Q.Draw a quadrilateral in the Cartesian plane, whose vertices are $(-4, 5)$, $(0, 7)$, $(5, -5)$ and $(-4, -2)$. Also, find its area.

Showing the 12 most recent of 60 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.$4x - y - 3 = 0$ and $x + 4y - 22 = 0$ represent two sides of a square. If $\left(\dfrac{7}{2}, \dfrac{5}{2}\right)$ is the point of intersection of its diagonals, then the sum of all the $Y$-coordinates of the vertices of that square is (A) $14$ (B) $10$ (C) $7$ (D) $12$

   ›Reveal solutionSolution

   Two perpendicular lines through a vertex of a square, plus the diagonal-intersection (centre), fully determine all four vertices; the answer is $10$.

   Concept and Intuition

   Two sides of a square meeting at a vertex are perpendicular lines. Their intersection is a vertex $A$; the opposite vertex $C$ is the reflection of $A$ in the centre $O$ (since diagonals bisect each other). The remaining two vertices lie on the two given side-lines at the same distance from $A$ as $A$ is from $O$'s reflection logic — found cleanly using the midpoint property of the other diagonal.

   Step-by-Step Solution

   1. Slopes of $4x-y-3=0$ and $x+4y-22=0$ are $4$ and $-1/4$; product $=-1$, confirming they are two perpendicular sides of the square, meeting at a vertex.
   2. Solve simultaneously: from $y=4x-3$, substitute into $x+4y=22$: $x+16x-12=22 \Rightarrow 17x=34 \Rightarrow x=2,\ y=5$. So $A=(2,5)$.
   3. Centre $O=(7/2,5/2)$ is the midpoint of both diagonals. Opposite vertex $C = 2O-A = (7-2,\,5-5) = (5,0)$.
   4. $B$ lies on line $4x-y-3=0$ at the same distance from $A$ as the side length; parametrizing along direction $(1,4)$ (unit length $\sqrt{17}$, matching side length $\sqrt{17}$ from $A$ to $O$'s geometry) gives candidates $A\pm(1,4)$. Similarly $D$ on $x+4y-22=0$ gives candidates $A\pm(4,-1)$.
   5. Requiring the midpoint of $BD$ to equal $O=(7/2,5/2)$ selects $B=(1,1)$ and $D=(6,4)$ (their midpoint is exactly $(7/2,5/2)$).
   6. Vertices: $A=(2,5),B=(1,1),C=(5,0),D=(6,4)$ — check all sides $=\sqrt{17}$ and adjacent sides perpendicular: confirmed a genuine square.

   Common Mistakes

   • Forgetting that the two given lines are adjacent sides (not a side and a diagonal) — checking their slopes' product is essential.
   • Picking $B,D$ without enforcing that $O$ is also the midpoint of $BD$, which can give an incorrect (non-square) quadrilateral.

   ✓Final answer

   The correct option is (B) — $10$.

   ANSWER: B

2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Let A = (1, 3), B = (3, 1) be two fixed points. If C is a point on the line $x + y + 1 = 0$, then the locus of the centroid of $\triangle ABC$ is (A) $3x + 3y - 7 = 0$ (B) $x + y = 10$ (C) $y = 4x$ (D) $x + y = 9$

   ›Reveal solutionSolution

   A locus problem: parametrize the moving point $C$ on the given line, write the centroid in terms of the parameter, then eliminate the parameter to get the locus equation.

   Concept and Intuition

   Since $C$ moves along a fixed line while $A, B$ are fixed, the centroid of $\triangle ABC$ also traces a straight line (an affine image of the original line). The standard technique is to parametrize $C$ using the line's equation, express the centroid's coordinates in terms of that parameter, and eliminate the parameter.

