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

The locus of a point whose sum of distances from two fixed points is constant defines an ellipse, provided that this sum is strictly greater than the distance between the fixed points. Given the perimeter of triangle $AOP$ is $4$ and the distance between the points $O(0,0)$ and $A(0,1)$ is $1$, the sum of the distances from $P$ to the two fixed foci $O$ and $A$ is $4 - 1 = 3$.

The center of this ellipse is the midpoint of the segment connecting the foci, which is $(0, 1/2)$, and the major axis length is $2a = 3$, giving $a = 3/2$. The distance between the foci is $2ae = 1$, which implies $ae = 1/2$. Using the relationship $b^2 = a^2 - (ae)^2$, we calculate $b^2 = (3/2)^2 - (1/2)^2 = 9/4 - 1/4 = 2$. Substituting these values into the standard ellipse equation $\frac{x^2}{b^2} + \frac{(y - 1/2)^2}{a^2} = 1$ leads to $\frac{x^2}{2} + \frac{(y - 1/2)^2}{9/4} = 1$. Multiplying the entire equation by 18 to clear the denominators results in $9x^2 + 8(y^2 - y + 1/4) = 18$, which simplifies to $9x^2 + 8y^2 - 8y = 16$.

Determining the slope of the line passing through $P(2,3)$ and intersecting $x + y = 7$ at a distance of 4 units relies on the geometry of the right triangle formed by the perpendicular distance from $P$ to the line, the distance along the line to the intersection, and the segment connecting $P$ to the intersection point. The perpendicular distance from $P(2,3)$ to the line $x + y - 7 = 0$ is $d = \frac{|2 + 3 - 7|}{\sqrt{1^2 + 1^2}}$, which simplifies to $\sqrt{2}$. Because the distance along the intersecting line is 4 units, the sine of the angle $\theta$ between the two lines satisfies $\sin\theta = \frac{\sqrt{2}}{4} = \frac{1}{2\sqrt{2}}$. Using the identity $\tan\theta = \frac{\sin\theta}{\sqrt{1 - \sin^2\theta}}$, this evaluates to $\tan\theta = \frac{1}{\sqrt{7}}$.

The angle between two lines with slopes $m$ and $-1$ is given by $\tan\theta = \left| \frac{m - (-1)}{1 + m(-1)} \right|$. Equating this to the calculated tangent value results in the equation:
$\left| \frac{m + 1}{1 - m} \right| = \frac{1}{\sqrt{7}}$
Solving this equation for $m$ provides the possible slopes, one of which is $m = \frac{1 - \sqrt{7}}{1 + \sqrt{7}}$.

The orthocenter of a triangle lies at the intersection of its altitudes, meaning the line segment connecting any vertex to the orthocenter is perpendicular to the side opposite that vertex. For a triangle with vertices $A(0, 2)$, $B(4, 3)$, and orthocenter $H(0, 0)$, the line segment $CH$ is the altitude to side $AB$, while the segment $BH$ is the altitude to side $AC$. Because the slope of $AB$ is $m_{AB} = \frac{3-2}{4-0} = \frac{1}{4}$, the perpendicular altitude $CH$ must have a slope of $-4$. This establishes the relationship $b = -4a$ for the coordinates $(a, b)$ of the third vertex $C$.

Similarly, the altitude $BH$ passes through $(4, 3)$ and $(0, 0)$ with a slope of $\frac{3}{4}$, so the side $AC$ must have a negative reciprocal slope of $-\frac{4}{3}$. Expressing this as $\frac{b-2}{a-0} = -\frac{4}{3}$ results in the linear equation $4a + 3b = 6$. Substituting $b = -4a$ into this equation yields $4a + 3(-4a) = 6$, which simplifies to $-8a = 6$, giving $a = -\frac{3}{4}$ and $b = 3$. Since the $x$-coordinate is negative and the $y$-coordinate is positive, the third vertex lies in the second quadrant.

