Among the statements: (S1): If A(5,-1) and B(-2,3) are two vertices of a triangle, whose orthocentre is (0,0), then its third vertex is (-4,-7) and (S2): If positive numbers 2a, b, c are three consecutive terms of an A.P., then the lines ax + by + c = 0 are concurrent at (2,-2).

Orthocenter property: Altitude from A is perpendicular to BC, Altitude from B is perpendicular to AC. Concurrency of lines: ax+by+c=0 passes through (x0, y0) if ax0+by0+c=0 holds. Statement 1: Let third vertex be C(h,k) and orthocenter O(0,0). - Slope AO cdot Slope BC = -1 ((-1/5)) cdot ((k-3/h+2)) = -1 5h - k + 13 = 0 - Slope BO cdot Slope AC = -1 ((3/-2)) cdot ((k+1/h-5)) = -1 4k = 7h Solving simultaneously gives h = -4, k = -7 C(-4,-7). Statement 1 is correct. Statement 2: 2a, b, c in A.P. 2b = 2a + c 2a - 2b + c = 0. Comparing with ax + by + c = 0 gives x = 2, y = -2. Statement 2 is correct. Both (S1) and (S2) are correct statements. Orthocenter coordinates yield perpendicularity conditions via slopes. Concurrency follows directly from linear relation between line coefficients. Class 11 Maths: Straight Lines

Among the statements: (S1): If A(5,-1) and B(-2,3) are two vertices of a triangle, whose orthocentre is (0,0), then its third vertex is (-4,-7) and (S2): If positive numbers 2a, b, c are three consecu

Orthocenter property: Altitude from A is perpendicular to BC, Altitude from B is perpendicular to AC. Concurrency of lines: ax+by+c=0 passes through (x0, y0) if ax0+by0+c=0 holds. Statement 1: Let third vertex be C(h,k) and orthocenter O(0,0). - Slope AO cdot Slope BC = -1 ((-1/5)) cdot ((k-3/h+2)) = -1 5h - k + 13 = 0 - Slope BO cdot Slope AC = -1 ((3/-2)) cdot ((k+1/h-5)) = -1 4k = 7h Solving simultaneously gives h = -4, k = -7 C(-4,-7). Statement 1 is correct. Statement 2: 2a, b, c in A.P. 2b

Among the statements: (S1): If A(5,-1) and B(-2,3) are two vertices of a triangle, whose orthocentre is (0,0), then its third vertex is (-4,-7) and (S2): If positive numbers 2a, b, c are three consecutive terms of an A.P., then the lines ax + by + c = 0 are concurrent at (2,-2).

Orthocenter property: Altitude from A is perpendicular to BC, Altitude from B is perpendicular to AC. Concurrency of lines: ax+by+c=0 passes through (x0, y0) if ax0+by0+c=0 holds. Statement 1: Let third vertex be C(h,k) and orthocenter O(0,0). - Slope AO cdot Slope BC = -1 ((-1/5)) cdot ((k-3/h+2)) = -1 5h - k + 13 = 0 - Slope BO cdot Slope AC = -1 ((3/-2)) cdot ((k+1/h-5)) = -1 4k = 7h Solving simultaneously gives h = -4, k = -7 C(-4,-7). Statement 1 is correct. Statement 2: 2a, b, c in A.P. 2b = 2a + c 2a - 2b + c = 0. Comparing with ax + by + c = 0 gives x = 2, y = -2. Statement 2 is correct. Both (S1) and (S2) are correct statements. Orthocenter coordinates yield perpendicularity conditions via slopes. Concurrency follows directly from linear relation between line coefficients. Class 11 Maths: Straight Lines

Among the statements: (S1): If A(5,-1) and B(-2,3) are two vertices of a triangle, whose orthocentre is (0,0), then its third vertex is (-4,-7) and (S2): If positive numbers 2a, b, c are three consecu

