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IIT JEE Main Maths -Unit 10 -Straight line- Exam Style Questions- New Syllabus

IIT JEE Main Maths - Exam Style Practice Questions - All Chapters
Question 
Let the triangle PQR be the image of the triangle with vertices (1,3), (3,1) and (2, 4) in the line \( x + 2y = 2 \). If the centroid of \(\Delta\)PQR is the point \((\alpha, \beta)\), then \(15(\alpha – \beta)\) is equal to :
\((1)\ 24\)
\((2)\ 19\)
\((3)\ 21\)
\((4)\ 22\)
▶️ Answer/Explanation
The centroid of the original triangle is \( G \equiv \left( 2, \frac{8}{3} \right) \). The image of \( G \) with respect to the line \( x + 2y – 2 = 0 \) is calculated as:
\(\alpha = \frac{-2}{15}, \quad \beta = \frac{-24}{15}\)
\(15(\alpha – \beta) = -2 + 24 = 22\)
The correct answer is \((4)\ 22\).
Question 
Let $A(6, 8)$, $B(10 \cos \alpha, -10 \sin \alpha)$ and $C(-10 \sin \alpha, 10 \cos \alpha)$, be the vertices of a triangle. If $L(a, 9)$ and $G(h, k)$ be its orthocenter and centroid respectively, then $(5a – 3h + 6k + 100 \sin^2 \alpha)$ is equal to___________
▶ Answer/Explanation
Ans. (145)
Sol. All the three points $A$, $B$, $C$ lie on the circle $x^2 + y^2 = 100$ so circumcentre is $(0, 0)$

$a = 3h$
and $9 = 0 + \frac{k}{3}$ $\Rightarrow k = 3$
also centroid $6 + \frac{10\cos \alpha – 10\sin \alpha}{3} = h$
$\Rightarrow 10(\cos \alpha – \sin \alpha) = 3h – 6$ $\cdots$(i)
and $8 + \frac{10\cos \alpha – 10\sin \alpha}{3} = k$
$\Rightarrow 10(\cos \alpha – \sin \alpha) = 3k – 8 = 9 – 8 = 1$ $\cdots$(ii)
on squaring $100(1 – \sin^2 \alpha) = 1$
$\Rightarrow 100\sin^2 \alpha = 99$
from equ. (i) and (ii) we get $h = \frac{7}{3}$
Now $5a – 3h + 6k + 100 \sin^2 \alpha$
= $15h – 3h + 6k + 100 \sin^2 \alpha$
= $12 \times \frac{7}{3} + 18 + 99$
= 145
Question 
Let the distance between two parallel lines be 5 units and a point $P$ lie between the lines at a unit distance from one of them. An equilateral triangle $PQR$ is formed such that $Q$ lies on one of the parallel lines, while $R$ lies on the other. Then $(QR)^2$ is equal to ________.
▶ Answer/Explanation
Ans. (28)
Sol.
$PR = \csc \theta$, $PQ = 4\sec(30 + \theta)$
For equilateral $d = PR = PQ$
$\Rightarrow \cos(\theta + 30^\circ) = 4\sin \theta$
$\Rightarrow \frac{\sqrt{3} – 1}{2} = \frac{1 – \tan \theta}{2\sqrt{3}}$
$\Rightarrow \tan \theta = \frac{1 – \sqrt{3}}{3}$
$QR^2 = d^2 = \csc^2 \theta = 28$
Question 

Let the area of a \(\Delta PQR\) with vertices \(P(5, 4)\), \(Q(-2, 4)\) and \(R(a, b)\) be 35 square units. If its orthocenter and centroid are \(O\left(\frac{14}{3}, 5\right)\) and \(C(c, d)\) respectively, then \(c + 2d\) is equal to

(1) \(\frac{7}{3}\)
(2) 3
(3) 2
(4) \(\frac{8}{3}\)

▶️ Answer/Explanation

Ans. (2)

Sol.

Equation of lines \(QR = 5x + 2y + 2 = 0\)
Equation of lines \(PR = 10x – 3y – 38 = 0\)
\(\therefore\) Point \(R (2, -6)\)
Centroid = \(\left(\frac{5-2+2}{3}, \frac{4+4-6}{3}\right)\)
\(= \left(\frac{5}{3}, \frac{2}{3}\right)\)
\(c + 2d = \frac{5}{3} + 2 \cdot \frac{2}{3} = 3\)

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