   Step-by-Step Solution

   1. Let $C = (t, -1-t)$ since $C$ lies on $x+y+1=0 \Rightarrow y=-1-x$.
   2. Centroid of $\triangle ABC$ with $A=(1,3)$, $B=(3,1)$, $C=(t,-1-t)$:

   $G = \left(\frac{1+3+t}{3}, \frac{3+1+(-1-t)}{3}\right) = \left(\frac{4+t}{3}, \frac{3-t}{3}\right)$

   3. Let $x = \dfrac{4+t}{3}$ and $y = \dfrac{3-t}{3}$. Then $3x = 4+t \Rightarrow t = 3x-4$, and $3y = 3-t \Rightarrow t = 3-3y$.
   4. Equating: $3x-4 = 3-3y \Rightarrow 3x+3y = 7 \Rightarrow 3x+3y-7=0$.

   Common Mistakes

   • Sign error while substituting $y=-1-t$ from the line equation (easy to write $y=1-t$ instead).
   • Forgetting to divide the sum of coordinates by 3 for the centroid.

   ✓Final answer

   The correct option is (A) — $3x + 3y - 7 = 0$.

   ANSWER: A

3. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If the orthocenter of the triangle having the lines $x - 2y + 5 = 0$ and $x + y + 2 = 0$ as two of its sides is $\left(-\dfrac{2}{3}, \dfrac{4}{3}\right)$, then the centroid of the triangle is (A) $(0, 0)$ (B) $(1, 3)$ (C) $(2, -4)$ (D) $(-2, -1)$

   ›Reveal solutionSolution

   Reconstruct the triangle's vertices from the two given side-lines and the orthocenter (using the perpendicularity of altitudes to opposite sides), then compute the centroid.

   Concept and Intuition

   Each altitude of a triangle passes through the opposite vertex, is perpendicular to the opposite side, and passes through the orthocenter. Since we're given two full side-lines (meeting at one vertex) and the orthocenter, we can find the other two vertices as intersections of "altitude through $H$" with the corresponding given side-line, then compute the centroid from all three vertices.

   Step-by-Step Solution

   1. The two given lines $x-2y+5=0$ and $x+y+2=0$ meet at vertex $C$: solving simultaneously, $x=2y-5$; substitute: $2y-5+y+2=0\Rightarrow 3y=3\Rightarrow y=1, x=-3$. So $C=(-3,1)$.
   2. Let side $CA$ be the line $x-2y+5=0$ (slope $\frac12$) and side $CB$ be the line $x+y+2=0$ (slope $-1$).
   3. The altitude from $A$ is perpendicular to $CB$ (slope $-1$), so it has slope $1$, and passes through $H=\left(-\frac23,\frac43\right)$: $y-\frac43 = 1\cdot\left(x+\frac23\right) \Rightarrow y = x+2$.
   4. $A$ lies on both this altitude and line $CA$ ($x-2y+5=0$): substitute $y=x+2$: $x-2(x+2)+5=0 \Rightarrow -x+1=0 \Rightarrow x=1, y=3$. So $A=(1,3)$.
   5. The altitude from $B$ is perpendicular to $CA$ (slope $\frac12$), so slope $-2$, through $H$: $y-\frac43=-2\left(x+\frac23\right)\Rightarrow y=-2x$.
   6. $B$ lies on this altitude and on line $CB$ ($x+y+2=0$): substitute $y=-2x$: $x-2x+2=0\Rightarrow -x+2=0\Rightarrow x=2,y=-4$. So $B=(2,-4)$.
   7. Verify: the altitude from $C$ should be perpendicular to $AB$ (slope $\frac{-4-3}{2-1}=-7$) i.e. slope $\frac17$, through $C=(-3,1)$: at $x=-\frac23$, $y=1+\frac17\left(-\frac23+3\right)=1+\frac17\cdot\frac73=1+\frac13=\frac43$ — matches $H$. Confirms the construction.
   8. Centroid $=\left(\dfrac{1+2-3}{3},\dfrac{3-4+1}{3}\right) = (0,0)$.

   Common Mistakes

   • Assigning the wrong side-line to the wrong vertex's opposite side (mixing up which line is opposite to $A$ vs $B$).
   • Sign errors while finding the perpendicular slope (negative reciprocal).

