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Properties of Triangle question

2025 · 23 Jan · Shift 1 · Q43
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  5. /2025 · 23 Jan · Shift 1 · Q43

Properties of Triangle question

2025 · 23 Jan · Shift 1 · Q43

JEE MainMathematicsProperties of TriangleMCQ+4 / −1
Let the area of a △PQR\triangle P Q R△PQR with vertices P(5,4),Q(−2,4)P(5,4), Q(-2,4)P(5,4),Q(−2,4) and R(a,b)R(a, b)R(a,b) be 35 square units. If its orthocenter and centroid are O(2,145)O\left(2, \frac{14}{5}\right)O(2,514​) and C(c,d)C(c, d)C(c,d) respectively, then c+2dc+2 dc+2d is equal to
  1. A
    333
  2. B
    73\frac{7}{3}37​
  3. C
    222
  4. D
    83\frac{8}{3}38​
View written solutionFree

Correct answer: A

  1. Use the area condition to locate point R(a,b)R(a,b)R(a,b)

Points P(5,4)P(5,4)P(5,4) and Q(−2,4)Q(-2,4)Q(−2,4) lie on the horizontal line y=4y=4y=4. So base PQ=7PQ = 7PQ=7.

Given area of △PQR\triangle PQR△PQR is 353535:

12⋅7⋅∣b−4∣=35\frac{1}{2}\cdot 7 \cdot |b-4| = 3521​⋅7⋅∣b−4∣=35 ∣b−4∣=10|b-4| = 10∣b−4∣=10

Hence,

b=14orb=−6b=14 \quad \text{or} \quad b=-6b=14orb=−6
  1. Use the orthocenter condition

The orthocenter is O(2,145)O\left(2,\frac{14}{5}\right)O(2,514​).

Since PQPQPQ is horizontal, the altitude from RRR is vertical. Therefore the orthocenter must lie on the vertical line through RRR. So,

a=2a=2a=2

Thus,

R=(2,b)R=(2,b)R=(2,b)
  1. Find which value of bbb is valid

Slope of QRQRQR:

mQR=b−42−(−2)=b−44m_{QR} = \frac{b-4}{2-(-2)} = \frac{b-4}{4}mQR​=2−(−2)b−4​=4b−4​

So slope of altitude from PPP is

m⊥=−4b−4m_{\perp} = -\frac{4}{b-4}m⊥​=−b−44​

Equation of altitude from P(5,4)P(5,4)P(5,4):

y−4=−4b−4(x−5)y-4 = -\frac{4}{b-4}(x-5)y−4=−b−44​(x−5)

Since orthocenter lies on this altitude, substitute O(2,145)O\left(2,\frac{14}{5}\right)O(2,514​):

145−4=−4b−4(2−5)\frac{14}{5}-4 = -\frac{4}{b-4}(2-5)514​−4=−b−44​(2−5) −65=12b−4-\frac{6}{5} = \frac{12}{b-4}−56​=b−412​ −65(b−4)=12-\frac{6}{5}(b-4)=12−56​(b−4)=12 b−4=−10b-4=-10b−4=−10 b=−6b=-6b=−6

Hence,

R=(2,−6)R=(2,-6)R=(2,−6)
  1. Find the centroid C(c,d)C(c,d)C(c,d)

Centroid of triangle with vertices (x1,y1),(x2,y2),(x3,y3)(x_1,y_1),(x_2,y_2),(x_3,y_3)(x1​,y1​),(x2​,y2​),(x3​,y3​) is

(x1+x2+x33,y1+y2+y33)\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)(3x1​+x2​+x3​​,3y1​+y2​+y3​​)

So,

C=(5+(−2)+23,4+4+(−6)3)=(53,23)C=\left(\frac{5+(-2)+2}{3},\frac{4+4+(-6)}{3}\right) =\left(\frac{5}{3},\frac{2}{3}\right)C=(35+(−2)+2​,34+4+(−6)​)=(35​,32​)

Thus,

c=53,d=23c=\frac{5}{3}, \qquad d=\frac{2}{3}c=35​,d=32​
  1. Compute c+2dc+2dc+2d
c+2d=53+2⋅23=53+43=93=3c+2d = \frac{5}{3}+2\cdot \frac{2}{3} =\frac{5}{3}+\frac{4}{3} =\frac{9}{3}=3c+2d=35​+2⋅32​=35​+34​=39​=3
  1. Match with options
333

corresponds to Option A.

  1. Compare with stored correct answer

Stored correct answer: A

This matches our derived answer.

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