Orthocenter property: Altitude from A is perpendicular to BC, Altitude from B is perpendicular to AC. Concurrency of lines: ax+by+c=0 passes through (x0, y0) if ax0+by0+c=0 holds. Statement 1: Let third vertex be C(h,k) and orthocenter O(0,0). - Slope AO cdot Slope BC = -1 ((-1/5)) cdot ((k-3/h+2)) = -1 5h - k + 13 = 0 - Slope BO cdot Slope AC = -1 ((3/-2)) cdot ((k+1/h-5)) = -1 4k = 7h Solving simultaneously gives h = -4, k = -7 C(-4,-7). Statement 1 is correct. Statement 2: 2a, b, c in A.P. 2b

Among the statements: (S1): If A(5,-1) and B(-2,3) are two vertices of a triangle, whose orthocentre is (0,0), then its…

Year	2026	2025	2024	Total
Questions	7	13	9	29

Q jee_main_2024_29_january_evening Distance of a Point From a Line

The distance of the point (2, 3) from the line

2x - 3y + 28 = 0

, measured parallel to the line

\sqrt(3) x - y + 1 = 0

, is equal to

A. $4\sqrt(2)$ $4\sqrt{2}$
B. $6\sqrt(3)$ $6\sqrt{3}$
C. $3 + 4\sqrt(2)$ $3 + 4\sqrt{2}$
D. $4 + 6\sqrt(3)$ $4 + 6\sqrt{3}$

Solution

Related Formula

x = x₁ + r θ, y = y₁ + r θ

Core Logic

The line is measured parallel to $\sqrt(3)x - y + 1 = 0$ $\sqrt{3}x - y + 1 = 0$, which has a slope $θ = \sqrt(3) θ = 60^°$ . Thus, $θ = (1)/(2)$ $\cos \theta = \frac{1}{2}$ and $θ = \sqrt(3)2$ $\sin \theta = \frac{\sqrt{3}}{2}$.

Writing any point $P$ $P$ along this direction passing through $(2,3)$ $(2,3)$ in parametric coordinates:

P = (2 + r 60^°, 3 + r 60^°) = (2 + (r)/(2), 3 + \sqrt(3)r2)

Step 1: Finding Intersection Point

Since $P$ $P$ must lie on the given line $2x - 3y + 28 = 0$ $2x - 3y + 28 = 0$:

2(2 + (r)/(2)) - 3(3 + \sqrt(3)r2) + 28 = 0

4 + r - 9 - 3\sqrt(3)r2 + 28 = 0

23 + r(1 - 3\sqrt(3)2) = 0

r( 3\sqrt(3) - 22) = 23 r = 463\sqrt(3) - 2

Rationalizing the denominator:

r = 46(3\sqrt(3) + 2)(3\sqrt(3))² - 2² = 46(3\sqrt(3) + 2)27 - 4 = 46(3\sqrt(3) + 2)23 = 2(3\sqrt(3) + 2) = 4 + 6\sqrt(3)

Pattern Recognition

Distance measured parallel to a given direction is always resolved most efficiently using parametric equations of lines rather than perpendicular metrics.

Chapter Mix

Class 11 Mathematics: Straight Lines

Q jee_main_2024_29_january_evening Intersection of Lines and Distance

Let

A

be the point of intersection of the lines

3x + 2y = 14, 5x - y = 6

and

B

be the point of intersection of the lines

4x + 3y = 8, 6x + y = 5

. The distance of the point

P(5, -2)

from the line AB is

A. $(13)/(2)$ $\frac{13}{2}$
B. 8
C. $(5)/(2)$ $\frac{5}{2}$
D. 6

Solution

Related Formula

Perpendicular distance d = |ax₀ + by₀ + c|\sqrt(a² + b²)

Core Logic

Let us find coordinates of point $A$ $A$ by solving:

$3x + 2y = 14$ $3x + 2y = 14$
$5x - y = 6 y = 5x - 6$ $5x - y = 6 \implies y = 5x - 6$

Substituting $y$ $y$ in equation 1:

3x + 2(5x - 6) = 14 13x - 12 = 14 13x = 26 x = 2

y = 5(2) - 6 = 4 A = (2, 4)