   ✓Final answer

   The correct option is (A) — $(0, 0)$.

   ANSWER: A

4. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If the equation of the line passing through the orthocenter and circumcenter of $\triangle ABC$, whose vertices are $A(3,1), B(3,3), C(6,1)$ is $2x + by + c = 0$, then $b + c =$ (A) $-6$ (B) $6$ (C) $0$ (D) $2$

   ›Reveal solutionSolution

   The triangle is right-angled at $A$ (since $AB$ is vertical and $AC$ is horizontal), so the orthocenter is $A$ itself and the circumcenter is the midpoint of the hypotenuse $BC$. The line joining them gives $b+c=-6$ — option (A).

   Concept and Intuition

   For any right triangle:

   • The orthocenter (intersection of altitudes) is exactly the right-angle vertex, because the two legs of the right angle are themselves two of the altitudes.
   • The circumcenter (equidistant from all three vertices) is the midpoint of the hypotenuse, since the hypotenuse subtends a right angle and the circle with the hypotenuse as diameter (Thales' theorem) passes through all three vertices.

   Recognizing the right angle first turns this into simple coordinate geometry rather than needing to solve for altitudes and perpendicular bisectors from scratch.

   Step-by-Step Solution

   1. Vertices: $A(3,1)$, $B(3,3)$, $C(6,1)$.
   2. $AB$ has both x-coordinates equal to 3 → $AB$ is the vertical line $x=3$.
   3. $AC$ has both y-coordinates equal to 1 → $AC$ is the horizontal line $y=1$.
   4. Since $AB\perp AC$, the triangle has a right angle at vertex $A$.
   5. Orthocenter = the right-angle vertex = $A=(3,1)$.
   6. Circumcenter = midpoint of hypotenuse $BC$ (the side opposite the right angle):

   $\text{Midpoint of }BC = \left(\frac{3+6}{2}, \frac{3+1}{2}\right) = (4.5, 2).$

   7. Line through orthocenter $(3,1)$ and circumcenter $(4.5,2)$: slope $=\dfrac{2-1}{4.5-3}=\dfrac{1}{1.5}=\dfrac23$.
   8. Equation: $y-1=\dfrac23(x-3) \implies 3(y-1)=2(x-3) \implies 3y-3=2x-6 \implies 2x-3y-3=0$.
   9. Compare with the given form $2x+by+c=0$: $b=-3$, $c=-3$.
   10. $b+c = -3+(-3) = -6$.
   11. Sanity check both points satisfy $2x-3y-3=0$: at $(3,1)$: $6-3-3=0$ ✓; at $(4.5,2)$: $9-6-3=0$ ✓.

   Common Mistakes

   • Not recognizing the right angle at $A$ and instead grinding through the general (much longer) construction of altitudes and perpendicular bisectors.
   • Sign slip when rearranging $3y-3=2x-6$ into the form $2x+by+c=0$.

   ✓Final answer

   The correct option is (A) — $-6$.

   ANSWER: A

5. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.The circumcenter of a triangle lies at the origin and its centroid is the midpoint of the line segment joining the points $(a^2+1, a^2+1)$ and $(2a, -2a)$, $a \neq 0$. Then for any $a$, the line that passes through its orthocentre is (A) $y - 2ax = 0$ (B) $y - (a^2+1)x = 0$ (C) $x + y = 0$ (D) $(a-1)^2 x - (a+1)^2 y = 0$

   ›Reveal solutionSolution

   Using $O$ at the origin, the Euler line gives $H=3G$; substituting the computed $G$ shows only option (D) is satisfied identically for every $a$. Answer: (D).

   Concept and Intuition

   For any triangle, the circumcenter $O$, centroid $G$, and orthocentre $H$ are collinear on the Euler line, with $\vec{OG}:\vec{GH}=1:2$. That means $H-O = 3(G-O)$. Since $O$ is the origin here, this simplifies beautifully to $H=3G$ — we don't even need to know the triangle's actual vertices, just its centroid.

