The orthocentre of the triangle formed by the lines $x + y = 1$,$2x + 3y = 6$,and $4x - y + 4 = 0$ lies in which quadrant?

The distance of the origin from the centroid of the triangle whose two sides have the equations $x - 2y + 1 = 0$ and $2x - y - 1 = 0$ and whose orthocenter is $\left(\frac{7}{3}, \frac{7}{3}\right)$ is

Let $C(\alpha, \beta)$ be the circumcenter of the triangle formed by the lines $4x + 3y = 69$,$4y - 3x = 17$,and $x + 7y = 61$. Then $(\alpha - \beta)^2 + \alpha + \beta$ is equal to $.........$.

$A$ triangle has a vertex at $(1, 2)$ and the midpoints of the two sides through it are $(-1, 1)$ and $(2, 3)$. Then the centroid of this triangle is

Find the distance of the line $4x - y = 0$ from the point $P(4, 1)$ measured along the line making an angle of $135^{\circ}$ with the positive $x$-axis.

If the vertices of a triangle are $(a, 1), (b, 3),$ and $(4, c),$ then the centroid of the triangle will lie on the $x$-axis if