If the portion of a line intercepted between the coordinate axes is divided by the point $(2, -1)$ in the ratio $3:2$,then the equation of that line is:

If the coordinates of the points $A, B, C, D$ are $(a, b), (a', b'), (-a, b)$ and $(a', -b')$ respectively,then the equation of the line bisecting the line segments $AB$ and $CD$ is

Let $u = \hat{i} - 2\hat{j}$ and $v = -3\hat{i} + 5\hat{j}$. Consider three points $P, Q,$ and $R$ having position vectors $-\frac{1}{7}\hat{i}$,$-\frac{1}{4}\hat{j}$,and $-2\hat{i} + 3\hat{j}$ respectively. Among these,the points on the line segment passing through $u$ and $v$ are

The ends of the base of an isosceles triangle are at $(2a, 0)$ and $(0, a)$. The equation of one side is $x = 2a$. The equation of the other side is

$a$ and $b$ are the intercepts made by a line on the coordinate axes. If $3a = b$ and the line passes through $(1, 3)$,then the equation of the line is

If the transversals $y = m_r x; r = 1, 2, 3$ cut off equal intercepts on the line $x + y = 1$,then $1 + m_1, 1 + m_2, 1 + m_3$ are in