For any three segments to form a triangle, the sum of the lengths of any two sides must be strictly greater than the length of the third side. Applying this to sides of lengths 5, 5r, and 5r^2, we establish three simultaneous inequalities: 5 + 5r > 5r^2, 5 + 5r^2 > 5r, and 5r + 5r^2 > 5. Dividing each inequality by 5 simplifies the conditions to r^2 - r - 1 < 0, r^2 - r + 1 > 0, and r^2 + r - 1 > 0.

The inequality r^2 - r + 1 > 0 holds true for all positive values of r because the quadratic expression has no real roots and remains positive. Solving the quadratic inequality r^2 - r - 1 < 0 provides the upper bound r < $\frac{1+\sqrt{5}}{2}$, while solving r^2 + r - 1 > 0 yields the lower bound r > $\frac{\sqrt{5}-1}{2}$. Combining these constraints requires r to lie within the interval $(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2})$. Since the lower bound is approximately 0.618 and the upper bound is approximately 1.618, the value r must fall between 0.618 and 1.618. Evaluating the provided options, 1.5, 0.75, and 1.25 all fall within this valid range, whereas 7/4, which is 1.75, exceeds the upper bound and is therefore impossible.

When two lines are perpendicular, the product of their slopes must be equal to $-1$. The first line, given by the equation $2x - 3y + 17 = 0$, can be rearranged into slope-intercept form as $3y = 2x + 17$, or $y = \frac{2}{3}x + \frac{17}{3}$, identifying its slope as $\frac{2}{3}$. The slope of the second line, which passes through the points $(7, 17)$ and $(15, \beta)$, is calculated using the change in $y$ over the change in $x$, expressed as $\frac{\beta - 17}{15 - 7}$, which simplifies to $\frac{\beta - 17}{8}$. Multiplying these two slopes together and setting the result to $-1$ produces the equation $\frac{2}{3} \times \frac{\beta - 17}{8} = -1$, which simplifies to $\frac{\beta - 17}{12} = -1$. Solving this final equation leads to $\beta - 17 = -12$, establishing that $\beta = 5$.

A parallelogram is defined by two pairs of parallel lines, and we begin by determining the coordinates of the vertex where the two initially given lines meet. Solving the simultaneous equations $x + y = 3$ and $x - y + 3 = 0$ reveals that this intersection point lies at $(0, 3)$. Since the diagonals of a parallelogram bisect each other, the given intersection point of the diagonals, $(2, 4)$, acts as the midpoint between this vertex and its opposite corner. Applying the midpoint formula, we find that the coordinates of the opposite vertex must be $(4, 5)$ because $\frac{0 + x}{2} = 2$ and $\frac{3 + y}{2} = 4$.

With this opposite vertex established, the remaining two sides of the parallelogram can be identified by constructing lines parallel to the original ones that pass through $(4, 5)$. The side parallel to $x + y = 3$ must satisfy the equation $x + y = 9$, as the sum of the coordinates at the vertex $(4, 5)$ is $9$. Similarly, the other side parallel to $x - y + 3 = 0$ must satisfy the equation $x - y = -1$. One of the parallelogram's vertices is found at the intersection of the side originating from $(0, 3)$ along the line $x - y + 3 = 0$ and the parallel side extending from $(4, 5)$ along the line $x + y = 9$. Solving these two equations, $x - y = -3$ and $x + y = 9$, simultaneously results in the coordinate $(3, 6)$.