   Step-by-Step Solution

   1. The centroid $G$ is the midpoint of $(a^2+1,a^2+1)$ and $(2a,-2a)$:

   $G=\left(\frac{a^2+1+2a}{2},\frac{a^2+1-2a}{2}\right)=\left(\frac{(a+1)^2}{2},\frac{(a-1)^2}{2}\right)$

   2. Since $O=(0,0)$ is the circumcenter, Euler's relation gives $H=3G-2O=3G$:

   $H=\left(\frac{3(a+1)^2}{2},\frac{3(a-1)^2}{2}\right)$

   3. Test option (D): $(a-1)^2 x-(a+1)^2y=0$ at $H$:

   $(a-1)^2\cdot\frac{3(a+1)^2}{2}-(a+1)^2\cdot\frac{3(a-1)^2}{2}=0$

   This holds identically for every value of $a$ — the line always passes through $H$.

   4. The other options (linear in $a$ or fixed lines) do not vanish identically when $H$'s coordinates (quadratic in $a$) are substituted, except for special values of $a$, so they fail the "for any $a$" requirement.

   Common Mistakes

   • Forgetting the direction of the Euler-line ratio (using $H=\tfrac{3}{2}G$ instead of $3G$).
   • Trying to find the actual triangle vertices instead of using the O–G–H relation directly.

   ✓Final answer

   The correct option is (D) — $(a-1)^2x-(a+1)^2y=0$.

   ANSWER: D

6. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.A variable plane which is at a distance of 4 units from the origin meets the X, Y, Z coordinate axes at $P, Q, R$ respectively. Equation of the locus of the centroid of $\triangle PQR$ is (A) $\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=1$ (B) $x^2+y^2+z^2=9$ (C) $\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=\dfrac{9}{16}$ (D) $\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=2$

   ›Reveal solutionSolution

   The centroid locus of the axis-triangle cut by a plane at fixed distance $4$ from the origin is $\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=\dfrac{9}{16}$.

   Concept and Intuition

   A plane cutting intercepts $p,q,r$ on the three axes can be written $\dfrac{x}{p}+\dfrac{y}{q}+\dfrac{z}{r}=1$, and its perpendicular distance from the origin has the clean formula $\dfrac{1}{\sqrt{1/p^2+1/q^2+1/r^2}}$. Relating the centroid coordinates back to $p,q,r$ (each is one-third of the intercept) converts the distance condition into a locus equation.

   Step-by-Step Solution

   1. Plane: $\dfrac{x}{p}+\dfrac{y}{q}+\dfrac{z}{r}=1$, intercepts $P=(p,0,0)$, $Q=(0,q,0)$, $R=(0,0,r)$.
   2. Distance from origin to this plane: $d=\dfrac{1}{\sqrt{\frac{1}{p^2}+\frac{1}{q^2}+\frac{1}{r^2}}}=4$, so $\dfrac{1}{p^2}+\dfrac{1}{q^2}+\dfrac{1}{r^2}=\dfrac{1}{16}$.
   3. Centroid of $\triangle PQR$: $(X,Y,Z)=\left(\dfrac{p}{3},\dfrac{q}{3},\dfrac{r}{3}\right)$, so $p=3X,\,q=3Y,\,r=3Z$.
   4. Substitute: $\dfrac{1}{9X^2}+\dfrac{1}{9Y^2}+\dfrac{1}{9Z^2}=\dfrac{1}{16}$.
   5. Multiply both sides by $9$: $\dfrac{1}{X^2}+\dfrac{1}{Y^2}+\dfrac{1}{Z^2}=\dfrac{9}{16}$.

   Common Mistakes

   • Forgetting to square the factor of $3$ when substituting $p=3x$ into $1/p^2$ (giving a wrong constant like $1/16$ or $2$).
   • Confusing the plane's intercept form with the centroid coordinates directly (they differ by the factor $3$).

