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

The abscissa of the intersection point of two lines is determined by solving their equations simultaneously, and for this coordinate to be an integer, the denominator of the final expression must be a factor of the numerator. Substituting $y=mx+1$ into $3x+4y=9$ results in $3x+4(mx+1)=9$. Simplifying this expression leads to $x(3+4m)=5$, which identifies the abscissa as $x = \frac{5}{3+4m}$. Since $x$ must be an integer, $3+4m$ must be a divisor of 5, restricting the possible values for $3+4m$ to 1, -1, 5, and -5.

Setting the expression $3+4m$ equal to these four divisors yields distinct values for $m$. Equating $3+4m=1$ results in $4m=-2$, or $m=-0.5$. Equating $3+4m=-1$ gives $4m=-4$, so $m=-1$. Setting $3+4m=5$ results in $4m=2$, or $m=0.5$. Finally, setting $3+4m=-5$ leads to $4m=-8$, so $m=-2$. Evaluating these results, only $m=-1$ and $m=-2$ are integers, confirming there are 2 such values.

A triangle is classified as isosceles when at least two of its sides are equal in length, a property we verify by identifying the vertices of the shape and calculating the distances between them. Solving the system of equations $x-y=0$ and $x+2y=3$, $x-y=0$ and $2x+y=6$, and $x+2y=3$ and $2x+y=6$ provides the intersection points $(1,1)$, $(2,2)$, and $(3,0)$. Applying the distance formula between these vertices reveals side lengths of $\sqrt{(2-1)^{2}+(2-1)^{2}} = \sqrt{2}$ for the first segment, followed by $\sqrt{(3-1)^{2}+(0-1)^{2}} = \sqrt{5}$ and $\sqrt{(3-2)^{2}+(0-2)^{2}} = \sqrt{5}$ for the remaining two sides. Because two of the side lengths are identical, the triangle is defined as isosceles.

The angle bisectors of any pair of lines represented by the homogeneous equation $ax^2 + 2hxy + by^2 = 0$ satisfy the relation $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$. For the equation $x^2 - 4xy - 5y^2 = 0$, the coefficients are identified as $a = 1$, $b = -5$, and $2h = -4$, which means $h = -2$. Substituting these values into the ratio results in $\frac{x^2 - y^2}{1 - (-5)} = \frac{xy}{-2}$, which simplifies to $\frac{x^2 - y^2}{6} = \frac{xy}{-2}$. Rearranging the terms by cross-multiplying gives $x^2 - y^2 = -3xy$, which simplifies to $x^2 + 3xy - y^2 = 0$.

Identifying the geometric nature of triangle $ABC$ reveals that it is a right-angled triangle, as the side lengths satisfy the condition $AB^2 + BC^2 = AC^2$, with $45 + 5 = 50$. In any right-angled triangle, the circumcenter is located at the midpoint of the hypotenuse, which corresponds to segment $AC$. Computing the midpoint of $AC$ with vertices $A(-2,3)$ and $C(3,8)$ yields the circumcenter coordinate $(\frac{1}{2}, \frac{11}{2})$.

The midpoint of side $BC$ with vertices $B(1,9)$ and $C(3,8)$ is determined to be $(2, \frac{17}{2})$. Since line $L$ passes through the circumcenter $(\frac{1}{2}, \frac{11}{2})$ and the midpoint $(2, \frac{17}{2})$, the slope is calculated as $m = \frac{\frac{17}{2} - \frac{11}{2}}{2 - \frac{1}{2}}$, which simplifies to $\frac{3}{1.5} = 2$. Writing the equation of the line using the point-slope form $y - \frac{11}{2} = 2(x - \frac{1}{2})$ results in $y = 2x + \frac{9}{2}$. Setting $x$ to zero to find the $y$-axis intersection point gives $y = \frac{9}{2}$, meaning $\frac{\alpha}{2} = \frac{9}{2}$ and thus $\alpha = 9$.

