Q.11. In an equilateral triangle PQR, let the vertex P be at (3, 5) and the side QR be along the lin

Q: What is the correct answer for JEE Main Mathematics Q11?

The correct answer is D. Sol. Vertex P(3,5). Side QR is $x+y=4$. Since it's an equilateral triangle, the orthocentre, centroid, and circumcentre coincide. The altitude from P to QR is perpendicular to $x+y=4$. The slope of $x+y=4$ is -1, so the slope of the altitude is 1. The equation of the altitude from P is $y-5 = 1(x-3) \Rightarrow y-5 = x-3 \Rightarrow x-y+2 = 0$. To find the foot of the altitude (let's call it M) on

Q: Q.11. In an equilateral triangle PQR, let the vertex P be at (3, 5) and the side QR be along the line . If the orthocentre of the triangle PQR is (, ), then is equal to :

The correct answer is D. Sol. Vertex P(3,5). Side QR is $x+y=4$. Since it's an equilateral triangle, the orthocentre, centroid, and circumcentre coincide. The altitude from P to QR is perpendicular to $x+y=4$. The slope of $x+y=4$ is -1, so the slope of the altitude is 1. The equation of the altitude from P is $y-5 = 1(x-3) \Rightarrow y-5 = x-3 \Rightarrow x-y+2 = 0$. To find the foot of the altitude (let's call it M) on

Q: What are the options for this JEE Main Mathematics question?

A. 16 · B. 27 · C. 36 · D. 48

Question

Q.11. In an equilateral triangle PQR, let the vertex P be at (3, 5) and the side QR be along the line . If the orthocentre of the triangle PQR is (, ), then  is equal to :

FirstInTest · Accepted Answer

Sol. Vertex P(3,5). Side QR is $x+y=4$.
Since it's an equilateral triangle, the orthocentre, centroid, and circumcentre coincide.
The altitude from P to QR is perpendicular to $x+y=4$. The slope of $x+y=4$ is -1, so the slope of the altitude is 1.
The equation of the altitude from P is $y-5 = 1(x-3) \Rightarrow y-5 = x-3 \Rightarrow x-y+2 = 0$.
To find the foot of the altitude (let's call it M) on QR, solve $x+y=4$ and $x-y+2=0$.
Adding the two equations: $2x+2=4 \Rightarrow 2x=2 \Rightarrow x=1$.
Substituting $x=1$ into $x+y=4$: $1+y=4 \Rightarrow y=3$.
So, M(1,3) is the midpoint of QR.
For an equilateral triangle, the orthocentre (H) divides the median PM in the ratio 2:1.
Let H be ($\alpha$, $\beta$). Using the section formula:
$\alpha = \frac{2(1) + 1(3)}{2+1} = \frac{2+3}{3} = \frac{5}{3}$
$\beta = \frac{2(3) + 1(5)}{2+1} = \frac{6+5}{3} = \frac{11}{3}$
So the orthocentre is $(\frac{5}{3}, \frac{11}{3})$.
We need to find $9(\alpha + \beta) = 9(\frac{5}{3} + \frac{11}{3}) = 9(\frac{16}{3}) = 3 	imes 16 = 48$.

Answer

16

Answer

27

Answer

36

Answer

48

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