Distance of a Point from a Line

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JEE Main 2010

The line $L$ given by $\frac{x}{5} + \frac{y}{b} = 1$ passes through the point $(13, 32)$ . The line $K$ is parallel to $L$ and has the equation $\frac{x}{c} + \frac{y}{3} = 1$ . Then the distance between $L$ and $K$ is

17

(B)

\frac{17}{15}

(C)

\frac{23}{17}

(D)

\frac{23}{15}

JEE Main 2004

LEVELJEE Main

In a right angle $△ A B C$ , $∠ A = 9 0^{\circ}$ and sides a, b, c are respectively, 5 cm, 4 cm and 3 cm. If a force $F$ has moments 0, 9 and 16 in N cm. units respectively about vertices A, B and C, then magnitude of $F$ is

A straight line through the origin $O$ meets the parallel lines $4 x + 2 y = 9$ and $2 x + y + 6 = 0$ at points $P$ and $Q$ respectively. Then the point $O$ divides the segment $P Q$ in the ratio

If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is

two intersecting lines

JEE Advanced 1990

LEVELJEE Main

Line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a given angle, keeping the origin fixed, the same line $L$ has intercepts $p$ and $q$ , then

(A)

a^{2} + b^{2} = p^{2} + q^{2}

(B)

\frac{1}{a ^{2}} + \frac{1}{b ^{2}} = \frac{1}{p ^{2}} + \frac{1}{q ^{2}}

(C)

a^{2} + p^{2} = b^{2} + q^{2}

(D)

\frac{1}{a ^{2}} + \frac{1}{p ^{2}} = \frac{1}{b ^{2}} + \frac{1}{q ^{2}}

JEE Advanced 2014

LEVELJEE Main

For a point $P$ in the plane, let $d_{1} (P)$ and $d_{2} (P)$ be the distance of the point $P$ from the lines $x - y = 0$ and $x + y = 0$ respectively. The area of the region $R$ consisting of all points $P$ lying in the first quadrant of the plane and satisfying $2 \leq d_{1} (P) + d_{2} (P) \leq 4$ , is .........

JEE Main 2018 (15 April Evening)

LEVELJEE Main

The foot of the perpendicular drawn from the origin, on the line, $3 x + y = λ (λ \neq = 0)$ is P. If the line meets x-axis at A and y-axis at B, then the ratio BP : PA is :-

JEE Main 2019 (10 April Shift 2)

LEVELJEE Main

Lines are drawn parallel to the line $4 x - 3 y + 2 = 0$ , at a distance $3/5$ from the origin. Then which one of the following points lies on any of these lines ?

JEE Main 2020 (8 January Shift 2)

LEVELJEE Main

The length of the perpendicular from the origin, on the normal to the curve, $x^{2} + 2 x y - 3 y^{2} = 0$ at the point $(2, 2)$ is:

(A)

(B)

22

(C)

42

(D)

2

JEE Main 2020 - 8 Jan (Evening)

LEVELJEE Main

The length of the perpendicular from the origin on the normal to the curve, $x^{2} + 2 x y - 3 y^{2} = 0$ at the point $(2, 2)$ is

(A)

42

(B)

(C)

22

(D)

JEE Main 2020 - 5 Sep (Morning)

LEVELJEE Main

If the line, $2 x - y + 3 = 0$ is at a distance $\frac{1}{5}$ and $\frac{2}{5}$ from the lines $4 x - 2 y + α = 0$ and $6 x - 3 y + β = 0$ , respectively, then the sum of all possible values of $α$ and $β$ is

JEE Main 2020 - 6 Sep (Evening)

LEVELJEE Main

Let $L$ denote the line in the $x y$ -plane with $x$ and $y$ intercepts as 3 and 1 respectively. Then the image of the point $(- 1, - 4)$ in this line is :

(A)

(\frac{29}{5}, \frac{8}{5})

(B)

(\frac{29}{5}, \frac{11}{5})

(C)

(\frac{8}{5}, \frac{29}{5})

(D)

(\frac{11}{5}, \frac{28}{5})

JEE Main 2021 (31 Aug Shift 2)

LEVELJEE Main

Let $A$ be the set of all points $(α, β)$ such that the area of triangle formed by the points $(5, 6), (3, 2)$ and $(α, β)$ is 12 square units. Then the least possible length of a line segment joining the origin to a point in $A$ , is :

(A)

\frac{4}{5}

(B)

\frac{16}{5}

(C)

\frac{8}{5}

(D)

\frac{12}{5}

JEE Main 2021 (31 Aug Shift 1)

LEVELJEE Main

If $p$ and $q$ are the lengths of the perpendiculars from the origin on the lines, $x cosec α - y sec α = k cot 2 α$ and $x sin α + y cos α = k sin 2 α$ respectively, then $k^{2}$ is equal to:

(A)

4 p^{2} + q^{2}

(B)

2 p^{2} + q^{2}

(C)

p^{2} + 2 q^{2}

(D)

p^{2} + 4 q^{2}

JEE Main 2021 (25 February Shift 1)