Let us find coordinates of point $B$ $B$ by solving:

$4x + 3y = 8$ $4x + 3y = 8$
$6x + y = 5 y = 5 - 6x$ $6x + y = 5 \implies y = 5 - 6x$

Substituting $y$ $y$ in equation 3:

4x + 3(5 - 6x) = 8 4x + 15 - 18x = 8 -14x = -7 x = (1)/(2)

y = 5 - 6((1)/(2)) = 2 B = ((1)/(2), 2)

Step 1: Equation of line AB

Slope m = (4 - 2)/(2 - 1/2) = (2)/(3/2) = (4)/(3)

Equation of line AB:

y - 4 = (4)/(3)(x - 2) 3y - 12 = 4x - 8 4x - 3y + 4 = 0

Step 2: Distance Estimation

Perpendicular distance from point $P(5, -2)$ $P(5, -2)$ to line $4x - 3y + 4 = 0$ $4x - 3y + 4 = 0$:

d = |4(5) - 3(-2) + 4|\sqrt(4² + (-3)²) = |20 + 6 + 4|\sqrt(25) = (30)/(5) = 6

Pattern Recognition

Verify calculation metrics step-by-step. Finding straight intersections correctly upfront avoids scaling mistakes down the track.

Chapter Mix

Class 11 Mathematics: Straight Lines

Q jee_main_2024_27_jan_morning Angle Between Two Lines

The portion of the line

4x+5y=20

in the first quadrant is trisected by the lines

L₁

and

L₂

passing through the origin. The tangent of an angle between the lines

L₁

and

L₂

is:

A. $(8)/(5)$ $$\frac{8}{5}$$
B. $(25)/(41)$ $$\frac{25}{41}$$
C. $(2)/(5)$ $$\frac{2}{5}$$
D. $(30)/(41)$ $$\frac{30}{41}$$

Solution

Related Formula

θ = | (m₁ - m₂)/(1 + m₁ m₂) |

Core Logic

Find the intercepts of the line $4x + 5y = 20$ $4x + 5y = 20$ in the first quadrant. Put $y=0 \Rightarrow x=5$ $y=0 \Rightarrow x=5$. Point $X(5, 0)$ $X(5, 0)$. Put $x=0 \Rightarrow y=4$ $x=0 \Rightarrow y=4$. Point $Y(0, 4)$ $Y(0, 4)$. The line segment $XY$ $XY$ is trisected by two points, say $A$ $A$ and $B$ $B$. Point $A$ $A$ divides $YX$ $YX$ in the ratio $2:1$ $2:1$, and $B$ $B$ divides it in $1:2$ $1:2$.

Step 1: Trisection Points Calculation

Using the section formula for $A$ $A$ (closer to Y-axis, ratio 1:2 from Y to X):

A = ( (1(5) + 2(0))/(3), (1(0) + 2(4))/(3) ) = ( (5)/(3), (8)/(3) )

Using the section formula for $B$ $B$ (closer to X-axis, ratio 2:1 from Y to X):

B = ( (2(5) + 1(0))/(3), (2(0) + 1(4))/(3) ) = ( (10)/(3), (4)/(3) )

Step 2: Finding Line Slopes

The lines $L₁$ $L_1$ and $L₂$ $L_2$ pass through the origin $(0,0)$ $(0,0)$ to points $A$ $A$ and $B$ $B$. Slope of $OA$ $OA$ ( $m₁$ $m_1$):

m₁ = (8/3 - 0)/(5/3 - 0) = (8)/(5)

Slope of $OB$ $OB$ ( $m₂$ $m_2$):

m₂ = (4/3 - 0)/(10/3 - 0) = (4)/(10) = (2)/(5)

Step 3: Calculating Tangent of the Angle

Substitute the slopes into the angle formula:

θ = | (8/5 - 2/5)/(1 + (8/5)(2/5)) |

θ = (6/5)/(1 + 16/25)

θ = (6/5)/((25+16)/25) = (6/5)/(41/25)

θ = (6)/(5) \times (25)/(41) = (30)/(41)

Pattern Recognition

For trisection or specific division of an intercepted segment, identify the axis intercepts first, rapidly apply the internal section formula, compute origin-centered slopes (which equal just the y/x ratio of the points), and pass them into the tan formula.