   ✓Final answer

   The correct option is (C) — $\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=\dfrac{9}{16}$.

   ANSWER: C

7. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $(\alpha, \beta)$ is the image of the centroid of the triangle formed by the points $(1,3), (3,1)$ and $(2,4)$ with respect to the line $2x + 3y - 2 = 0$, then $\alpha + \beta =$ (A) $-\frac{118}{39}$ (B) $\frac{15}{43}$ (C) $\frac{16}{51}$ (D) $\frac{32}{39}$

   ›Reveal solutionSolution

   This tests finding a centroid, then reflecting a point about a line using the standard foot-of-perpendicular / image formula. Answer: $\alpha+\beta=-\dfrac{118}{39}$.

   Concept and Intuition

   The image (reflection) of a point $(x_1,y_1)$ about the line $ax+by+c=0$ is obtained by moving twice the signed perpendicular distance along the line's normal direction $(a,b)$. The formula $\dfrac{x'-x_1}{a}=\dfrac{y'-y_1}{b}=\dfrac{-2(ax_1+by_1+c)}{a^2+b^2}$ packages this in one step.

   Step-by-Step Solution

   1. Centroid of the triangle: $\left(\dfrac{1+3+2}{3},\dfrac{3+1+4}{3}\right)=\left(2,\dfrac83\right)$.
   2. Line: $2x+3y-2=0$, so $a=2,b=3,c=-2$; $a^2+b^2=13$.
   3. $d = a x_1+by_1+c = 2(2)+3\left(\tfrac83\right)-2 = 4+8-2=10$.
   4. Common ratio: $t=-\dfrac{2d}{13}=-\dfrac{20}{13}$.
   5. $\alpha = x_1+at = 2+2\left(-\tfrac{20}{13}\right)=2-\tfrac{40}{13}=-\tfrac{14}{13}$.
   6. $\beta = y_1+bt = \tfrac83+3\left(-\tfrac{20}{13}\right)=\tfrac83-\tfrac{60}{13}$. Using denominator 39: $\tfrac{104}{39}-\tfrac{180}{39}=-\tfrac{76}{39}$.
   7. $\alpha+\beta=-\tfrac{42}{39}-\tfrac{76}{39}=-\tfrac{118}{39}$ (converting $\alpha=-\tfrac{14}{13}=-\tfrac{42}{39}$).

   Common Mistakes

   • Sign error in $d=ax_1+by_1+c$ (forgetting the constant $c$).
   • Adding fractions with different denominators (13 vs 3) without converting to a common denominator (39) before summing.

   ✓Final answer

   The correct option is (A) — $-\dfrac{118}{39}$.

   ANSWER: A

8. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$O(0,0,0)$ and $A(2,1,-3)$ are vertices of a triangle $OAB$. If $(-1, 2, 1)$ is the mid point of the side AB, and the perimeter of the triangle is $\sqrt{2}(k + l\sqrt{7} + m\sqrt{13})$ then $k + l + m =$ (A) $7$ (B) $8$ (C) $5$ (D) $6$

   ›Reveal solutionSolution

   This tests 3D distance formula and the midpoint relation to recover a vertex. Computing all three side lengths and matching the perimeter's given form gives $k+l+m=8$.

   Concept and Intuition

   Given the midpoint of a segment and one endpoint, the other endpoint is found from $M=\frac{A+B}{2}\Rightarrow B=2M-A$. The triangle's perimeter is then just the sum of the three pairwise 3D distances, which we match to the given surd form term by term.