Lines parallel to a given line $Ax + By + C = 0$ must share the same coefficients for $x$ and $y$, taking the general form $Ax + By + k = 0$. The perpendicular distance from the origin $(0,0)$ to any line defined by this equation is calculated using the formula $\frac{|k|}{\sqrt{A^2 + B^2}}$. For the original line $4x - 3y + 2 = 0$, the parallel lines follow the structure $4x - 3y + k = 0$, and their distance from the origin is given by $\frac{|k|}{\sqrt{4^2 + (-3)^2}} = \frac{|k|}{5}$. Setting this distance equal to $\frac{3}{5}$ leads to $|k| = 3$, which yields the two specific line equations $4x - 3y + 3 = 0$ and $4x - 3y - 3 = 0$. By substituting the point $\left(-\frac{1}{4}, \frac{2}{3}\right)$ into the equation $4x - 3y + 3 = 0$, we get $4\left(-\frac{1}{4}\right) - 3\left(\frac{2}{3}\right) + 3 = -1 - 2 + 3 = 0$, which confirms that this point satisfies the equation and lies on one of the lines.

The defining characteristic of an orthocenter is that it marks the intersection point of a triangle's altitudes. Because an altitude is inherently perpendicular to the triangle's side, any line acting as an altitude will have a slope that is the negative reciprocal of the slope of the side it intersects. With the orthocenter at (1, 1), this geometric property allows us to determine the equations of the altitudes corresponding to the given sides.

Given the side equations $3x - 2y + 6 = 0$ and $4x + 5y - 20 = 0$, we find their respective altitudes. The first side has a slope of $3/2$, so the altitude passing through (1, 1) follows the equation $2x + 3y - 5 = 0$. Similarly, the second side has a slope of $-4/5$, so the corresponding altitude passing through (1, 1) must be $5x - 4y - 1 = 0$. By calculating the intersection points of these altitudes with the original sides, we identify the coordinates of the vertices of the triangle. Specifically, intersecting the first side with the second altitude gives one vertex, while intersecting the second side with the first altitude gives the other. With the coordinates for these two vertices now established, we can derive the linear equation for the third side passing through them, which results in the equation:

$26x - 122y - 1675 = 0$

Calculating the slope of $L_1$ using the points (1, 2) and (-3, 4) results in $m = -\frac{1}{2}$, leading to the linear equation $x + 2y = 5$, which implies $h + 2k = 5$ since the point (h, k) lies on this line. Because line $L_2$ passes through (h, k) and (4, 3) and is perpendicular to $L_1$, its slope must be 2, providing the relationship $\frac{3 - k}{4 - h} = 2$, which rearranges to $2h - k = 5$. Solving the system of equations $h + 2k = 5$ and $2h - k = 5$ simultaneously yields $h = 3$ and $k = 1$. Consequently, the value of the ratio $\frac{k}{h}$ is $\frac{1}{3}$.

A point is equidistant from the coordinate axes when its absolute coordinates are equal, meaning it must satisfy the condition $|x| = |y|$. This implies the point sits on one of the lines $y = x$ or $y = -x$. By substituting $y = x$ into the given line equation $3x + 5y = 15$, we calculate $8x = 15$, which yields the point $(\frac{15}{8}, \frac{15}{8})$ located in the 1st quadrant. Conversely, substituting $y = -x$ into the same line equation results in $-2x = 15$, which identifies the point $(-\frac{15}{2}, \frac{15}{2})$ situated in the 2nd quadrant. Consequently, these points appear only in the 1st and 2nd quadrants.

The x-intercept A and y-intercept B of the line $3x + y = \lambda$ are obtained by setting $y=0$ and $x=0$ respectively, yielding $A(\lambda/3, 0)$ and $B(0, \lambda)$. The line passing through the origin and perpendicular to $3x + y = \lambda$ has a slope of $1/3$, which gives the equation $x - 3y = 0$. Substituting $x=3y$ into the original line equation $3(3y) + y = \lambda$ results in $10y = \lambda$, determining the coordinates of the foot of the perpendicular $P$ as $P(3\lambda/10, \lambda/10)$.

