JEE Main 2025 23 January Morning Shift
Exam JEE Main
Year 2025
Date 23 January
Shift Morning Shift
Subject Mathematics
Chapter Straight Lines JEE Main 2025 · 23 January · Morning Shift · Mathematics · Straight Lines
Let the area of a P Q R with vertices P(5,4), Q(-2,4) and R(a, b)...
Let the area of a
△ P Q R \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)
b e 35 s q u a r e u n i t s . I f i t s o r t h o c e n t e r a n d c e n t r o i d a r e be 35 square units. If its orthocenter and centroid are b e 35 s q u a re u ni t s . I f i t sor t h oce n t er an d ce n t ro i d a re O ( 2 , 14 5 ) O\left(2, \frac{14}{5}\right) O ( 2 , 5 14 ) and
C ( c , d ) C(c, d) C ( c , d ) respectively, then
c + 2 d c+2 d c + 2 d is equal to
Options A 8 3 \frac{8}{3} 3 8 B 7 3 \frac{7}{3} 3 7 C 2 2 2 D 3 3 3
Step-by-step solution
Equation of lines Q R = 5 x + 2 y + 2 = 0 Equation of lines P R = 10 x − 3 y − 38 = 0 ∴ Point R ( 2 , − 6 ) Centroid = ( 5 − 2 + 2 3 , 4 + 4 − 6 3 ) = ( 5 3 , 2 3 ) c + 2 d = 5 3 + 4 3 = 3 \begin{aligned} & \text{ Equation\text{ of } lines } Q R=5 x+2 y+2=0 \\ & \text{ Equation of lines } P R=10 x-3 y-38=0 \\ & \therefore \text{ Point } R(2,-6) \\ & \text{ Centroid }=\left(\frac{5-2+2}{3}, \frac{4+4-6}{3}\right) \\ & =\left(\frac{5}{3}, \frac{2}{3}\right) \\ & c+2 d=\frac{5}{3}+\frac{4}{3}=3\end{aligned} Equation of lines QR = 5 x + 2 y + 2 = 0 Equation of lines PR = 10 x − 3 y − 38 = 0 ∴ Point R ( 2 , − 6 ) Centroid = ( 3 5 − 2 + 2 , 3 4 + 4 − 6 ) = ( 3 5 , 3 2 ) c + 2 d = 3 5 + 3 4 = 3
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