   Step-by-Step Solution

   1. $B=2(-1,2,1)-(2,1,-3)=(-2,4,2)-(2,1,-3)=(-4,3,5)$.
   2. $OA=\sqrt{2^2+1^2+(-3)^2}=\sqrt{4+1+9}=\sqrt{14}=\sqrt2\cdot\sqrt7$.
   3. $OB=\sqrt{(-4)^2+3^2+5^2}=\sqrt{16+9+25}=\sqrt{50}=5\sqrt2$.
   4. $AB=\sqrt{(2-(-4))^2+(1-3)^2+(-3-5)^2}=\sqrt{36+4+64}=\sqrt{104}=2\sqrt{26}=2\sqrt2\cdot\sqrt{13}$.
   5. Perimeter $=OA+OB+AB=\sqrt2\sqrt7+5\sqrt2+2\sqrt2\sqrt{13}=\sqrt2\left(5+\sqrt7+2\sqrt{13}\right)$.
   6. Matching with $\sqrt2(k+l\sqrt7+m\sqrt{13})$: $k=5,l=1,m=2$, so $k+l+m=8$.

   Common Mistakes

   • Sign errors computing $B=2M-A$ (subtracting in the wrong order).
   • Forgetting to factor out the common $\sqrt2$ before reading off $k,l,m$.

   ✓Final answer

   The correct option is (B) — $8$.

   ANSWER: B

9. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.A(−4 , 9 , k), B(−1 , k , k), C(0 , 7 , 10) form an isosceles right-angled triangle. If AB = BC and AC is an integer then the perimeter of $\triangle ABC$ is (A) $4(1 + \sqrt{2})$ (B) $\sqrt{14}(2 + \sqrt{2})$ (C) $\sqrt{10}(2 + \sqrt{2})$ (D) $6(1 + \sqrt{2})$

   ›Reveal solutionSolution

   This tests using AB = BC (isosceles) plus the integer-hypotenuse condition to pin down $k$, then verifying the right angle at $B$ to get the perimeter $6(1+\sqrt2)$.

   Concept and Intuition

   In an isosceles right triangle with the equal legs meeting at the right angle, if AB = BC are the legs then AC is the hypotenuse and $AC=\sqrt2\,AB$. We first use $AB=BC$ to solve for $k$, using the integer-AC condition to choose between the two algebraic roots, and then confirm the right angle actually sits at $B$.

   Step-by-Step Solution

   1. $AB^2 = (-1+4)^2+(k-9)^2+0^2 = 9+(k-9)^2$.
   2. $BC^2 = (0+1)^2+(7-k)^2+(10-k)^2 = 1+(7-k)^2+(10-k)^2$.
   3. Setting $AB^2=BC^2$: $9+k^2-18k+81 = 1+49-14k+k^2+100-20k+k^2$ $\Rightarrow k^2-18k+90 = 2k^2-34k+150 \Rightarrow k^2-16k+60=0 \Rightarrow k=10\text{ or }6$.
   4. $AC^2 = 16+4+(10-k)^2 = 20+(10-k)^2$. For $k=10$: $AC^2=20$ (not a perfect square). For $k=6$: $AC^2=20+16=36\Rightarrow AC=6$ (integer) — so $k=6$.
   5. Check the right angle: $B=(-1,6,6)$, $\vec{BA}=(-3,3,0)$, $\vec{BC}=(1,1,4)$. $\vec{BA}\cdot\vec{BC}=-3+3+0=0$ — confirmed right angle at $B$.
   6. $AB^2=9+9=18\Rightarrow AB=3\sqrt2$; $BC^2=1+1+16=18\Rightarrow BC=3\sqrt2$.
   7. Perimeter $=AB+BC+AC = 3\sqrt2+3\sqrt2+6 = 6+6\sqrt2 = 6(1+\sqrt2)$.

   Common Mistakes

   • Picking $k=10$ without checking the integer-AC condition.
   • Forgetting to verify that the right angle is genuinely at $B$.

   ✓Final answer

   The correct option is (D) — $6(1+\sqrt2)$.

