Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
Let the mid points of the sides of a triangle ABC be $\left(\frac{5}{2}, 7\right)$, $\left(\frac{5}{2}, 3\right)$ and $(4, 5)$. If its incentre is $(h, k)$, then $3h + k$ is equal to :
(1) 11
(2) 12
(3) 13
(4) 14
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Answer: Option (3)
Key idea: Finding vertices from midpoints and calculating incentre coordinates
Step 1: Use midpoint formulas to set up equations: If midpoints are $D(\frac{5}{2}, 7)$, $E(\frac{5}{2}, 3)$, $F(4, 5)$, then $x_2 + x_3 = 5$, $x_1 + x_3 = 5$, $x_1 + x_2 = 8$ (and similar for $y$-coordinates).
Step 2: Solve the system by adding all equations: $2(x_1 + x_2 + x_3) = 18 \Rightarrow x_1 + x_2 + x_3 = 9$. Similarly, $y_1 + y_2 + y_3 = 15$.
Let a circle pass through the origin and its centre be the point of intersection of two mutually perpendicular lines $x + (k-1)y + 3 = 0$ and $2x + k^2y - 4 = 0$. If the line $x - y + 2 = 0$ intersects the circle at the points A and B, then $(AB)^2$ is equal to:
(1) 10
(2) 27
(3) 18
(4) 34
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Answer: Option (3)
Key idea: Circle passing through origin, perpendicular lines, chord length using perpendicular distance formula
Use perpendicularity condition $a_1a_2 + b_1b_2 = 0$ to get $2 + k^2(k-1) = 0$, which gives $k = -1$
Substitute $k = -1$ to get lines $x - 2y + 3 = 0$ and $2x + y - 4 = 0$, solve to find center $(1, 2)$
Since circle passes through origin, radius $r^2 = 1^2 + 2^2 = 5$
Find perpendicular distance from center $(1, 2)$ to line $x - y + 2 = 0$: $d = \frac{|1-2+2|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$
Let the line $L_1: x + 3 = 0$ intersect the lines $L_2: x - y = 0$ and $L_3: 3x + y = 0$ at the points A and B, respectively. Let the bisector of the obtuse angle between the lines $L_2$ and $L_3$ intersect the line $L_1$ at the point C. Then $BC^2: AC^2$ is equal to:
(1) $5:1$
(2) $1:5$
(3) $2:3$
(4) $3:2$
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Answer: Option (1)
Key idea: Angle bisectors of two lines and distance ratios
Find point $A$ by solving $L_1: x = -3$ and $L_2: y = x$, giving $A = (-3, -3)$
Find point $B$ by solving $L_1: x = -3$ and $L_3: y = -3x$, giving $B = (-3, 9)$
Find the angle bisectors using $\frac{x - y}{\sqrt{2}} = \pm \frac{3x + y}{\sqrt{10}}$
Identify the obtuse angle bisector as $y(\sqrt{10} - \sqrt{2}) = x(\sqrt{10} + 3\sqrt{2})$ by checking slopes
Find point $C$ on $L_1$ where $x = -3$, giving $C = (-3, -6 - 3\sqrt{5})$
Let the vertex A of a triangle ABC be $(1, 2)$, and the mid-point of the side AB be $(5, -1)$. If the centroid of this triangle is $(3, 4)$ and its circumcenter is $(\alpha, \beta)$, then $21(\alpha + \beta)$ is equal to:
(1) 309
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Answer: Option (497)
Key idea: Centroid, Circumcenter, and Coordinate Geometry
Let A, B be points on the two half-lines $x - \sqrt{3}|y| = a$, $a > 0$ at a distance of $a$ from their point of intersection P. The line segment AB meets the angle bisector of the given half-lines at the point Q. If $PQ = \frac{9}{2}$ and R is the radius of the circumcircle of $\triangle PAB$, then $\frac{a^2}{R}$ is equal to _________
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Answer: 9
Key idea: Coordinate Geometry: Half-lines, angle bisectors, and circumradius of triangles
The equation $x - \sqrt{3}|y| = a$ represents two half-lines meeting at $P(a, 0)$ making angles of $\pm 30°$ with the x-axis. The angle bisector is the x-axis.
Points $A$ and $B$ are at distance $a$ from $P$ on the two half-lines: $A = (a + \frac{a\sqrt{3}}{2}, \frac{a}{2})$ and $B = (a + \frac{a\sqrt{3}}{2}, -\frac{a}{2})$.
Line $AB$ is vertical at $x = a(1 + \frac{\sqrt{3}}{2})$, so $Q = (a(1 + \frac{\sqrt{3}}{2}), 0)$ and $PQ = \frac{a\sqrt{3}}{2}$.
Given $PQ = \frac{9}{2}$, we get $\frac{a\sqrt{3}}{2} = \frac{9}{2}$, so $a = 3\sqrt{3}$.