The distance $BP$ is calculated using the distance formula:
$BP = \sqrt{\left(\frac{3\lambda}{10} - 0\right)^2 + \left(\frac{\lambda}{10} - \lambda\right)^2} = \sqrt{\frac{9\lambda^2}{100} + \frac{81\lambda^2}{100}} = \frac{3\lambda}{\sqrt{10}}$
Similarly, the distance $PA$ is determined as:
$PA = \sqrt{\left(\frac{\lambda}{3} - \frac{3\lambda}{10}\right)^2 + \left(0 - \frac{\lambda}{10}\right)^2} = \sqrt{\left(\frac{\lambda}{30}\right)^2 + \left(\frac{\lambda}{10}\right)^2} = \frac{\lambda}{3\sqrt{10}}$
Dividing these distances gives the ratio $BP/PA = (3\lambda/\sqrt{10}) / (\lambda/(3\sqrt{10})) = 9/1$.

The centroid of a triangle is the point of intersection of its medians, which can be found by solving the given median equations $x+y=5$ and $x=4$ to obtain $G(4, 1)$. Given that the centroid coordinates are defined by $G = \left(\frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3}\right)$ and the vertex $A$ is $(1, 2)$, the sum of the coordinates of the vertices must satisfy $x_A+x_B+x_C = 12$ and $y_A+y_B+y_C = 3$.

Substituting the known coordinate $x_A=1$ and the fact that point $C$ lies on the line $x=4$ into the sum $1+x_B+x_C = 12$ yields $x_B=7$. Using the line $x+y=5$ for point $B$, we find $y_B = 5 - 7 = -2$. Applying these values to the y-coordinate sum $2 + (-2) + y_C = 3$ results in $y_C = 3$.

The area of the triangle with vertices $A(1, 2)$, $B(7, -2)$, and $C(4, 3)$ is determined by the formula $Area = \frac{1}{2} |x_A(y_B-y_C) + x_B(y_C-y_A) + x_C(y_A-y_B)|$. Substituting the coordinates gives $Area = \frac{1}{2} |1(-2-3) + 7(3-2) + 4(2-(-2))| = \frac{1}{2} |-5 + 7 + 16| = 9$ square units.

The intercept form of a line equation, $\frac{x}{a} + \frac{y}{b} = 1$, describes a line that intersects the coordinate axes at points $(a,0)$ and $(0,b)$. Since the line passes through the fixed point $(2,3)$, we substitute these coordinates into the equation to establish the constraint $\frac{2}{a} + \frac{3}{b} = 1$. In the rectangle $OPRQ$ formed with the origin $O$, the vertex $R$ is located at $(a,b)$, meaning we can define the locus of $R(x,y)$ by setting $x=a$ and $y=b$. Substituting these variables into our constraint yields $\frac{2}{x} + \frac{3}{y} = 1$. Multiplying this entire equation by $xy$ results in $2y + 3x = xy$, which is equivalent to $3x + 2y = xy$.

The sides of the rhombus are parallel to lines with slopes $m_1 = 1$ and $m_2 = 7$. Because the diagonals of a rhombus are the angle bisectors of the lines containing its sides, the slopes $m$ of the diagonals satisfy the equation $\frac{m - 1}{1 + m} = -\frac{m - 7}{1 + 7m}$.

Expanding and rearranging this equation yields $7m^2 - 6m - 1 = -m^2 + 6m + 7$, which simplifies to $8m^2 - 12m - 8 = 0$, or $2m^2 - 3m - 2 = 0$. Factoring the quadratic gives $(2m + 1)(m - 2) = 0$, so the slopes of the diagonals are $m = 2$ and $m = -1/2$. The diagonal $AP$ passes through $A(0, k)$ and $P(1, 2)$, which means its slope is $\frac{2 - k}{1 - 0} = 2 - k$. Since $A$ is not the origin, $k \neq 0$, so the slope cannot be $2$ (which would imply $k = 0$). Setting the slope to $-1/2$ gives $2 - k = -1/2$, which results in $k = 5/2$.

The area of a triangle with coordinates $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is calculated using the absolute value of the determinant expression

. Substituting the given vertices $(k, -3k), (5, k),$ and $(-k, 2)$ into this determinant yields the equation $\frac{1}{2} |5k^2 + 13k + 10| = 28$, which simplifies to $|5k^2 + 13k + 10| = 56$.