   ANSWER: D

10. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.P(6, 4) is a point on the line $x - y - 2 = 0$. If $A(\alpha, \beta)$ and $B(\gamma, \delta)$ are two points on this line lying on either side of P at a distance of 4 units from P, then $\alpha^2 + \beta^2 + \gamma^2 + \delta^2 =$ (A) $136$ (B) $\dfrac{85}{\sqrt{2}}$ (C) $23 + \dfrac{5}{\sqrt{2}}$ (D) $52$

    ›Reveal solutionSolution

    Two points 4 units from $P(6,4)$ along the line $x-y-2=0$ are found using the line's unit direction vector; their coordinates' squares sum to $136$ (the surd terms cancel).

    Concept and Intuition

    Points at a fixed distance $r$ from $P$ along a line with direction angle $\theta$ are $P\pm r(\cos\theta,\sin\theta)$. Since the line $x-y-2=0$ has slope $1$ (i.e. $\theta=45^\circ$), its unit direction vector is $\left(\dfrac{1}{\sqrt2},\dfrac{1}{\sqrt2}\right)$.

    Step-by-Step Solution

    1. Line: $x-y-2=0$, slope $=1$, so direction angle $\theta=45^\circ$, unit vector $=\left(\dfrac{1}{\sqrt2},\dfrac{1}{\sqrt2}\right)$.
    2. Points at distance 4 from $P(6,4)$: $\left(6\pm4\cdot\dfrac{1}{\sqrt2},\ 4\pm4\cdot\dfrac{1}{\sqrt2}\right)=(6\pm2\sqrt2,\ 4\pm2\sqrt2)$.
    3. Let $A=(6+2\sqrt2,\,4+2\sqrt2)=(\alpha,\beta)$ and $B=(6-2\sqrt2,\,4-2\sqrt2)=(\gamma,\delta)$.
    4. $\alpha^2=(6+2\sqrt2)^2=36+24\sqrt2+8=44+24\sqrt2$; $\beta^2=(4+2\sqrt2)^2=16+16\sqrt2+8=24+16\sqrt2$. $\alpha^2+\beta^2=68+40\sqrt2$.
    5. $\gamma^2=(6-2\sqrt2)^2=44-24\sqrt2$; $\delta^2=(4-2\sqrt2)^2=24-16\sqrt2$. $\gamma^2+\delta^2=68-40\sqrt2$.
    6. Total: $\alpha^2+\beta^2+\gamma^2+\delta^2=(68+40\sqrt2)+(68-40\sqrt2)=136$.

    Common Mistakes

    • Forgetting to check $P$ actually lies on the given line before proceeding (it does: $6-4-2=0$) — otherwise the direction/parametrization is invalid.
    • Losing track of which point is $A$ vs $B$; it doesn't matter here since we want $\alpha^2+\beta^2+\gamma^2+\delta^2$, which is symmetric, but it matters in other variants of this question.
    • Arithmetic errors expanding the surd terms — pairing $+2\sqrt2$ and $-2\sqrt2$ versions makes the surds cancel cleanly, which is a good check.

    ✓Final answer

    The correct option is (A) — $136$.

    ANSWER: A

11. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.G(1,0,1) is the centroid of the triangle ABC. If A = (1, -4, 2) and B = (3,1,0) then $AG^2+CG^2 =$ (A) $BG^2$ (B) $2BG^2$ (C) $6BG^2$ (D) $5BG^2$

    ›Reveal solutionSolution

    Find the third vertex $C$ from the centroid formula, compute the three squared distances $AG^2, BG^2, CG^2$ directly, and match $AG^2+CG^2$ against a multiple of $BG^2$.

    Concept and Intuition

    The centroid of a triangle with vertices $A,B,C$ satisfies $G=\frac{A+B+C}{3}$, so knowing $G$ and two vertices immediately gives the third: $C=3G-A-B$. Once all four points ($A,B,C,G$) are known explicitly, the required distances are just direct 3D distance computations — no special theorem is needed here, only correct algebra with coordinates.

    Step-by-Step Solution

    1. $G=(1,0,1)$, $A=(1,-4,2)$, $B=(3,1,0)$.
    2. $C=3G-A-B$:

    $C_x=3(1)-1-3=-1,\quad C_y=3(0)-(-4)-1=3,\quad C_z=3(1)-2-0=1.$

    So $C=(-1,3,1)$.

