equations can be expressed as a linear combination of the form $(a x+2 b y+3 b)+\lambda(b x-2 a y-3 a)=0$. Rearranging this equation to group the variables results in $(a+b \lambda) x+(2 b-2 a \lambda) y+(3 b-3 \lambda a)=0$. Because the line must be parallel to the $x$-axis, the coefficient of $x$ must be zero, which implies $\lambda = -a/b$. Substituting this value into the combined equation causes the $x$-term to vanish and simplifies the remaining expression to $y(2b + 2a^2/b) = -(3b + 3a^2/b)$. This simplifies further to $y = -3/2$, which means the line is situated 3/2 units below the $x$-axis.

The defining characteristic of a harmonic progression is that the reciprocals of its terms form an arithmetic progression. Given that $a, b, c$ are in harmonic progression, the values $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ must follow an arithmetic progression, which requires the middle term to be the average of the outer terms, expressed as $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$. Rearranging this equation into the form $\frac{1}{a} - \frac{2}{b} + \frac{1}{c} = 0$ provides a fixed relationship between these variables. By comparing this derived identity directly to the equation of the line $\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0$, it becomes clear that setting $x = 1$ and $y = -2$ allows the equation to hold true for any such set of numbers. Consequently, the line always passes through the fixed point $(1, -2)$.

The centroid of any triangle is found by averaging the coordinates of its vertices, meaning for vertex $C(x, y)$ and fixed points $A(2, -3)$ and $B(-2, 1)$, the centroid is situated at $\left( \frac{x}{3}, \frac{y - 2}{3} \right)$. Since this centroid must satisfy the linear equation $2x + 3y = 1$, substituting these specific coordinates leads to the condition $2\left( \frac{x}{3} \right) + 3\left( \frac{y - 2}{3} \right) = 1$. Multiplying the entire equation by 3 eliminates the denominators to give $2x + 3(y - 2) = 3$, which simplifies to $2x + 3y - 6 = 3$, identifying the locus of point $C$ as $2x + 3y = 9$.