This absolute value equation results in two potential quadratic scenarios: $5k^2 + 13k - 46 = 0$ and $5k^2 + 13k + 66 = 0$. Applying the discriminant to the second equation reveals $169 - 1320 < 0$, indicating no real solutions. The first equation, however, factors into $(k - 2)(5k + 23) = 0$, giving $k = 2$ and $k = -4.6$. Since the problem specifies that $k$ is an integer, $k = 2$ is the only valid value.

With $k = 2$, the vertices of the triangle are $A(2, -6), B(5, 2),$ and $C(-2, 2)$. Because the side $BC$ is a horizontal segment lying on $y = 2$, the altitude from $A$ to this side must be a vertical line, which is defined by the equation $x = 2$. To locate the orthocenter, one needs the intersection of this line with a second altitude, such as the one from vertex $B$ to side $AC$. The slope of side $AC$ is $\frac{2 - (-6)}{-2 - 2} = -2$, so the altitude from $B$ must have a perpendicular slope of $\frac{1}{2}$. The equation for this altitude is $y - 2 = \frac{1}{2}(x - 5)$, and substituting $x = 2$ into this equation gives $y - 2 = \frac{1}{2}(2 - 5)$, which simplifies to $y = \frac{1}{2}$. The intersection point is thus $(2, \frac{1}{2})$.

The vertices of a square with side length $s$ and one vertex at the origin are determined by the angle $\theta$ that one side makes with the positive x-axis. Given side length $s=2$ and $\theta = 30^\circ$, the first vertex is at $V_1 = (0,0)$, and the second vertex is at $V_2 = (2\cos 30^\circ, 2\sin 30^\circ) = (\sqrt{3}, 1)$. Since the adjacent side of the square is perpendicular to the first, it lies at an angle of $30^\circ + 90^\circ = 120^\circ$, placing the fourth vertex at $V_4 = (2\cos 120^\circ, 2\sin 120^\circ) = (-1, \sqrt{3})$. The third vertex, $V_3$, is the vector sum of these two adjacent sides, resulting in coordinates $(\sqrt{3}-1, 1+\sqrt{3})$. Summing the x-coordinates of all four vertices yields $0 + \sqrt{3} + (\sqrt{3}-1) + (-1)$, which simplifies to $2\sqrt{3}-2$.

A rhombus is geometrically defined by two pairs of parallel lines, meaning the four lines bounding the shape must share slopes in pairs. Given the two initial side equations $x - y + 1 = 0$ and $7x - y - 5 = 0$, the other two sides of the rhombus are identified as $x - y - 3 = 0$ and $7x - y + 15 = 0$. Because every vertex of a rhombus lies at the intersection of two adjacent sides, we can determine these coordinates by solving systems of equations formed by these intersecting lines. Intersecting the line $x - y - 3 = 0$ with the line $7x - y - 5 = 0$ results in the system $x - y = 3$ and $7x - y = 5$, where subtracting the first equation from the second yields $6x = 2$. This calculation gives $x = \frac{1}{3}$, which leads to $y = -\frac{8}{3}$, establishing $\left(\frac{1}{3}, -\frac{8}{3}\right)$ as one of the vertices of the rhombus.

The principle of reflection dictates that the angle between the incident ray and the mirror line must be equal to the angle between the reflected ray and the same mirror line. The mirror line is given by $7x-y+1=0$, which has a slope of $7$, and the reflected ray lies along the line $y+2x=1$, which has a slope of $-2$. If the incident ray has an unknown slope $m$, the angle $\theta$ between the incident ray and the mirror line must satisfy the same tangent relationship as the angle between the mirror line and the reflected ray. Calculating this tangent value, we have $\tan \theta = \left|\frac{-2 - 7}{1 + (-2)(7)}\right| = \left|\frac{-9}{-13}\right| = \frac{9}{13}$.