Chapter Mix

Class 11 Maths: Straight Lines

Q7 jee_main_2024_29_jan_morning Angle Bisector and Reflection

In a

Δ ABC

, suppose

y=x

is the equation of the bisector of the angle B and the equation of the side AC is

2x-y=2

. If

2AB=BC

and the point A and B are respectively

(4,6)

and

(α,β)

, then

α+2β

is equal to

A. $42$ $42$
B. $39$ $39$
C. $48$ $48$
D. $45$ $45$

Solution

Related Formula

Image of a point $(x₁, y₁)$ $(x_1, y_1)$ across line $y=x$ $y=x$ is $(y₁, x₁)$ $(y_1, x_1)$.

Angle Bisector Theorem: The angle bisector of a triangle divides the opposite side into segments proportional to the lengths of the adjacent sides:

(AB)/(BC) = (AD)/(DC)

Core Logic

Given $A(4,6)$ $A(4,6)$ and Angle bisector of B is $y=x$ $y=x$. Because $y=x$ $y=x$ bisects angle B, the geometric reflection of vertex A across the bisector line $y=x$ $y=x$ must lie exactly on the line containing the side $BC$ $BC$. Let the reflection of $A(4,6)$ $A(4,6)$ be $A'$ $A'$. Across $y=x$ $y=x$, the coordinates swap: $A' = (6,4)$ $A' = (6,4)$

Next, find the intersection point $D$ $D$ of the bisector $y=x$ $y=x$ and side $AC$ $AC$ ( $2x-y=2$ $2x-y=2$). Substitute $y=x$ $y=x$ into $2x-y=2$ $2x-y=2$:

2x - x = 2 \Rightarrow x = 2 \Rightarrow y = 2

So, point $D$ $D$ is $(2,2)$ $(2,2)$.

Step 1: Utilize Section Formula

By the internal angle bisector theorem:

(AD)/(DC) = (AB)/(BC)

Given $2AB = BC$ $2AB = BC$, so $(AB)/(BC) = (1)/(2)$ $\frac{AB}{BC} = \frac{1}{2}$. This means point $D(2,2)$ $D(2,2)$ divides the segment $AC$ $AC$ in the ratio $1:2$ $1:2$. Let $C$ $C$ have coordinates $(xc, yc)$ $(x_c, y_c)$. Applying the section formula for $D(2,2)$ $D(2,2)$ dividing $A(4,6)$ $A(4,6)$ and $C(xc, yc)$ $C(x_c, y_c)$ in ratio $1:2$ $1:2$:

2 = (1 \cdot xc + 2 \cdot 4)/(1 + 2) \Rightarrow 6 = xc + 8 \Rightarrow xc = -2

2 = (1 \cdot yc + 2 \cdot 6)/(1 + 2) \Rightarrow 6 = yc + 12 \Rightarrow yc = -6

So, $C$ $C$ is $(-2,-6)$ $(-2,-6)$.

Step 2: Find Equation of BC

The line $BC$ $BC$ passes through point $C(-2,-6)$ $C(-2,-6)$ and the reflection point $A'(6,4)$ $A'(6,4)$. Find the slope of $BC$ $BC$:

mBC = (4 - (-6))/(6 - (-2)) = (10)/(8) = (5)/(4)

Equation of $BC$ $BC$:

y - 4 = (5)/(4)(x - 6)

4y - 16 = 5x - 30

5x - 4y - 14 = 0

Step 3: Solve for Vertex B

Vertex $B(α, β)$ $B(\alpha, \beta)$ is the intersection of line $BC$ $BC$ and the angle bisector $y=x$ $y=x$. Substitute $y=x$ $y=x$ into $5x - 4y - 14 = 0$ $5x - 4y - 14 = 0$:

5x - 4x - 14 = 0 \Rightarrow x = 14

Thus, $y = 14$ $y = 14$. Therefore, $B$ $B$ is $(14, 14)$ $(14, 14)$, implying $α = 14$ $\alpha = 14$ and $β = 14$ $\beta = 14$.