The area of a triangle depends on the base and the perpendicular height from the third vertex to that base line. Since the two vertices $(5, 6)$ and $(3, 2)$ are fixed, the third vertex $(\alpha, \beta)$ must maintain a constant perpendicular distance from the line containing the base to keep the area fixed at 12 square units. Geometrically, this means the locus of the point $(\alpha, \beta)$ consists of two lines parallel to the base, one on each side. Using the determinant form for the area of a triangle, the condition $\frac{1}{2} | \alpha(6 - 2) + 5(2 - \beta) + 3(\beta - 6) | = 12$ simplifies to $| 4\alpha - 2\beta - 8 | = 24$. This yields two linear equations: $2\alpha - \beta - 16 = 0$ and $2\alpha - \beta + 8 = 0$, which define the set $A$.

To find the minimum distance from the origin $(0, 0)$ to this set, we use the formula for the perpendicular distance from a point to a line, $d = \frac{|C|}{\sqrt{A^2 + B^2}}$. For the first line $2\alpha - \beta - 16 = 0$, the distance from the origin is $\frac{|-16|}{\sqrt{2^2 + (-1)^2}} = \frac{16}{\sqrt{5}}$. For the second line $2\alpha - \beta + 8 = 0$, the distance is $\frac{|8|}{\sqrt{5}}$. Comparing these two values, the least possible length of the line segment joining the origin to a point in $A$ is $\frac{8}{\sqrt{5}}$.

The midpoint $M$ of the line segment connecting $A(0, 6)$ and $B(2t, 0)$ is determined by averaging the respective coordinates, resulting in $M(t, 3)$. Since the slope of segment $AB$ is $\frac{0-6}{2t-0} = -\frac{3}{t}$, the perpendicular bisector passing through $M$ must have a slope of $\frac{t}{3}$. The equation for this line is $y - 3 = \frac{t}{3}(x - t)$.

To identify point $C$, where this bisector intersects the Y-axis, we set $x = 0$ in the equation, yielding $y = 3 - \frac{t^2}{3}$, so $C$ is $(0, 3 - \frac{t^2}{3})$. The midpoint $P(h, k)$ of segment $MC$ is found by averaging the coordinates of $M$ and $C$, resulting in $h = \frac{t+0}{2} = \frac{t}{2}$ and $k = \frac{3 + (3 - \frac{t^2}{3})}{2} = 3 - \frac{t^2}{6}$.

To define the locus, we eliminate the parameter $t$ by substituting $t = 2h$ into the equation for $k$. This gives $k = 3 - \frac{(2h)^2}{6}$, which simplifies to $k = 3 - \frac{2h^2}{3}$. Multiplying by 3 results in $3k = 9 - 2h^2$, and rearranging the terms leads to the equation $2h^2 + 3k - 9 = 0$. Replacing the variables $h$ and $k$ with $x$ and $y$ provides the final locus $2x^2 + 3y - 9 = 0$.

The curve $x^2y^2 = 1$ is equivalent to two hyperbolas $xy = 1$ and $xy = -1$. For a square centered at the origin, vertices must be symmetrically positioned on these curves, so we can define $A(t_1, 1/t_1)$ on $xy = 1$ and $B(t_2, -1/t_2)$ on $xy = -1$. Because the diagonals of the square are perpendicular, the product of the slopes of segments $OA$ and $OB$ must be $-1$, which leads to the relation $(1/t_1^2) \cdot (-1/t_2^2) = -1$, or $t_1^2t_2^2 = 1$. The midpoint $P$ of side $AB$ has coordinates $(\frac{t_1+t_2}{2}, \frac{t_2-t_1}{2t_1t_2})$. Given that this midpoint lies on the curve $xy = 1$, the product of its coordinates must be $1$, which gives $(\frac{t_1+t_2}{2})(\frac{t_2-t_1}{2t_1t_2}) = 1$, simplifying to $t_2^2 - t_1^2 = 4t_1t_2$.

The square of the side length $AB$ is calculated using the distance formula $AB^2 = (t_1-t_2)^2 + (1/t_1 + 1/t_2)^2$. Applying the relation $t_1^2t_2^2 = 1$, the expression for the squared side length simplifies to $(t_1-t_2)^2 + (t_1+t_2)^2 = 2(t_1^2 + t_2^2)$. Squaring the midpoint condition $t_2^2 - t_1^2 = 4t_1t_2$ gives $(t_2^2 - t_1^2)^2 = 16t_1^2t_2^2 = 16$. Using the algebraic identity $(t_1^2 + t_2^2)^2 = (t_2^2 - t_1^2)^2 + 4t_1^2t_2^2$, we find $(t_1^2 + t_2^2)^2 = 16 + 4 = 20$, implying $t_1^2 + t_2^2 = \sqrt{20}$. Consequently, the area of the square, which equals $AB^2$, is $2\sqrt{20} = \sqrt{80}$. The square of this area is therefore $(\sqrt{80})^2 = 80$.