LEVELJEE Main

The image of the point $(3, 5)$ in the line $x - y + 1 = 0$ , lies on :

(A)

(x - 2)^{2} + (y - 4)^{2} = 4

(B)

(x - 4)^{2} + (y + 2)^{2} = 16

(C)

(x - 4)^{2} + (y - 4)^{2} = 8

(D)

(x - 2)^{2} + (y - 2)^{2} = 12

JEE Main 2022 (26 June Shift 1)

LEVELJEE Main

Let $R$ be the point $(3, 7)$ and let $P$ and $Q$ be two points on the line $x + y = 5$ such that $P QR$ is an equilateral triangle. Then the area of $Δ P QR$ is :

(A)

\frac{25}{4 3}

(B)

\frac{25 3}{2}

(C)

\frac{25}{3}

(D)

\frac{25}{2 3}

JEE Main 2022 (25 July Shift 2)

LEVELJEE Main

Let the point $P (α, β)$ be at a unit distance from each of the two lines $L_{1} : 3 x - 4 y + 12 = 0$ , and $L_{2} : 8 x + 6 y + 11 = 0$ . If $P$ lies below $L_{1}$ and above $L_{2}$ , then $100 (α + β)$ is equal to

(A)

- 14

(B)

42

(C)

- 22

(D)

14

JEE Main 2022 (29 July Shift 2)

LEVELJEE Advanced

Let $m_{1}, m_{2}$ be the slopes of two adjacent sides of a square of side $a$ such that $a^{2} + 11 a + 3 (m_{1}^{2} + m_{2}^{2}) = 220$ . If one vertex of the square is $(10 (cos α - sin α), 10 (sin α + cos α))$ , where $α \in [0, π /2]$ and the equation of one diagonal is $(cos α - sin α) x + (sin α + cos α) y = 10$ , then $72 (sin^{4} α + cos^{4} α) + a^{2} - 3 a + 13$ is equal to:

(A)

119

(B)

128

(C)

145

(D)

155

JEE Main 2023 (10 Apr Shift 2)

LEVELJEE Main

Let the equations of two adjacent sides of a parallelogram $A B C D$ be $2 x - 3 y = - 23$ and $5 x + 4 y = 23$ . If the equation of its one diagonal $A C$ is $3 x + 7 y = 23$ and the distance of $A$ from the other diagonal is $d$ , then $50 d^{2}$ is equal to

JEE Main 2024 (04 Apr Shift 1)

LEVELJEE Main

The vertices of a triangle are $A (- 1, 3)$ , $B (- 2, 2)$ and $C (3, - 1)$ . A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

(A)

x + y + (2 - 2) = 0

(B)

- x + y - (2 - 2) = 0

(C)

x + y - (2 - 2) = 0

(D)

x - y - (2 + 2) = 0

JEE Main 2024 (08 Apr Shift 2)

LEVELJEE Main

If the image of the point $(- 4, 5)$ in the line $x + 2 y = 2$ lies on the circle $(x + 4)^{2} + (y - 3)^{2} = r^{2}$ , then $r$ is equal to:

JEE Main 2024 (27 Jan Shift 2)

LEVELJEE Main

Let $R$ be the interior region between the lines $3 x - y + 1 = 0$ and $x + 2 y - 5 = 0$ containing the origin. The set of all values of $a$ , for which the points $(a^{2}, a + 1)$ lie in $R$ , is :

(A)

(- 3, - 1) \cup (- \frac{1}{3}, 1)

(B)

(- 3, 0) \cup (\frac{1}{3}, 1)

(C)

(- 3, 0) \cup (\frac{2}{3}, 1)

(D)

(- 3, - 1) \cup (\frac{1}{3}, 1)

JEE Main 2024 (29 Jan Shift 2)

LEVELJEE Main

Let $A$ be the point of intersection of the lines $3 x + 2 y = 14$ , $5 x - y = 6$ and $B$ be the point of intersection of the lines $4 x + 3 y = 8$ , $6 x + y = 5$ . The distance of the point $P (5, - 2)$ from the line $A B$ is

(A)

\frac{13}{2}

(B)

(C)

\frac{5}{2}

(D)

JEE Main 2024 (31 Jan Shift 2)

LEVELJEE Main

Let $A (a, b), B (3, 4)$ and $(- 6, - 8)$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $P (2 a + 3, 7 b + 5)$ from the line $2 x + 3 y - 4 = 0$ measured parallel to the line $x - 2 y - 1 = 0$ is

(A)

\frac{15 5}{7}

(B)

\frac{17 5}{6}

(C)

\frac{17 5}{7}

(D)

\frac{5}{17}

JEE Main 2025 (April)