Calculate $α + 2β$ $\alpha + 2\beta$:

α + 2β = 14 + 2(14) = 42

Pattern Recognition

Reflection properties drastically simplify angle bisector questions. If you know the bisector equation, reflecting one vertex over it gives a coordinate on the opposing extended ray. This paired with the angle bisector proportion theorem locks the entire geometric frame.

Chapter Mix

Class 11 Mathematics: Straight Lines

Q7 jee_main_2024_30_january_evening Angle Bisectors

If

x² - y² + 2hxy + 2gx + 2fy + c = 0

is the locus of a point, which moves such that it is always equidistant from the lines

x + 2y + 7 = 0

and

2x - y + 8 = 0

, then the value of

g + c + h - f

equals

A. $14$ $14$
B. $6$ $6$
C. $8$ $8$
D. $29$ $29$

Solution

Related Formula

Distance of (x, y) from ax+by+c=0 is d = |ax + by + c|\sqrt(a² + b²)

Core Logic

The locus of a point $P(x, y)$ $P(x, y)$ equidistant from lines $x + 2y + 7 = 0$ $x + 2y + 7 = 0$ and $2x - y + 8 = 0$ $2x - y + 8 = 0$ is the pair of angle bisectors:

|x + 2y + 7|\sqrt(1² + 2²) = |2x - y + 8|\sqrt(2² + (-1)²)

x + 2y + 7\sqrt(5) = \pm 2x - y + 8\sqrt(5)

Step 1: Generating the Combined Equation

Squaring both sides eliminates the $\pm$ $\pm$ and generates the combined equation of the bisectors:

(x + 2y + 7)² - (2x - y + 8)² = 0

Using $a² - b² = (a - b)(a + b)$ $a^2 - b^2 = (a - b)(a + b)$:

[ (x + 2y + 7) - (2x - y + 8) ] [ (x + 2y + 7) + (2x - y + 8) ] = 0

(-x + 3y - 1)(3x + y + 15) = 0

(x - 3y + 1)(3x + y + 15) = 0

Step 2: Expanding the Equation

Multiply out the terms:

3x² + xy + 15x - 9xy - 3y² - 45y + 3x + y + 15 = 0

3x² - 3y² - 8xy + 18x - 44y + 15 = 0

Step 3: Comparing Coefficients

The standard form given is $x² - y² + 2hxy + 2gx + 2fy + c = 0$ $x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0$. Divide our derived equation by 3 to match the leading coefficients:

x² - y² - (8)/(3)xy + 6x - (44)/(3)y + 5 = 0

Now, compare coefficients:

2h = -(8)/(3) \Rightarrow h = -(4)/(3)

2g = 6 \Rightarrow g = 3

2f = -(44)/(3) \Rightarrow f = -(22)/(3)

$c = 5$ $c = 5$

Step 4: Final Calculation

Substitute into the expression $g + c + h - f$ $g + c + h - f$:

3 + 5 - (4)/(3) - (-(22)/(3)) = 8 + (18)/(3) = 8 + 6 = 14

Pattern Recognition

Locus of equidistant points from two lines is their pair of angle bisectors. Equating squares $d₁² = d₂²$ $d_1^2 = d_2^2$ directly yields the joint equation of bisectors without needing explicit line separation.

Chapter Mix

Class 11 Maths: Straight Lines

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Straight Lines Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Finding Intersection Point

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Equation of line AB

Step 2: Distance Estimation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Trisection Points Calculation

Step 2: Finding Line Slopes

Step 3: Calculating Tangent of the Angle

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Utilize Section Formula

Step 2: Find Equation of BC

Step 3: Solve for Vertex B

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Generating the Combined Equation

Step 2: Expanding the Equation

Step 3: Comparing Coefficients

Step 4: Final Calculation

Pattern Recognition

Chapter Mix

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