The intercept form of a straight line equation is given by $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ and $b$ represent the intercepts on the $x$ and $y$ axes. We are told the arithmetic mean of the reciprocals of these intercepts is $\frac{1}{4}$, which gives the condition $\frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{1}{4}$, or equivalently $\frac{1}{a} + \frac{1}{b} = \frac{1}{2}$. To determine which stone lies on this path, we check the given coordinates $(x, y)$ by substituting them into the line equation. For point $B(2, 2)$, substituting these values results in $\frac{2}{a} + \frac{2}{b} = 1$, which simplifies to $2(\frac{1}{a} + \frac{1}{b}) = 1$, or $\frac{1}{a} + \frac{1}{b} = \frac{1}{2}$. This result perfectly matches the derived condition, confirming that the stone at $(2, 2)$ lies on the man's path, whereas the other points do not satisfy the equation.

Linear programming problems involving a non-linear objective function require substituting boundary constraints to reduce the expression to a single variable. For the objective function $z=6 x y+y^{2}$, the constraint $4 x+3 y \leq 75$ serves as the primary boundary for $x$ within the feasible region, allowing us to express it as $x \leq \frac{75-3 y}{4}$. Substituting this into the objective function yields $z \leq y \left( 6 \left( \frac{75-3 y}{4} \right) + y \right)$, which simplifies to the quadratic form $z \leq \frac{225 y-7 y^{2}}{2}$.

Since this parabola opens downward, its maximum value occurs at the vertex, which is calculated as $\frac{225^2}{56}$. Performing this calculation yields approximately 904.02, establishing 904 as the maximum value.

Finding the reflection of a point across a line relies on the geometric property that the segment connecting the point and its image is perpendicular to the line, with their midpoint lying directly on the line itself. For a point $(x_1, y_1)$ and a line defined by $ax+by+c=0$, the coordinates of the image $(x, y)$ are determined using the following relationship:

$\frac{x-x_1}{a} = \frac{y-y_1}{b} = -2\frac{ax_1+by_1+c}{a^2+b^2}$

Applying this formula to the point $(3,5)$ and the line $x-y+1=0$, we substitute the respective values:

$\frac{x-3}{1} = \frac{y-5}{-1} = -2\frac{3-5+1}{1^2+(-1)^2}$

This expression simplifies to:

$\frac{x-3}{1} = \frac{y-5}{-1} = -2\left(\frac{-1}{2}\right) = 1$

Solving the individual equations $\frac{x-3}{1}=1$ and $\frac{y-5}{-1}=1$ yields $x=4$ and $y=4$. The image of the point is therefore $(4,4)$. Testing this coordinate in the given options, the equation $(x-2)^{2}+(y-4)^{2}=4$ results in $(4-2)^{2}+(4-4)^{2} = 4+0=4$, confirming that the point lies on this curve.

The triangle formed by connecting the midpoints of the sides of any triangle is known as the medial triangle, which possesses exactly one-fourth the area of the original triangle. Consequently, identifying the area of $\triangle ABC$ involves first determining the vertices of this medial triangle by solving the systems of equations formed by the lines, computing the area of that smaller triangle, and finally scaling that result by a factor of four.

Solving the intersection of $x-y+1=0$ and $x-2y+3=0$ by subtracting the equations gives $y=2$ and $x=1$, establishing the first vertex at $(1,2)$. For the second intersection, substitute $x=y-1$ from the first line into $2x-5y+11=0$ to obtain $2(y-1)-5y+11=0$, which simplifies to $-3y+9=0$ where $y=3$ and $x=2$, identifying the second vertex as $(2,3)$. Finally, substitute $x=2y-3$ from $x-2y+3=0$ into $2x-5y+11=0$ to solve the third intersection, yielding $2(2y-3)-5y+11=0$, which simplifies to $-y+5=0$ where $y=5$ and $x=7$, providing the third vertex at $(7,5)$.