Triangle $PAB$ is isosceles with $PA = PB = a$ and angle $\angle APB = 60°$. By the law of cosines, $AB = a$, making it equilateral.
For an equilateral triangle with side $a$, the circumradius is $R = \frac{a}{\sqrt{3}}$.
In an equilateral triangle PQR, let the vertex P be at (3, 5) and the side QR be along the line $x + y = 4$. If the orthocentre of the triangle PQR is ($\alpha$, $\beta$), then 9($\alpha$ + $\beta$) is equal to:
(1) 16
(2) 27
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Orthocentre of equilateral triangle, perpendicular distance, section formula
From the point $(-1, -1)$, two rays are sent making angles of $45°$ with the line $x+y=0$. These rays get reflected from the mirror $x+2y=1$. If the equations of the reflected rays are $ax+by=9$ and $cx+dy=7$, a, b, c, d $\in$ Z, then the value of ad+bc is __________.
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Answer: 7
Key idea: Reflection of lines about a mirror line, angle between lines, coordinate geometry
Find the two incident rays from $(-1, -1)$ making $45°$ with $x+y=0$. Since $x+y=0$ has slope $-1$ (making $135°$ with x-axis), the rays at $45°$ to it are horizontal ($y=-1$) and vertical ($x=-1$).
For the horizontal ray $y=-1$, find its reflection about mirror $x+2y=1$. Use the reflection formula: from point $(x_0, -1)$, find the perpendicular to the mirror, locate the foot of perpendicular, and find the image point.
The reflected ray from $y=-1$ passes through $(3/5, 11/5)$ and $(18/5, -9/5)$, giving slope $-4/3$. The equation is $4x+3y=9$.
For the vertical ray $x=-1$, similarly find its reflection. The reflected ray passes through $(3/5, 11/5)$ and $(-17/5, -4/5)$, giving slope $3/4$. The equation is $-3x+4y=7$.
Let the line $x - y = 4$ intersect the circle $C: (x-4)^2 + (y+3)^2 = 9$ at the points Q and R. If $P(\alpha, \beta)$ is a point on C such that $PQ = PR$, then $(6\alpha + 8\beta)^2$ is equal to _________.
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Answer: 18
Key idea: Perpendicular bisector of a chord, intersection of line and circle
Circle $C$ has center $O(4, -3)$ and radius $3$. Line $x - y = 4$ intersects it at $Q$ and $R$.
Since $PQ = PR$, point $P$ lies on the perpendicular bisector of chord $QR$.
The perpendicular bisector of any chord passes through the center. Line $QR$ has slope $1$, so the perpendicular bisector has slope $-1$.
Perpendicular bisector through $O(4, -3)$ with slope $-1$: $y + 3 = -1(x - 4)$, which gives $x + y = 1$.
Point $P(\alpha, \beta)$ satisfies both $(x-4)^2 + (y+3)^2 = 9$ and $x + y = 1$. Substituting $y = 1 - x$: $(x-4)^2 + (4-x)^2 = 9$, giving $2(x-4)^2 = 9$.
If a straight line drawn through the point of intersection of the lines $4x + 3y - 1 = 0$ and $3x + 4y - 1 = 0$, meets the co-ordinate axes at the points P and Q, then the locus of the mid point of PQ is:
(1) $x + y - 7 = 0$
(2) $x + y - 14xy = 0$
(3) $2x + y + 14xy = 0$
(4) $x + 2y - 14xy = 0$
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Answer: Option (2)
Key idea: Locus of midpoint of intercepts of a line passing through a fixed point
Find the intersection point $A$ of lines $4x + 3y - 1 = 0$ and $3x + 4y - 1 = 0$. Solving gives $A = (\frac{1}{7}, \frac{1}{7})$.
Write the line through $A$ in intercept form: $\frac{x}{a} + \frac{y}{b} = 1$, where $P(a,0)$ and $Q(0,b)$ are the intercepts.
Since the line passes through $A(\frac{1}{7}, \frac{1}{7})$: $\frac{1}{7a} + \frac{1}{7b} = 1$, which gives $\frac{1}{a} + \frac{1}{b} = 7$.
Let midpoint of $PQ$ be $M(h,k)$. Then $h = \frac{a}{2}$ and $k = \frac{b}{2}$, so $a = 2h$ and $b = 2k$.
Substitute in the constraint: $\frac{1}{2h} + \frac{1}{2k} = 7 \Rightarrow \frac{1}{h} + \frac{1}{k} = 14 \Rightarrow h + k = 14hk$.
Replace $(h,k)$ with $(x,y)$ to get the locus: $x + y - 14xy = 0$.
FAQs
Frequently asked questions
How many questions come from Straight Lines in JEE Main?
In the 9 shifts of JEE Main 2026 Session 2 we found 9 questions on it, about 1.0 per shift.
Where can I practise Straight Lines PYQs for free?
On this page, with solutions. For more, use NTA Abhyas and our free mock tests.
