Q.Line through the points $(-2, 6)$ and $(4, 8)$ is perpendicular to the line through the points $(8, 12)$ and $(x, 24)$. Find the value of $x$.

Showing the 12 most recent of 26 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A line $L \equiv ax + 6y + 15 = 0$ is perpendicular to the line passing through the points $(-a, 3)$ and $(5, 6)$. If the line $bx + ay + k = 0$ is parallel to the line $L = 0$, then $a + 3b =$ (A) $0$ (B) $60$ (C) $20$ (D) $40$

   ›Reveal solutionSolution

   Use the perpendicularity condition to find $a$, then the parallelism condition between the two lines to find $b$.

   Concept and Intuition

   Two lines $A_1x+B_1y+C_1=0$ and $A_2x+B_2y+C_2=0$ are perpendicular iff $A_1A_2+B_1B_2=0$ (equivalently, product of slopes $=-1$), and parallel iff $A_1B_2=A_2B_1$ (equivalently, equal slopes, $A_1/A_2=B_1/B_2$). This problem chains both conditions: first perpendicularity pins down $a$, then parallelism (using that $a$) pins down $b$.

   Step-by-Step Solution

   1. Slope of the line through $(-a,3)$ and $(5,6)$: $\dfrac{6-3}{5-(-a)}=\dfrac{3}{5+a}$.
   2. Slope of $L\equiv ax+6y+15=0$: rewriting $y=-\dfrac a6x-\dfrac{15}{6}$, slope $=-\dfrac a6$.
   3. Perpendicularity: $\left(-\dfrac a6\right)\cdot\dfrac{3}{5+a}=-1\Rightarrow\dfrac{-3a}{6(5+a)}=-1\Rightarrow -3a=-6(5+a)=-30-6a$.
   4. $-3a+6a=-30\Rightarrow3a=-30\Rightarrow a=-10$.
   5. Now $L\equiv-10x+6y+15=0$. The line $bx+ay+k=0$ becomes $bx-10y+k=0$. …

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2. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A tangent $L_1$ with slope $m$ drawn to the parabola $y^2 = 8x$ is perpendicular to the normal $L_2$ drawn to the parabola $y^2 = 12x$. If $m = 1$ and the point of intersection of $L_1$ and $L_2$ is $(h, k)$, then $h + k =$ (A) $2$ (B) $4$ (C) $9$ (D) $6$

   ›Reveal solutionSolution

   Writing the tangent and normal in slope-form for their respective parabolas and using perpendicularity gives the intersection point $(3.5,5.5)$, so $h+k=9$.

   Concept and Intuition

   For $y^2=4ax$, the tangent of slope $m$ is $y=mx+a/m$, and the normal of slope $m$ is $y=mx-2am-am^3$. Perpendicularity between $L_1$ and $L_2$ links their slopes via $m_1 m_2=-1$.

   Step-by-Step Solution

   1. $y^2=8x\Rightarrow 4a=8\Rightarrow a=2$. Tangent $L_1$ with slope $m=1$: $y=1\cdot x+\dfrac21=x+2$.
   2. $L_1\perp L_2\Rightarrow$ slope of $L_2 = -1/m=-1$.
   3. $y^2=12x\Rightarrow 4a=12\Rightarrow a=3$. Normal $L_2$ with slope $-1$: $y=(-1)x-2(3)(-1)-3(-1)^3=-x+6+3=-x+9$. …

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3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $A(2,12), B(5,12+\sqrt{3}), C(3,12-\sqrt{3})$ are the vertices of a triangle $ABC$, then the ratio in which the perpendicular drawn through $A$ divides the side $BC$ is (A) $2:\sqrt{3}$ (B) $\sqrt{3}:4$ (C) $3:5$ (D) $1:3$

   ›Reveal solutionSolution

   Drop the altitude from $A$ onto $BC$; computing the foot $D$ directly gives $BD=3$, $DC=1$ — the division ratio is $1:3$.

   Concept and Intuition

   The perpendicular from a vertex to the opposite side is the altitude; its foot $D$ splits the opposite side into two segments whose ratio we can get either by the foot's coordinates or, more directly, from the perpendicular-foot formula applied to the line $BC$.