Applying the coordinate area formula for the vertices $(1,2)$, $(2,3)$, and $(7,5)$, the area of the medial triangle is $\frac{1}{2} |1(3-5) + 2(5-2) + 7(2-3)|$, which results in $\frac{1}{2} |-2 + 6 - 7| = 1.5$. Since the area of the original triangle is four times the area of the medial triangle, the final area is $4 \times 1.5 = 6$.

To minimize the travel time at a constant speed, the man must travel the shortest distance, which corresponds to the path $PR + RQ$ being a straight line. By reflecting point $Q(0,2)$ across the X-axis to create $Q'(0,-2)$, the shortest path lies along the segment $PQ'$, and point $R$ is the intersection of this line with the X-axis. The slope of the line passing through $P(-3,4)$ and $Q'(0,-2)$ is $m = \frac{-2 - 4}{0 - (-3)} = -2$, resulting in the equation $y = -2x - 2$, which intersects the X-axis at $R(-1,0)$. Computing the distance squares, we find $PR^{2} = (-1 - (-3))^{2} + (0 - 4)^{2} = 2^{2} + (-4)^{2} = 20$ and $RQ^{2} = (0 - (-1))^{2} + (2 - 0)^{2} = 1^{2} + 2^{2} = 5$. Substituting these values into the final expression yields $50(20 + 5) = 1250$.

The slope of a line making a specific angle with another line is determined by the tangent of the angle between them, which is calculated as the absolute difference of their slopes divided by one plus the product of their slopes. Given the known line $y+1=3\sqrt{2}x$ has a slope of $3\sqrt{2}$ and the required angle is $\tan^{-1}(\sqrt{2})$, the unknown slope $m$ satisfies the condition $\left| \frac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m} \right| = \sqrt{2}$.

Squaring both sides of this relationship yields the equation $\frac{m^2 - 6\sqrt{2}m + 18}{(1 + 3\sqrt{2}m)^2} = 2$, which simplifies to $m^2 - 6\sqrt{2}m + 18 = 2(1 + 18m^2 + 6\sqrt{2}m)$. Expanding and consolidating these terms results in the quadratic equation $35m^2 + 18\sqrt{2}m - 16 = 0$. Applying the quadratic formula provides two possible slopes, $m = \frac{2\sqrt{2}}{7}$ and $m = \frac{-4\sqrt{2}}{5}$.

Substituting $m = \frac{-4\sqrt{2}}{5}$ into the point-slope form $y - 3 = m(x - 1)$ gives $y - 3 = -\frac{4\sqrt{2}}{5}(x - 1)$. Multiplying this equation by five and rearranging the terms leads to $5y - 15 = -4\sqrt{2}x + 4\sqrt{2}$, which simplifies to $4\sqrt{2}x + 5y - (15 + 4\sqrt{2}) = 0$.

Applying successive geometric transformations relies on the composition of functions, where each operation maps the previous coordinate state to a new one. Starting with $P(a, b)$, reflection across the line $y=x$ swaps the coordinates to $(b, a)$, and a translation by 2 units along the positive $x$-axis shifts the point to $(b+2, a)$. To process the final rotation by $\frac{\pi}{4}$ in the anti-clockwise direction, the point is treated as the complex number $(b+2) + ia$ and multiplied by the rotation factor $e^{i\frac{\pi}{4}} = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$.

Executing this multiplication produces the complex value $\frac{1}{\sqrt{2}} ((b+2-a) + i(b+2+a))$. Equating the real and imaginary parts of this expression to the given coordinates $-\frac{1}{\sqrt{2}} + i\frac{7}{\sqrt{2}}$ leads to the following system of equations:
$b+2-a = -1$
$b+2+a = 7$
Solving these simultaneous equations gives $a=4$ and $b=1$, which makes the value of $2a+b$ equal to $2(4) + 1$, totaling $9$.

The circumcenter of a triangle is defined by the intersection point of the perpendicular bisectors of its sides. Since the equation for the perpendicular bisector of $PR$ is explicitly given as $2x - y + 2 = 0$, we simply need to find the equation of the perpendicular bisector for side $PQ$ and determine where it intersects the first line.