LEVELJEE Main

Let the equation $x (x + 2) (12 - k) = 2$ have equal roots. Then the distance of the point $(k, \frac{k}{2})$ from the line $3 x + 4 y + 5 = 0$ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The line L given by 5x​+by​=1 passes through the point (13,32). The line K is parallel to L and has the equation cx​+3y​=1. Then the distance between L and K is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A straight line through the origin O meets the parallel lines 4x+2y=9 and 2x+y+6=0 at points P and Q respectively. Then the point O divides the segment PQ in the ratio

If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Line L has intercepts a and b on the coordinate axes. When the axes are rotated through a given angle, keeping the origin fixed, the same line L has intercepts p and q, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The foot of the perpendicular drawn from the origin, on the line, 3x+y=λ(λ=0) is P. If the line meets x-axis at A and y-axis at B, then the ratio BP : PA is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lines are drawn parallel to the line 4x−3y+2=0, at a distance 3/5 from the origin. Then which one of the following points lies on any of these lines ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The length of the perpendicular from the origin, on the normal to the curve, x2+2xy−3y2=0 at the point (2,2) is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The length of the perpendicular from the origin on the normal to the curve, x2+2xy−3y2=0 at the point (2,2) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line, 2x−y+3=0 is at a distance 5​1​ and 5​2​ from the lines 4x−2y+α=0 and 6x−3y+β=0, respectively, then the sum of all possible values of α and β is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let L denote the line in the xy-plane with x and y intercepts as 3 and 1 respectively. Then the image of the point (−1,−4) in this line is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If p and q are the lengths of the perpendiculars from the origin on the lines, xcosecα−ysecα=kcot2α and xsinα+ycosα=ksin2α respectively, then k2 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The image of the point (3,5) in the line x−y+1=0, lies on :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let R be the point (3,7) and let P and Q be two points on the line x+y=5 such that PQR is an equilateral triangle. Then the area of ΔPQR is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the image of the point (−4,5) in the line x+2y=2 lies on the circle (x+4)2+(y−3)2=r2, then r is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let R be the interior region between the lines 3x−y+1=0 and x+2y−5=0 containing the origin. The set of all values of a, for which the points (a2,a+1) lie in R, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the equation x(x+2)(12−k)=2 have equal roots. Then the distance of the point (k,2k​) from the line 3x+4y+5=0 is

The line $L$ given by $\frac{x}{5} + \frac{y}{b} = 1$ passes through the point $(13, 32)$ . The line $K$ is parallel to $L$ and has the equation $\frac{x}{c} + \frac{y}{3} = 1$ . Then the distance between $L$ and $K$ is

A straight line through the origin $O$ meets the parallel lines $4 x + 2 y = 9$ and $2 x + y + 6 = 0$ at points $P$ and $Q$ respectively. Then the point $O$ divides the segment $P Q$ in the ratio

Line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a given angle, keeping the origin fixed, the same line $L$ has intercepts $p$ and $q$ , then

The foot of the perpendicular drawn from the origin, on the line, $3 x + y = λ (λ \neq = 0)$ is P. If the line meets x-axis at A and y-axis at B, then the ratio BP : PA is :-

Lines are drawn parallel to the line $4 x - 3 y + 2 = 0$ , at a distance $3/5$ from the origin. Then which one of the following points lies on any of these lines ?

The length of the perpendicular from the origin, on the normal to the curve, $x^{2} + 2 x y - 3 y^{2} = 0$ at the point $(2, 2)$ is:

The length of the perpendicular from the origin on the normal to the curve, $x^{2} + 2 x y - 3 y^{2} = 0$ at the point $(2, 2)$ is

If the line, $2 x - y + 3 = 0$ is at a distance $\frac{1}{5}$ and $\frac{2}{5}$ from the lines $4 x - 2 y + α = 0$ and $6 x - 3 y + β = 0$ , respectively, then the sum of all possible values of $α$ and $β$ is

Let $L$ denote the line in the $x y$ -plane with $x$ and $y$ intercepts as 3 and 1 respectively. Then the image of the point $(- 1, - 4)$ in this line is :

If $p$ and $q$ are the lengths of the perpendiculars from the origin on the lines, $x cosec α - y sec α = k cot 2 α$ and $x sin α + y cos α = k sin 2 α$ respectively, then $k^{2}$ is equal to:

The image of the point $(3, 5)$ in the line $x - y + 1 = 0$ , lies on :

Let $R$ be the point $(3, 7)$ and let $P$ and $Q$ be two points on the line $x + y = 5$ such that $P QR$ is an equilateral triangle. Then the area of $Δ P QR$ is :

If the image of the point $(- 4, 5)$ in the line $x + 2 y = 2$ lies on the circle $(x + 4)^{2} + (y - 3)^{2} = r^{2}$ , then $r$ is equal to:

Let $R$ be the interior region between the lines $3 x - y + 1 = 0$ and $x + 2 y - 5 = 0$ containing the origin. The set of all values of $a$ , for which the points $(a^{2}, a + 1)$ lie in $R$ , is :

Let the equation $x (x + 2) (12 - k) = 2$ have equal roots. Then the distance of the point $(k, \frac{k}{2})$ from the line $3 x + 4 y + 5 = 0$ is