   Step-by-Step Solution

   1. Slope of $BC$: $\dfrac{(12-\sqrt3)-(12+\sqrt3)}{3-5}=\dfrac{-2\sqrt3}{-2}=\sqrt3$.
   2. Line $BC$: $y-(12+\sqrt3)=\sqrt3(x-5)\Rightarrow \sqrt3x-y+(12-4\sqrt3)=0$.
   3. Foot of perpendicular from $A(2,12)$: using $x=x_0-\dfrac{a(ax_0+by_0+c)}{a^2+b^2}$, with $a=\sqrt3,b=-1,c=12-4\sqrt3$: $ax_0+by_0+c=2\sqrt3-12+12-4\sqrt3=-2\sqrt3$, and $a^2+b^2=4$.
   4. $x=2-\sqrt3\cdot\dfrac{-2\sqrt3}{4}=2+\dfrac{6}{4}=3.5$; $y=12-(-1)\cdot\dfrac{-2\sqrt3}{4}=12-\dfrac{\sqrt3}{2}$. So $D=(3.5,\,12-\tfrac{\sqrt3}2)$.
   5. $BD=\sqrt{(5-3.5)^2+(\sqrt3-(-\tfrac{\sqrt3}2))^2}=\sqrt{1.5^2+(1.5\sqrt3)^2}=\sqrt{2.25+6.75}=\sqrt9=3$. …

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4. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The equation of the normal to the parabola $y^2=16x$ which is perpendicular to the line $2x-y+5=0$ is (A) $x+2y+9=0$ (B) $x+2y-9=0$ (C) $x+2y+16=0$ (D) $x+2y-16=0$

   ›Reveal solutionSolution

   Find the normal to $y^2=16x$ perpendicular to a given line; answer is $x+2y-9=0$.

   Concept and Intuition

   A normal to $y^2=4ax$ at parameter $t$ has a definite slope $-t$; matching that slope to the perpendicularity condition pins down $t$, and then the standard normal-line formula gives the equation.

   Step-by-Step Solution

   1. Line $2x-y+5=0\Rightarrow y=2x+5$ has slope $m=2$.
   2. "Perpendicular to this line" means the normal's slope is $-\dfrac{1}{2}$.
   3. For $y^2=16x=4ax$, $a=4$. The normal at parameter $t$: $y=-tx+2at+at^3$, whose slope is $-t$.
   4. Set $-t=-\dfrac12\Rightarrow t=\dfrac12$. …

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5. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The locus of the point which forms a right angled triangle with the fixed points (2, 3) and (5, 1) represents (A) a circle or a pair of parallel lines (B) a pair of parallel lines which are parallel to the line passing through the given points (C) a circle having the line joining the given points as a chord (D) the perpendicular bisector of the line joining the given points

   ›Reveal solutionSolution

   This tests recognizing that "forms a right triangle" doesn't fix where the right angle is — it could be at any of the three vertices, giving a compound locus. Answer: a circle (Thales) union a pair of parallel lines (the two perpendiculars at the fixed points).

   Concept and Intuition

   A triangle is "right-angled" if any one of its three angles is $90°$. With two vertices fixed at $A=(2,3)$ and $B=(5,1)$ and the third vertex $P$ varying, we must consider each possible location of the right angle separately, then union the resulting loci — the full locus is the set of all $P$ for which some vertex has a right angle.

   Step-by-Step Solution

   1. Right angle at $P$: $\vec{PA}\cdot\vec{PB}=0$. This is exactly Thales' theorem: $P$ traces the circle having segment $AB$ as its diameter.
   2. Right angle at $A$: $\vec{AP}\cdot\vec{AB}=0$, i.e. $AP\perp AB$. So $P$ lies anywhere on the line through $A$ perpendicular to $AB$.
   3. Right angle at $B$: similarly, $P$ lies on the line through $B$ perpendicular to $AB$.
   4. The two lines from steps 2–3 are both perpendicular to the same line $AB$, hence parallel to each other. …

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6. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If the image of the point P(2 , 3) in a line L is Q(4 , 5), then the image of the point R(0 , 0) in the same line is (A) (4 , 5) (B) (3 , 4) (C) (2 , 2) (D) (7 , 7)

   ›Reveal solutionSolution

   The unknown mirror line $L$ is recovered as the perpendicular bisector of $P$ and its image $Q$; reflecting the origin in that recovered line gives $(7,7)$.