The side $PQ$ spans from $P(-2, 4)$ to $Q(4, -2)$. First, we locate the midpoint of $PQ$, which is $\left(\frac{-2+4}{2}, \frac{4-2}{2}\right) = (1,1)$. Next, we find the slope of the segment $PQ$, which is $\frac{-2-4}{4-(-2)} = -1$. Because the perpendicular bisector must be perpendicular to this segment, its slope must be the negative reciprocal, which is $1$. The equation of this bisector, passing through $(1,1)$ with a slope of $1$, is $y - 1 = 1(x - 1)$, which simplifies to $y = x$.

Solving for the intersection of the two bisector equations $2x - y + 2 = 0$ and $y = x$ requires substituting the second into the first. This yields $2x - x + 2 = 0$, which simplifies to $x = -2$. Plugging this value back into $y = x$, we find that $y = -2$. Thus, the coordinates of the circumcenter are $(-2, -2)$.

The perpendicular distance from the origin to any line $Ax + By + C = 0$ is determined by the formula $d = \frac{|C|}{\sqrt{A^2 + B^2}}$. By applying this geometric principle to the two given lines, we can express the perpendicular lengths $p$ and $q$ in terms of $\alpha$ and $k$.

For the first equation, $x \csc \alpha - y \sec \alpha = k \cot 2\alpha$, rewriting it in general form gives $x \csc \alpha - y \sec \alpha - k \cot 2\alpha = 0$. The perpendicular distance $p$ is:
$p = \frac{k \cot 2\alpha}{\sqrt{\csc^2 \alpha + \sec^2 \alpha}}$
The denominator simplifies as $\sqrt{\csc^2 \alpha + \sec^2 \alpha} = \sqrt{\frac{1}{\sin^2 \alpha} + \frac{1}{\cos^2 \alpha}} = \frac{1}{\sin \alpha \cos \alpha} = \frac{2}{\sin 2\alpha}$. Substituting this into the expression for $p$ yields:
$p = \frac{k \cos 2\alpha / \sin 2\alpha}{2 / \sin 2\alpha} = \frac{k \cos 2\alpha}{2}$
For the second equation, $x \sin \alpha + y \cos \alpha = k \sin 2\alpha$, the perpendicular distance $q$ is:
$q = \frac{k \sin 2\alpha}{\sqrt{\sin^2 \alpha + \cos^2 \alpha}} = k \sin 2\alpha$

With $p = \frac{k \cos 2\alpha}{2}$ and $q = k \sin 2\alpha$, we isolate the trigonometric terms to find $\cos 2\alpha = \frac{2p}{k}$ and $\sin 2\alpha = \frac{q}{k}$. Applying the identity $\sin^2 2\alpha + \cos^2 2\alpha = 1$, we substitute these expressions to obtain:
$\left( \frac{q}{k} \right)^2 + \left( \frac{2p}{k} \right)^2 = 1$
Multiplying by $k^2$ leads to $q^2 + 4p^2 = k^2$, which confirms that $4p^2 + q^2 = k^2$.

The ratio of the areas of two triangles sharing a common angle at vertex $C$ is determined by the product of the ratios of their adjacent sides, given by $\frac{Area(\triangle PQC)}{Area(\triangle ABC)} = \frac{CP}{CA} \cdot \frac{CQ}{CB}$. Since $Area(\triangle ABC) = 3 Area(\triangle PQC)$, the required product of these side ratios is equal to $\frac{1}{3}$.

The line $AC$ passing through $A(-1,1)$ and $C(2,0)$ has the equation $x+3y=2$, and the line $BC$ through $B(3,4)$ and $C(2,0)$ follows the equation $y=4x-8$. Solving the intersection of these lines with the line $y=mx$ yields the coordinates $P\left(\frac{2}{3m+1}, \frac{2m}{3m+1}\right)$ and $Q\left(\frac{8}{4-m}, \frac{8m}{4-m}\right)$.