    3. $AG^2=(1-1)^2+(0-(-4))^2+(1-2)^2=0+16+1=17$.

    4. $CG^2=(1-(-1))^2+(0-3)^2+(1-1)^2=4+9+0=13$.

    5. $BG^2=(1-3)^2+(0-1)^2+(1-0)^2=4+1+1=6$.

    6. $AG^2+CG^2=17+13=30$.

    7. Compare to multiples of $BG^2=6$: $5\times6=30$. So $AG^2+CG^2=5BG^2$.

    Common Mistakes

    • Sign errors when solving $C=3G-A-B$ component-wise (especially with the negative coordinates in $A$), which throws off every subsequent distance.
    • Assuming a general theorem like $AG^2+BG^2+CG^2=\text{const}$ applies directly without actually computing — here the question specifically wants a relation between $AG^2+CG^2$ and $BG^2$, which must be checked numerically.

    ✓Final answer

    The correct option is (D) — $5BG^2$.

    ANSWER: D

12. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.If A(0, 1, 2), B(2, -1, 3) and C(1, -3, 1) are the vertices of a triangle, then the distance between its circumcentre and orthocentre is (A) $\dfrac{3}{\sqrt{2}}$ (B) $\dfrac{3}{2}$ (C) 3 (D) $\dfrac{9}{2}$

    ›Reveal solutionSolution

    Recognize the triangle is right-angled (via the side lengths), which immediately locates both the orthocenter (the right-angle vertex) and the circumcenter (midpoint of the hypotenuse) without any further computation. Answer: $\dfrac{3}{\sqrt2}$.

    Concept and Intuition

    Rather than computing the circumcenter and orthocenter from scratch (which is messy in 3D), check first whether the triangle is a right triangle — this simplifies both centers dramatically: the orthocenter of a right triangle is the vertex at the right angle, and its circumcenter is the midpoint of the hypotenuse (since the hypotenuse subtends a right angle, it's a diameter of the circumcircle).

    Step-by-Step Solution

    1. $A(0,1,2)$, $B(2,-1,3)$, $C(1,-3,1)$.
    2. $\vec{AB}=(2,-2,1) \Rightarrow AB^2=4+4+1=9 \Rightarrow AB=3$.
    3. $\vec{BC}=(-1,-2,-2) \Rightarrow BC^2=1+4+4=9 \Rightarrow BC=3$.
    4. $\vec{AC}=(1,-4,-1) \Rightarrow AC^2=1+16+1=18 \Rightarrow AC=3\sqrt2$.
    5. Check Pythagoras: $AB^2+BC^2=9+9=18=AC^2$ ✓ — right angle at $B$ (the vertex between the two legs $AB,BC$; $AC$ is the hypotenuse).
    6. Orthocenter of a right triangle = the right-angle vertex: $H=B=(2,-1,3)$.
    7. Circumcenter of a right triangle = midpoint of the hypotenuse: $O=\text{mid}(A,C)=\left(\dfrac{0+1}{2},\dfrac{1-3}{2},\dfrac{2+1}{2}\right)=(0.5,-1,1.5)$.
    8. Distance $OH=\sqrt{(2-0.5)^2+(-1+1)^2+(3-1.5)^2}=\sqrt{1.5^2+0+1.5^2}=\sqrt{4.5}=\sqrt{\dfrac92}=\dfrac{3}{\sqrt2}$.

    Common Mistakes

    • Not checking for a right angle first and instead trying to solve the general (messy) 3D circumcenter/orthocenter system.
    • Misidentifying which vertex holds the right angle — it's the vertex common to the two legs whose squares sum to the third side's square, i.e. $B$ here (between $AB$ and $BC$), not $A$ or $C$.

    ✓Final answer

    The correct option is (A) — $\dfrac{3}{\sqrt2}$.

    ANSWER: A