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### Consider a triangle $\Delta$ whose two sides ...

Consider a triangle $\Delta$ whose two sides lie on the x-axis and the line x+y+1=0. If the orthocentre of $\Delta$ is (1, 1), then the equation of the circle passing through the vertices of the triangle $\Delta$ is

(A) $x^2+y^2-3x+y=0$
(B) $x^2+y^2+x+3y=0$
(C) $x^2+y^2+2y-1=0$
(D) $x^2+y^2+x+y=0$

Solution

Reflection of orthocentre about side of a triangle lies on the circumcircle. (Refer https://www.123iitjee.com/2022/03/for-triangle-show-that-point-where.html )

Reflection of orthocentre (1, 1) about side of triangle lying on x-axis is the point (1, -1). Out of 4 options, only option (B) satisfies this.

Let (h, k) be the reflection of orthocentre (1, 1) about side of triangle lying on the line x+y+1=0.

Then, the point $\left( {\frac{{h + 1}}{2},\frac{{k + 1}}{2}} \right)$ lies on the line x+y+1=0.

$\Rightarrow \frac{{h + 1}}{2} + \frac{{k + 1}}{2} + 1 = 0$

Or, h+k+4=0 ....(*)

Also, $\frac{{k - 1}}{{h - 1}} \times - 1 = - 1$

Or h = k

From (*), h = k = -2

As expected, (-2, -2) satisfies option (B).

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$