   Concept and Intuition

   If $Q$ is the reflection of $P$ in a line $L$, then $L$ must be the perpendicular bisector of segment $PQ$ — this is exactly the defining property of a mirror reflection (the mirror line is equidistant from a point and its image, and perpendicular to the segment joining them). Once we know $L$ explicitly, we can reflect any other point in it using the standard reflection formula.

   Step-by-Step Solution

   1. $P(2,3)$, $Q(4,5)$. Midpoint of $PQ = (3,4)$; this point lies on $L$.
   2. Slope of $PQ = \dfrac{5-3}{4-2}=1$. Since $L \perp PQ$, slope of $L = -1$.
   3. $L: y-4 = -1(x-3) \Rightarrow y = -x+7 \Rightarrow x+y-7=0$.
   4. Sanity check: reflect $P(2,3)$ in $x+y-7=0$ using $x' = x_0 - \dfrac{2a(ax_0+by_0+c)}{a^2+b^2}$ with $a=b=1,\ c=-7$: $ax_0+by_0+c = 2+3-7=-2$, so $x' = 2 - \dfrac{2(-2)}{2} = 2+2=4$, $y' = 3+2=5$ — matches $Q(4,5)$. ✓ …

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7. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The points (h , k) , (1 , 2) and (−3 , 4) lie on the line $L_1$. If a line $L_2$ passing through the points (h , k) and (4 , 3) is perpendicular to the line $L_1$, then $\dfrac{k}{h} =$ (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{3}$ (C) $-\dfrac{1}{7}$ (D) $-\dfrac{1}{5}$

   ›Reveal solutionSolution

   Using collinearity with $L_1$ and perpendicularity of $L_2$ gives two linear equations in $h,k$; solving them yields $h=3,\ k=1$, so $k/h=1/3$.

   Concept and Intuition

   Three points are collinear iff any two of the pairwise slopes match. Here $(h,k)$ lies on the same line $L_1$ as $(1,2)$ and $(-3,4)$, so its slope with either of those points must equal $L_1$'s slope. Separately, $L_2$ (through $(h,k)$ and $(4,3)$) being perpendicular to $L_1$ pins down another relation between $h$ and $k$ — two equations, two unknowns.

   Step-by-Step Solution

   1. Slope of $L_1$ from $(1,2)$ and $(-3,4)$: $m_1 = \dfrac{4-2}{-3-1} = \dfrac{2}{-4} = -\dfrac12$.
   2. Since $(h,k)$ lies on $L_1$ too, using $(h,k)$ and $(1,2)$: $\dfrac{k-2}{h-1} = -\dfrac12 \Rightarrow k = 2 - \dfrac{h-1}{2} = \dfrac52 - \dfrac{h}{2}$. — (i)
   3. $L_2$ through $(h,k)$ and $(4,3)$ is perpendicular to $L_1$, so slope of $L_2 = \dfrac{-1}{m_1} = 2$.
   4. $\dfrac{3-k}{4-h} = 2 \Rightarrow 3-k = 8-2h \Rightarrow k = 2h-5$. — (ii) …

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8. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.If the lines $3x+y-4=0$, $x-\alpha y+10=0$, $\beta x+2y+4=0$ and $3x+y+k=0$ represent the sides of a square, then $\alpha\beta(k+4)^2=$ (A) $-256$ (B) $-512$ (C) $-128$ (D) $-1024$

   ›Reveal solutionSolution

   Identify the two parallel pairs forming the square, match slopes for perpendicularity, then equate the two side-lengths.

   Concept and Intuition

   A square's sides come in two parallel pairs, each pair perpendicular to the other, and all four sides equal in length. Matching slopes pins down $\alpha,\beta$; matching perpendicular distances pins down $k$.