Setting the tangent of the angle between the incident line and the mirror line equal to this value gives the equation $\left|\frac{m - 7}{1 + 7m}\right| = \frac{9}{13}$. Taking the negative case to find the incident slope, we solve $\frac{m - 7}{1 + 7m} = -\frac{9}{13}$, which simplifies to $13(m - 7) = -9(1 + 7m)$. Expanding this yields $13m - 91 = -9 - 63m$, so $76m = 82$, resulting in a slope of $m = \frac{41}{38}$. Since the incident ray must pass through the point $(0, 1)$, we use the point-slope form $y - 1 = \frac{41}{38}(x - 0)$. Multiplying by $38$ gives $38y - 38 = 41x$, which rearranges to the final equation of the incident line, $41x - 38y + 38 = 0$.

To determine the path traced by the midpoint of a variable line, we utilize the concept of a family of lines passing through the intersection of the two given lines, $\frac{x}{3}+\frac{y}{4}=1$ and $\frac{x}{2}+\frac{y}{3}=1$. By converting these to the standard form $4x+3y-12=0$ and $3x+4y-12=0$, the family is expressed as $(4x+3y-12) + \lambda(3x+4y-12) = 0$, where $\lambda$ is an arbitrary constant. Collecting terms, this equation becomes $(4+3\lambda)x + (3+4\lambda)y = 12(1+\lambda)$.

Setting $y=0$ identifies the x-intercept of the line at coordinate $A = (\frac{12(1+\lambda)}{4+3\lambda}, 0)$, while setting $x=0$ identifies the y-intercept at $B = (0, \frac{12(1+\lambda)}{3+4\lambda})$. Let $(h, k)$ represent the midpoint of segment $AB$. Applying the midpoint formula, we find $h = \frac{6(1+\lambda)}{4+3\lambda}$ and $k = \frac{6(1+\lambda)}{3+4\lambda}$.

To find the locus, we eliminate $\lambda$ by observing the reciprocal relationship between $h$ and $k$. Since $\frac{1}{h} = \frac{4+3\lambda}{6(1+\lambda)}$ and $\frac{1}{k} = \frac{3+4\lambda}{6(1+\lambda)}$, adding these two fractions results in $\frac{1}{h} + \frac{1}{k} = \frac{4+3\lambda+3+4\lambda}{6(1+\lambda)} = \frac{7(1+\lambda)}{6(1+\lambda)} = \frac{7}{6}$. This simplifies to $6(h+k) = 7hk$, which reveals the final locus equation as $7xy = 6(x+y)$.

Any line passing through the origin can be represented using parametric coordinates, where a point on the line is defined by $x = r \cos \theta$ and $y = r \sin \theta$ for some distance $r$ from the origin and angle $\theta$. Substituting these expressions into the first line equation $4x + 3y = 10$ yields:

$r_A(4 \cos \theta + 3 \sin \theta) = 10$

This allows us to express the distance $OA$ as $r_A = \frac{10}{4 \cos \theta + 3 \sin \theta}$. Performing the same substitution for the second line $8x + 6y + 5 = 0$ results in the equation:

$r_B(8 \cos \theta + 6 \sin \theta) = -5$

Because $8 \cos \theta + 6 \sin \theta$ is equal to $2(4 \cos \theta + 3 \sin \theta)$, the equation simplifies to $2r_B(4 \cos \theta + 3 \sin \theta) = -5$, giving $r_B = \frac{-5}{2(4 \cos \theta + 3 \sin \theta)}$. Since the term $(4 \cos \theta + 3 \sin \theta)$ is common to both distance expressions, the ratio of the lengths $OA$ to $OB$ is:

$\left| \frac{r_A}{r_B} \right| = \left| \frac{10}{-2.5} \right| = 4$

The magnitude of the ratio is 4, confirming that the origin divides the segment $AB$ in the ratio $4:1$.