The lengths of the segments from $C$ to the vertices are $CA = \sqrt{10}$ and $CB = \sqrt{17}$, while the distances to the intersection points are $CP = \frac{2m\sqrt{10}}{3m+1}$ and $CQ = \frac{2m\sqrt{17}}{4-m}$. Substituting these into the ratio formula $\frac{CP}{CA} \cdot \frac{CQ}{CB} = \frac{1}{3}$ simplifies the expression to $\frac{2m\sqrt{10}}{3m+1} \cdot \frac{1}{\sqrt{10}} \cdot \frac{2m\sqrt{17}}{4-m} \cdot \frac{1}{\sqrt{17}} = \frac{1}{3}$.

This relationship reduces to the equation $\frac{4m^2}{(3m+1)(4-m)} = \frac{1}{3}$, which cross-multiplies to $12m^2 = -3m^2 + 11m + 4$. Rearranging terms results in the quadratic equation $15m^2 - 11m - 4 = 0$, which factors as $(15m+4)(m-1)=0$. Given the condition that $m>0$, the value must be $m=1$.

To determine the angle $\theta$, we first need the coordinates of vertex $C$. Let $C = (a, b)$. Since the median from $B$ passes through the midpoint of $AC$, the point $E = \left(\frac{a-3}{2}, \frac{b+1}{2}\right)$ must satisfy the equation of the median $2x+y-3=0$. Substituting these coordinates yields $2\left(\frac{a-3}{2}\right) + \left(\frac{b+1}{2}\right) - 3 = 0$, which simplifies to $2a+b=11$. Furthermore, because point $C$ lies on the angle bisector $7x-4y-1=0$, its coordinates must satisfy $7a-4b=1$. Solving this system of equations leads to $a=3$ and $b=5$, establishing the location of vertex $C$.

With vertex $C$ identified, we calculate the slope of line $AC$ using the coordinates $A(-3, 1)$ and $C(3, 5)$, which results in $m_{AC} = \frac{5-1}{3-(-3)} = \frac{2}{3}$. The line $CD$, defined by $7x-4y-1=0$, has a slope of $m_{CD} = \frac{7}{4}$. Because $CD$ is the angle bisector, the angle between $AC$ and $CD$ is exactly half of $\theta$. Applying the tangent difference formula, $\tan\left(\frac{\theta}{2}\right) = \left| \frac{7/4 - 2/3}{1 + (7/4)(2/3)} \right| = \frac{13/12}{26/12} = \frac{1}{2}$. Using the double-angle identity $\tan\theta = \frac{2\tan(\theta/2)}{1 - \tan^2(\theta/2)}$, we find that $\tan\theta = \frac{2(1/2)}{1 - (1/2)^2} = \frac{1}{3/4} = \frac{4}{3}$.

The diagonals of a parallelogram always bisect each other, meaning the point of intersection of one diagonal serves as the midpoint for the other. The two lines forming the parallelogram sides, $4x + 5y = 0$ and $7x + 2y = 0$, intersect at the origin $(0, 0)$, which acts as one vertex. Solving for the intersection of the given diagonal $11x + 7y = 9$ with the first side line yields vertex $D \left(\frac{5}{3}, -\frac{4}{3}\right)$, and its intersection with the second side line provides vertex $B \left(-\frac{2}{3}, \frac{7}{3}\right)$.

Averaging the coordinates of these two vertices $B$ and $D$ gives the midpoint $\left(\frac{1}{2}, \frac{1}{2}\right)$. Because the other diagonal must pass through this midpoint and the origin vertex at $(0, 0)$, the line representing this diagonal is determined by the equation $y = x$. Checking the provided coordinates against this line reveals that the point $(2, 2)$ satisfies the condition.

Since the circumcentre of $\triangle ABC$ is the origin, the vertices $A, B,$ and $C$ lie on a circle centered at $(0,0)$. With the slopes of the line segments $OA, OB,$ and $OC$ defined as $\tan \alpha, \tan \beta,$ and $\tan \gamma$, we can express the coordinates of the vertices as $(\cos \alpha, \sin \alpha)$, $(\cos \beta, \sin \beta)$, and $(\cos \gamma, \sin \gamma)$. Because the circumcentre, orthocentre, and centroid of any triangle are collinear along the Euler line, and we are given that both the circumcentre and orthocentre lie on the $y$-axis (the former being the origin), the centroid must also lie on the $y$-axis. The $x$-coordinate of the centroid is the average of the $x$-coordinates of the vertices, meaning $\cos \alpha + \cos \beta + \cos \gamma = 0$.