   Step-by-Step Solution

   1. Line 1: $3x+y-4=0$ (slope $-3$); Line 4: $3x+y+k=0$ (slope $-3$) — these are one parallel pair.
   2. Line 2: $x-\alpha y+10=0$ (slope $1/\alpha$); Line 3: $\beta x+2y+4=0$ (slope $-\beta/2$) must be perpendicular to slope $-3$, so both equal $1/3$: $1/\alpha=1/3\Rightarrow\alpha=3$; $-\beta/2=1/3\Rightarrow\beta=-2/3$.
   3. With $\alpha=3$: Line 2 is $x-3y+10=0$. With $\beta=-2/3$: Line 3 becomes (after clearing fractions) $x-3y-6=0$.
   4. Distance between lines 2 & 3: $\dfrac{|10-(-6)|}{\sqrt{1+9}}=\dfrac{16}{\sqrt{10}}$ — this is the square's side length. …

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9. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.If $4x-3y-5=0$ is a normal to the ellipse $3x^2+8y^2=k$, then the equation of the tangent drawn to this ellipse at the point $(-2,m)$ $(m>0)$ is (A) $3x+4y-14=0$ (B) $3x-4y+10=0$ (C) $3x-4y+1=0$ (D) $4x+3y-3=0$

   ›Reveal solutionSolution

   This tests the normal-to-ellipse condition (to fix $k$) followed by the tangent-to-ellipse formula at a specific point. The answer is $3x-4y+10=0$.

   Concept and Intuition

   A normal line to a conic at a point is perpendicular to the tangent there. If we know one normal line (as an equation) and the conic's shape up to the unknown $k$, we can pin down both $k$ and the point of contact by requiring: (i) the slope condition (normal slope from the conic matches the given line's slope), and (ii) the point actually lies on the given line.

   Step-by-Step Solution

   1. Differentiate $3x^2+8y^2=k$ implicitly: $6x+16yy'=0 \Rightarrow y'=-\dfrac{3x}{8y}$.
   2. At the point of tangency $(x_1,y_1)$, the normal's slope is the negative reciprocal: $\dfrac{8y_1}{3x_1}$.
   3. The given normal is $4x-3y-5=0$, i.e. $y=\dfrac{4}{3}x-\dfrac53$, slope $\dfrac43$. So $\dfrac{8y_1}{3x_1}=\dfrac43 \Rightarrow 8y_1=4x_1 \Rightarrow y_1=\dfrac{x_1}{2}$.
   4. Since $(x_1,y_1)$ lies on the line: $4x_1-3y_1-5=0$. Substituting $y_1=x_1/2$: $4x_1-\dfrac{3x_1}{2}=5 \Rightarrow \dfrac{5x_1}{2}=5 \Rightarrow x_1=2,\ y_1=1$.
   5. Since $(2,1)$ lies on the ellipse: $3(4)+8(1)=k \Rightarrow k=20$. So the ellipse is $3x^2+8y^2=20$. …

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10. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.If the line $5x-2y-6=0$ is a tangent to the hyperbola $5x^2-ky^2=12$, then the equation of the normal to this hyperbola at the point $(\sqrt6,b)$ $(b<0)$ is (A) $\sqrt{6}x+2y=0$ (B) $2\sqrt{6}x+3y=3$ (C) $\sqrt{6}x-5y=21$ (D) $3\sqrt{6}x-y=21$

    ›Reveal solutionSolution

    Using the tangency (discriminant=0) condition pins down $k$, then standard implicit differentiation gives the normal at the required point. The answer is $\sqrt6x-5y=21$.

    Concept and Intuition

    A line is tangent to a conic exactly when substituting it into the conic's equation produces a quadratic with a repeated root, i.e. discriminant zero. This is the most reliable way to extract an unknown parameter from a tangency statement, more robust than guessing a point first.

