Triangles and Centres
Orthocentre, Circumcentre and their Properties
GRB_1000_MCQ
Grade Class 12
Question:
Consider $\triangle ABC$, $A(5,-1)$, $B(\alpha,-7)$, $C(-2,\beta)$. Let $(-6,-4)$ is image of orthocentre of $\triangle ABC$ in the point mirror $M$ which is mid-point of the side $BC$. Also $(p,q)$ is circumcentre of triangle $ABC$, then:
the value of $\beta^2 - \alpha^2 + 5\beta - \alpha$ is 12.
the value of $2p+1$ is 0.
the value of $2q+5$ is $-6$.
the value of $q^2 - \dfrac{p}{2}$ is $\dfrac{13}{2}$.
Step-by-Step Solution
Step 1: The image of the orthocentre $H$ in the midpoint $M$ of side $BC$ is a point $H'$ on the circumcircle. This point $H'$ is diametrically opposite to vertex $A$.
Given $A(5,-1)$ and $H'(-6,-4)$.
Step 2: The circumcentre $(p,q)$ is the midpoint of the diameter $AH'$.
$$p = \frac{5+(-6)}{2} = -\frac{1}{2}$$
$$q = \frac{-1+(-4)}{2} = -\frac{5}{2}$$
Step 3: The value of $2p+1$ is:
$$2p+1 = 2\left(-\frac{1}{2}\right)+1 = -1+1 = 0$$
Step 4: The value of $2q+5$ is:
$$2q+5 = 2\left(-\frac{5}{2}\right)+5 = -5+5 = 0$$
Step 5: The circumcentre $(p,q) = \left(-\frac{1}{2}, -\frac{5}{2}\right)$ is equidistant from all vertices $A, B, C$.
The square of the circumradius $R^2$ is $PA^2$:
$$R^2 = \left(5 - \left(-\frac{1}{2}\right)\right)^2 + \left(-1 - \left(-\frac{5}{2}\right)\right)^2$$
$$R^2 = \left(\frac{11}{2}\right)^2 + \left(\frac{3}{2}\right)^2 = \frac{121}{4} + \frac{9}{4} = \frac{130}{4}$$
Using $PB^2 = R^2$ for $B(\alpha,-7)$:
$$\left(\alpha - \left(-\frac{1}{2}\right)\right)^2 + \left(-7 - \left(-\frac{5}{2}\right)\right)^2 = \frac{130}{4}$$
$$\left(\alpha+\frac{1}{2}\right)^2 + \left(-\frac{9}{2}\right)^2 = \frac{130}{4}$$
$$\left(\alpha+\frac{1}{2}\right)^2 + \frac{81}{4} = \frac{130}{4}$$
$$\left(\alpha+\frac{1}{2}\right)^2 = \frac{49}{4}$$
$$\alpha+\frac{1}{2} = \pm\frac{7}{2}$$
$$\alpha = -\frac{1}{2} \pm \frac{7}{2} \implies \alpha = 3 \text{ or } \alpha = -4$$
Using $PC^2 = R^2$ for $C(-2,\beta)$:
$$\left(-2 - \left(-\frac{1}{2}\right)\right)^2 + \left(\beta - \left(-\frac{5}{2}\right)\right)^2 = \frac{130}{4}$$
$$\left(-\frac{3}{2}\right)^2 + \left(\beta+\frac{5}{2}\right)^2 = \frac{130}{4}$$
$$\frac{9}{4} + \left(\beta+\frac{5}{2}\right)^2 = \frac{130}{4}$$
$$\left(\beta+\frac{5}{2}\right)^2 = \frac{121}{4}$$
$$\beta+\frac{5}{2} = \pm\frac{11}{2}$$
$$\beta = -\frac{5}{2} \pm \frac{11}{2} \implies \beta = 3 \text{ or } \beta = -8$$
Step 6: The value of $\beta^2 - \alpha^2 + 5\beta - \alpha$ using $\alpha = 3$ and $\beta = -8$:
$$(-8)^2 - (3)^2 + 5(-8) - 3 = 64 - 9 - 40 - 3 = 12$$
Step 7: The value of $q^2 - \dfrac{p}{2}$ is:
$$q^2 - \frac{p}{2} = \left(-\frac{5}{2}\right)^2 - \frac{-1/2}{2} = \frac{25}{4} + \frac{1}{4} = \frac{26}{4} = \frac{13}{2}$$
Correct Answer: 1, 2, 3, 4