    Step-by-Step Solution

    1. From $5x-2y-6=0$, $y=\dfrac{5x-6}{2}$.
    2. Substitute into $5x^2-ky^2=12$: $5x^2-k\left(\dfrac{5x-6}{2}\right)^2=12$. Multiply by 4: $20x^2-k(25x^2-60x+36)=48$.
    3. Rearranged: $(20-25k)x^2+60kx-(36k+48)=0$.
    4. Tangency $\Rightarrow$ discriminant $=0$: $(60k)^2+4(20-25k)(36k+48)=0$.
    5. Expand $(20-25k)(36k+48)=-900k^2-480k+960$. So $3600k^2+4(-900k^2-480k+960)=0 \Rightarrow -1920k+3840=0\Rightarrow k=2$.
    6. Hyperbola is $5x^2-2y^2=12$. At $(\sqrt6,b)$: $5(6)-2b^2=12\Rightarrow 2b^2=18\Rightarrow b^2=9$; since $b<0$, $b=-3$. …

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11. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.The point $(a,b)$ is the foot of the perpendicular drawn from the point $(3,1)$ to the line $x+3y+4=0$. If $(p,q)$ is the image of $(a,b)$ with respect to the line $3x-4y+11=0$, then $\dfrac{p}{a}+\dfrac{q}{b}=$ (A) $-3$ (B) $-5$ (C) $3$ (D) $7$

    ›Reveal solutionSolution

    First find the foot of perpendicular $(a,b)=(2,-2)$ from $(3,1)$ to the given line, then reflect it across the second line to get $(p,q)=(-4,6)$; the requested expression evaluates to $-5$.

    Concept and Intuition

    Two standard line-geometry formulas are chained: (1) the foot of perpendicular from a point to a line, and (2) the reflection (image) of a point in a line. Both use the same building block $\dfrac{Ax_1+By_1+C}{A^2+B^2}$.

    Step-by-Step Solution

    1. Foot of perpendicular from $(3,1)$ to $x+3y+4=0$ ($A=1,B=3,C=4$): $Ax_1+By_1+C=3+3+4=10$, $A^2+B^2=10$, ratio $=1$. $(a,b)=(3,1)-1\cdot(1,3)=(2,-2)$.
    2. Check: $2+3(-2)+4=2-6+4=0$ ✓, so $(2,-2)$ indeed lies on the line.
    3. Reflect $(a,b)=(2,-2)$ in $3x-4y+11=0$ ($A=3,B=-4,C=11$): $Ax_1+By_1+C=6+8+11=25$, $A^2+B^2=25$, ratio $=1$. $(p,q)=(2,-2)-2\cdot1\cdot(3,-4)=(2-6,\,-2+8)=(-4,6)$. …

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12. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.If the line $2x-3y+5=0$ is the perpendicular bisector of the line segment joining $(1,-2)$ and $(\alpha,\beta)$, then $\alpha+\beta =$ (A) $7$ (B) $1$ (C) $-1$ (D) $-7$

    ›Reveal solutionSolution

    A perpendicular bisector gives two conditions: the midpoint lies on the line, and the segment is perpendicular to it. Solving both gives $\alpha+\beta=1$.

    Concept and Intuition

    'Line $L$ is the perpendicular bisector of segment $PQ$' means two things simultaneously: (1) the midpoint of $PQ$ lies on $L$, and (2) $PQ\perp L$. Both give linear equations in the unknown endpoint.

    Step-by-Step Solution

    1. Midpoint of $(1,-2)$ and $(\alpha,\beta)$ is $\left(\dfrac{1+\alpha}{2}, \dfrac{-2+\beta}{2}\right)$; it lies on $2x-3y+5=0$: $2\cdot\dfrac{1+\alpha}{2} - 3\cdot\dfrac{\beta-2}{2} + 5 = 0 \Rightarrow (1+\alpha) - \dfrac32(\beta-2) + 5 = 0$.
    2. $(1+\alpha) - 1.5\beta+3+5=0 \Rightarrow \alpha - 1.5\beta + 9 = 0 \Rightarrow \alpha-1.5\beta=-9$. — (I)
    3. Line $2x-3y+5=0$ has slope $2/3$; the perpendicular segment has slope $-3/2$.
    4. Slope of segment joining $(1,-2)$ and $(\alpha,\beta)$: $\dfrac{\beta+2}{\alpha-1} = -\dfrac32 \Rightarrow 2(\beta+2) = -3(\alpha-1) \Rightarrow 2\beta+4=-3\alpha+3 \Rightarrow 3\alpha+2\beta=-1$. — (II) …

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